One AA battery can light a bulb, spin a toy fan, or ring a buzzer. The same stored energy comes out three different ways. Tracking what energy turns INTO is the skill of this unit's first lesson.
Energy comes in several distinct forms. This lesson sharpens one skill: naming the form of energy before a change and the form after it.
The idea
A battery-powered flashlight starts with chemical energy stored in the battery.
When you switch the flashlight on, the battery drives a current, turning the chemical energy into electrical energy.
The bulb then turns the electrical energy into light energy.
A change of energy from one form to another is called an 'energy transformation'.
To identify an energy transformation, name the form of energy before the change and the form after it.
The forms you will name in this course are chemical, electrical, light, thermal, kinetic, and sound energy.
Forms of energy you will name
Form
Where you meet it
chemical energy
stored in batteries, fuels, and food
electrical energy
carried by an electric current
light energy
carried by light
thermal energy
what hot objects have more of
kinetic energy
energy of motion
sound energy
carried by sound waves
The six forms of energy named in this course.
Worked examples
Worked example 1. An electric kettle plugs into the wall, and its element makes the water hot. Identify the energy transformation in the kettle.
Step 1
Before the change: electrical energy carried by the current.
Step 2
After the change: thermal energy in the hot element and water.
Step 3
The kettle transforms electrical energy into thermal energy.
Worked example 2. Wind spins a turbine's blades, and the turbine sends current into the power grid. Identify the energy transformation in the turbine.
Step 1
Before the change: kinetic energy of the moving air and spinning blades.
Step 2
After the change: electrical energy carried by the current.
Step 3
The turbine transforms kinetic energy into electrical energy.
You can now identify the energy transformation in an everyday device or process by naming the form of energy before the change and the form after it.
Check your understanding
A solar panel sits in sunlight and sends current to a house. Which energy transformation happens in the panel?
ALight energy → electrical energycorrect
BElectrical energy → light energy
This option is wrong — you reversed the before and after — the panel takes light IN and sends current OUT.
CThermal energy → electrical energy
This option is wrong — you named the sun's warmth instead of its light — the panel converts the light that lands on it.
DChemical energy → electrical energy
This option is wrong — you described a battery, not a solar panel — no stored chemical fuel is used up in the panel.
Name the form before the change: light energy arriving from the Sun. Name the form after the change: electrical energy in the current. The panel transforms light energy into electrical energy.
Check your understanding
An electric fan is plugged in and switched on. Which energy transformation spins its blades?
AElectrical energy → kinetic energycorrect
BKinetic energy → electrical energy
This option is wrong — you reversed the before and after — current goes IN and motion comes OUT; the reverse describes a generator.
CElectrical energy → sound energy
This option is wrong — you named the fan's hum instead of the blades' motion — the transformation asked for is the one that spins the blades.
DChemical energy → kinetic energy
This option is wrong — you gave the fan a fuel tank — a plugged-in fan runs on electrical energy, not stored chemical energy.
Name the form before the change: electrical energy from the outlet. Name the form after the change: kinetic energy of the spinning blades. The fan transforms electrical energy into kinetic energy.
Check your understanding
A glow stick is snapped, its contents react, and it shines for hours. Which energy transformation happens in the glow stick?
AChemical energy → light energycorrect
BLight energy → chemical energy
This option is wrong — you reversed the before and after — the stored reactants are used up and light comes out.
CElectrical energy → light energy
This option is wrong — you treated the glow stick like a bulb — nothing plugs in; the energy starts out stored in the chemicals.
DKinetic energy → light energy
This option is wrong — you credited the snap — snapping only mixes the chemicals; the light's energy comes from the reaction of the stored reactants.
Name the form before the change: chemical energy stored in the glow stick's reactants. Name the form after the change: light energy shining out. The glow stick transforms chemical energy into light energy.
Lesson 2 of 40 · THM-002
Temperature
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Did You Know?
You have seen that in a gas, a higher temperature means the particles move faster on average. That link is not just for gases — and it is exactly what a temperature reading measures.
The idea
Temperature measures the average kinetic energy of the particles in a sample.
Kinetic energy is energy of motion, so a higher temperature means the sample's particles move faster on average.
This holds for every kind of matter — solid, liquid, or gas.
The water molecules in 80 °C water move faster on average than the water molecules in 20 °C water.
The word average matters: every sample has some faster and some slower particles, and temperature reports the average, not any single particle.
Same substance, two temperatures: the 80 °C sample's molecules move faster on average.
Worked examples
Worked example 1. One iron block sits at 150 °C and another at 50 °C. In which block do the iron particles move faster on average?
Step 1
Answer: the 150 °C block — higher temperature means higher average kinetic energy, so its particles move faster on average.
Worked example 2. A thermometer in a sample reads 35 °C. What does that number report about the sample's particles?
Step 1
Answer: the average kinetic energy of the sample's particles — 35 °C is a measure of how fast they move on average.
You can now state that temperature measures the average kinetic energy of the particles in a sample, so the particles of a hotter sample move faster on average.
Check your understanding
What does temperature measure?
AThe average kinetic energy of the particles in a sample.correct
BThe total kinetic energy of all the particles in a sample.
This option is wrong — you reported the total instead of the average — temperature reports the average per-particle picture, not a sum over the whole sample.
CThe number of particles in a sample.
This option is wrong — you measured amount instead of motion — a huge cold sample and a tiny cold sample read the same temperature.
DThe energy a sample transfers to a colder object.
This option is wrong — you described an energy transfer between two objects — temperature is a property of one sample by itself.
Temperature measures the average kinetic energy of the particles in a sample. Higher temperature means the particles move faster on average — for solids, liquids, and gases alike.
Check your understanding
A carton of juice sits at 5 °C and a second carton of the same juice sits at 25 °C. Which statement about their particles is correct?
AThe particles in the 25 °C juice move faster on average.correct
BThe particles in the 5 °C juice move faster on average.
This option is wrong — you flipped the link — higher temperature means higher average kinetic energy, so the warmer juice's particles are the faster ones on average.
CThe particles in both cartons move at the same average speed.
This option is wrong — you separated temperature from particle motion — different temperatures mean different average kinetic energies.
DThe particles in the 5 °C juice have stopped moving.
This option is wrong — you treated cold as motionless — particles at 5 °C still move; they are just slower on average than at 25 °C.
Temperature measures the average kinetic energy of the particles in a sample. 25 °C is the higher temperature, so the 25 °C juice's particles move faster on average.
Check your understanding
A steel bolt is warmed from 20 °C to 90 °C. What happens to its particles?
ATheir average kinetic energy increases — they move faster on average.correct
BNothing happens to them, because the particles of a solid cannot move.
This option is wrong — you froze the solid's particles — particles in a solid still move, and warming makes them move faster on average.
CEvery particle ends up moving at exactly the same, higher speed.
This option is wrong — you dropped the word average — a sample always has some faster and some slower particles; temperature reports the average.
DThe number of particles in the bolt increases.
This option is wrong — you turned added energy into added matter — warming changes the particles' motion, not how many there are.
Temperature measures the average kinetic energy of the particles in a sample — solid, liquid, or gas. Raising the bolt's temperature from 20 °C to 90 °C raises its particles' average kinetic energy, so they move faster on average.
Lesson 3 of 40 · THM-003
Thermal energy
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Did You Know?
You have seen that temperature reports the average kinetic energy of a sample's particles. An average says nothing about how many particles there are — and the number matters for the sample's total.
The idea
Add up the kinetic energy of every particle in a sample and you get the sample's total.
That total is called the sample's 'thermal energy'.
Thermal energy depends on the sample's temperature, because at a higher temperature each particle carries more kinetic energy on average.
Thermal energy also depends on how much matter the sample contains, because more particles means more kinetic energies to add up.
Two liters of water at 40 °C hold twice the thermal energy of one liter of water at 40 °C.
The temperatures match, but the totals do not — same average, twice the particles.
Worked examples
Worked example 1. A pitcher holds 500 g of milk at 60 °C, and a glass holds 250 g of milk at 60 °C. Which sample holds more thermal energy?
Step 1
Answer: the 500 g sample — the average kinetic energy per particle is the same, but it has twice as many particles to add up.
Worked example 2. A gold ring sits at 25 °C, and later the same ring sits at 75 °C. When does the ring hold more thermal energy?
Step 1
Answer: at 75 °C — the number of particles is unchanged, but each particle carries more kinetic energy on average.
You can now state that thermal energy is the total kinetic energy of all the particles in a sample, so it depends on the sample's temperature and on how much matter the sample contains.
Check your understanding
What is the thermal energy of a sample?
AThe total kinetic energy of all the particles in the sample.correct
BThe average kinetic energy of the particles in the sample.
This option is wrong — you gave the average instead of the total — the average is what temperature measures; thermal energy is the sum over every particle.
CThe kinetic energy of the sample's single fastest particle.
This option is wrong — you picked one particle — thermal energy adds up the kinetic energy of ALL the particles.
DThe energy the sample passes to a colder object.
This option is wrong — you described a transfer between two objects — thermal energy is what one sample holds, counted across its own particles.
Thermal energy is the total kinetic energy of all the particles in a sample. It rises with temperature and with the amount of matter, because both raise the total.
Check your understanding
Two identical 100 g blocks of aluminum sit on a bench — one at 30 °C and one at 70 °C. Which block holds more thermal energy?
AThe 70 °C block, because its particles carry more kinetic energy on average and the particle counts are equal.correct
BThe 30 °C block, because cooler samples store energy more tightly.
This option is wrong — you flipped the temperature link — hotter particles carry more kinetic energy on average, so the hotter block's total is larger.
CBoth hold the same thermal energy, because the blocks have the same mass and therefore the same number of particles.
This option is wrong — you used only the amount — with equal particle counts, the higher temperature tips the total.
DNeither holds thermal energy, because aluminum is a solid.
This option is wrong — you froze the solid's particles — a solid's particles move too, so every solid holds thermal energy.
Thermal energy is the total kinetic energy of all the particles in a sample. Equal masses mean equal particle counts, so the sample whose particles move faster on average — the 70 °C block — holds the larger total.
Check your understanding
An aquarium and a drinking glass are both filled with 22 °C water. Which statement is correct?
AThe aquarium's water holds more thermal energy, because it has far more particles at the same average kinetic energy.correct
BThe two hold the same thermal energy, because they are at the same temperature.
This option is wrong — you used temperature alone — matching averages do not mean matching totals when the particle counts differ.
CThe glass's water holds more thermal energy, because energy is more concentrated in a small sample.
This option is wrong — you inverted the size effect — fewer particles means fewer kinetic energies to add up, so the small sample's total is smaller.
DThe aquarium's water is at a higher temperature, because bigger samples are hotter.
This option is wrong — you let size raise the temperature — both read 22 °C; size changes the total, not the average.
Thermal energy depends on temperature and on how much matter the sample contains. Both samples share one temperature, so the sample with far more particles — the aquarium — holds the larger total.
Lesson 4 of 40 · THM-004
Heat
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Did You Know?
You have seen what temperature and thermal energy say about one sample on its own. Put two objects at different temperatures together and something new happens between them.
The idea
Set a cool pan on a hot stove burner and energy flows from the burner into the pan.
Energy transferred from a hotter object to a colder object because of the temperature difference between them is called 'heat'.
Heat is energy in transfer — it is the flow between two objects, not something one object contains.
The flow always runs from the hotter object to the colder one.
Once the energy has arrived, it becomes part of the colder object's thermal energy; the word heat names only the transfer.
Worked examples
Worked example 1. A blacksmith drops a red-hot horseshoe into a barrel of cool water. Between which objects does heat flow, and in which direction?
Step 1
Answer: heat flows from the hot horseshoe into the cooler water — from hotter to colder, driven by the temperature difference.
Worked example 2. You wrap your warm hand around a glass of iced lemonade. In which direction does heat flow?
Step 1
Answer: from your warmer hand into the colder glass — the glass feels cold because your hand is losing energy, not because cold flows in.
You can now state that heat is energy transferred from a hotter object to a colder object because of the temperature difference between them.
Check your understanding
What is heat?
AEnergy transferred from a hotter object to a colder object because of their temperature difference.correct
BThe total kinetic energy of all the particles moving inside a single sample of matter.
This option is wrong — you gave thermal energy's definition — that total belongs to one sample; heat is the energy flowing between two objects.
CThe average kinetic energy of the particles in a sample.
This option is wrong — you gave temperature's definition — that average belongs to one sample; heat is the energy flowing between two objects.
DA substance stored inside hot objects that leaks out of them.
This option is wrong — you made heat a stored substance — heat is not contained in anything; it is energy in transfer between a hotter and a colder object.
Heat is energy transferred from a hotter object to a colder object because of the temperature difference between them. It names the flow between two objects, never the contents of one object.
Check your understanding
You cup your hands around a cold soda can. In which direction does energy flow?
AFrom your warmer hands into the colder can.correct
BFrom the colder can into your warmer hands.
This option is wrong — you sent the flow the wrong way — heat always flows from the hotter object to the colder one, so your hands are the givers.
CColdness flows from the can into your hands.
This option is wrong — you moved 'cold' as if it were a substance — only energy flows, and it flows from your warmer hands into the colder can.
DNo energy flows, because the can and your hands are separate objects.
This option is wrong — you required the objects to merge — contact plus a temperature difference is exactly what drives heat flow.
Heat is energy transferred from a hotter object to a colder object because of the temperature difference between them. Your hands are hotter than the can, so the energy flows from your hands into the can — the chill you feel is your hands losing energy.
Check your understanding
A casserole sits in a 200 °C oven. Which statement uses the word heat correctly?
AHeat flows from the hot oven air into the cooler casserole.correct
BThe oven air contains a large amount of heat.
This option is wrong — you stored heat inside one object — the oven air holds thermal energy; heat names only the energy being transferred.
CHeat flows from the casserole into the hotter oven air.
This option is wrong — you sent the flow the wrong way — heat flows from the hotter object to the colder one, so the oven air is the giver.
DThe casserole's heat rises as it cooks.
This option is wrong — you treated heat as something the casserole owns — its temperature and thermal energy rise; heat is the incoming transfer.
Heat is energy in transfer from a hotter object to a colder one. The oven air is hotter than the casserole, so heat flows from the air into the casserole — and once it arrives, it counts as the casserole's thermal energy.
Lesson 5 of 40 · THM-005
Heat, temperature, or thermal energy
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Did You Know?
'Shut the door — you're letting all the warmth out!' Everyday talk blurs three ideas that you have now seen separately: temperature, thermal energy, and heat. Chemists keep them apart, and one question does the sorting.
The idea
Ask what the statement reports.
If it reports how hot a sample is — the average kinetic energy of its particles — it is a statement about temperature.
If it reports the total kinetic energy of all the particles in a sample, it is a statement about thermal energy.
If it reports energy transferred from a hotter object to a colder one, it is a statement about heat.
'The burner transferred energy to the pot' reports an energy transfer from hotter to colder, so it is a statement about heat.
Sorting energy statements
Quantity
What the statement reports
How many objects
temperature
average kinetic energy of a sample's particles
one
thermal energy
total kinetic energy of all a sample's particles
one
heat
energy transferred from a hotter object to a colder one
two
Temperature and thermal energy describe one sample; heat is a transfer between two objects.
A quick test: temperature and thermal energy each belong to one sample, while heat always involves two objects — an energy giver and an energy receiver.
Worked examples
Worked example 1. 'The bathwater is at 38 °C.' Is this a statement about temperature, thermal energy, or heat?
Step 1
The statement reports how hot one sample is — a reading of the particles' average kinetic energy.
Step 2
One sample, no transfer.
Step 3
It is a statement about temperature.
Worked example 2. 'Taken as a whole, the lake's particles hold far more kinetic energy than the pond's, even though both sit at 15 °C.' Is this a statement about temperature, thermal energy, or heat?
Step 1
The statement reports the kinetic energy of ALL the particles in each sample, added up.
Step 2
It compares two totals — no energy moves between the lake and the pond.
Step 3
It is a statement about thermal energy.
You can now classify a statement about a sample as describing its temperature, its thermal energy, or heat.
Check your understanding
'Energy flowed from the grill into the burgers.' Which quantity does this statement describe?
AHeatcorrect
BTemperature
This option is wrong — you picked the one-sample average — this statement names two objects and an energy flow from the hotter to the colder, which is heat.
CThermal energy
This option is wrong — you picked the one-sample total — this statement is about energy moving between two objects, not the energy one sample holds.
DNone of the above
This option is wrong — you rejected all three — energy transferred from a hotter object to a colder one is exactly the definition of heat.
Ask what the statement reports: energy moving from one object into another. Two objects, one flow from hotter to colder — that is heat.
Check your understanding
'The oven air is at 220 °C.' Which quantity does this statement describe?
ATemperaturecorrect
BHeat
This option is wrong — you looked for a transfer — this statement involves one sample and no flow; a °C reading reports the particles' average kinetic energy.
CThermal energy
This option is wrong — you picked the total — a °C reading is the per-particle average, not the sum over all the oven air's particles.
DNone of the above
This option is wrong — you rejected all three — a reading of how hot one sample is reports its temperature.
Ask what the statement reports: how hot one sample is, as a °C reading. One sample, an average per particle — that is temperature.
Check your understanding
'Counting every particle, the full kettle's water holds twice the kinetic energy of the half-full kettle's.' Which quantity does this statement describe?
AThermal energycorrect
BTemperature
This option is wrong — you picked the average — this statement adds up the kinetic energy of every particle, and that total is thermal energy.
CHeat
This option is wrong — you looked for a transfer — nothing flows between the kettles; each total belongs to one sample on its own.
DNone of the above
This option is wrong — you rejected all three — the total kinetic energy of all a sample's particles is exactly the definition of thermal energy.
Ask what the statement reports: the kinetic energy of ALL the particles, added up. A total belonging to each sample, with no flow between them — that is thermal energy.
Lesson 6 of 40 · THM-006
Conduction
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Did You Know?
You have seen that heat flows from a hotter object into a colder one. This lesson shows how that flow travels when two objects touch — and why it runs the way it does.
The idea
Stand a metal spoon in hot soup and the spoon's handle soon warms, though only the bowl of the spoon touches the soup.
Heat transfer between objects in direct contact, in which energy passes from particle to particle where the objects touch, is called 'conduction'.
The energy passes along because when faster-moving particles collide with slower-moving ones, energy passes from the faster particles to the slower ones.
In the soup, fast-moving soup particles collide with the slower particles of the spoon, and those particles collide with their own neighbors farther up the handle.
Conduction: energy passes from particle to particle by collisions; the particles themselves stay roughly in place.
No particle travels up the spoon — each particle stays roughly in place and hands energy on by collision.
Worked examples
Worked example 1. A cast-iron skillet sits on a flame, and after a few minutes its handle is too hot to hold, though the handle never touched the flame. How did the energy reach the handle?
Step 1
The pan and handle are one piece of solid in direct contact with the flame's hot gases.
Step 2
Energy passes from particle to particle along the iron, because when faster-moving particles collide with slower-moving ones, energy passes from the faster particles to the slower ones.
Step 3
The energy reached the handle by conduction — collision after collision along the iron, with no particle leaving its place.
Worked example 2. You step barefoot onto cold bathroom tile and your feet feel cold at once. Describe the transfer that is happening where your skin touches the tile.
Step 1
Your feet are hotter than the tile, and the two are in direct contact.
Step 2
Where they touch, the faster-moving particles of your skin collide with the slower-moving particles of the tile, and energy passes from the faster particles to the slower ones.
Step 3
Heat flows from your feet into the tile by conduction — your feet feel cold because they are losing energy, not because cold climbs in.
You can now state that conduction is heat transfer between objects in direct contact, in which energy passes from particle to particle where the objects touch, because when faster-moving particles collide with slower-moving ones, energy passes from the faster particles to the slower ones.
Check your understanding
What is conduction?
AHeat transfer between objects in direct contact, in which energy passes from particle to particle where the objects touch.correct
BHeat transfer in which the fastest particles travel out of the hot object and into the cold one, carrying the energy.
This option is wrong — you moved the particles instead of the energy — in conduction each particle stays roughly in place and passes energy on by collision.
CHeat transfer by the bulk movement of a liquid or gas.
This option is wrong — you described matter carrying energy from place to place — conduction works through contact, with the particles staying put.
DHeat transfer that crosses the gap between objects that are not touching.
This option is wrong — you removed the contact — conduction happens exactly where two objects touch, particle to particle.
Conduction is heat transfer between objects in direct contact, energy passing from particle to particle where they touch. It works because when faster-moving particles collide with slower-moving ones, energy passes from the faster particles to the slower ones.
Check your understanding
A metal rod is held with one end in a flame, and the far end slowly warms. How does the energy move along the rod?
AFaster-moving particles collide with slower-moving neighbors, passing energy along while each particle stays roughly in place.correct
BHot particles from the flame end travel through the rod all the way to the cool end, carrying the energy with them.
This option is wrong — you sent the particles on a journey — the rod's particles stay in place; only the energy moves, collision by collision.
CHot air flows through tiny channels inside the metal, carrying the energy.
This option is wrong — you invented a moving fluid inside the solid — the rod's own particles pass the energy on by colliding with their neighbors.
DThe energy jumps from the flame end to the cool end without involving the particles between.
This option is wrong — you skipped the middle of the rod — conduction is a relay: every particle along the way passes energy to the next by collision.
In conduction, energy passes from particle to particle through the points of contact. It moves because when faster-moving particles collide with slower-moving ones, energy passes from the faster particles to the slower ones — a relay along the whole rod.
Check your understanding
On a summer afternoon, a car's metal seatbelt buckle burns your hand the moment you grip it. What is happening where your skin meets the buckle?
AThe buckle's faster-moving particles collide with your skin's slower-moving particles, passing energy into your hand.correct
BYour hand's particles collide with the buckle's particles and pass energy into the metal, which is why the buckle stays hot.
This option is wrong — you sent the flow the wrong way — the buckle is hotter, so its faster particles pass energy to your slower skin particles.
CHot particles leave the buckle and burrow into your skin.
This option is wrong — you moved the particles instead of the energy — the buckle's particles stay in the buckle; the energy crosses by collisions at the contact.
DThe energy crosses a small air gap as the buckle and your skin never really touch.
This option is wrong — you removed the contact — the burn happens through direct touch, energy passing particle to particle where skin meets metal.
Grip the buckle and the two surfaces are in direct contact. Because when faster-moving particles collide with slower-moving ones, energy passes from the faster particles to the slower ones — from the hot buckle into your hand. That particle-to-particle transfer through contact is conduction.
Lesson 7 of 40 · THM-007
Convection
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Did You Know?
You have seen conduction pass energy through a solid while every particle stays in place. In liquids and gases the particles are free to move — and that freedom opens a second way for heat to travel.
The idea
Heat a pot of water from below and the water soon starts to circulate on its own.
Heat transfer by the bulk movement of a liquid or gas is called 'convection'.
Convection in a heated pot: warmer, less dense water rises while cooler, denser water sinks, forming a loop that carries energy through the whole pot.
The water at the bottom warms first, expands slightly, and becomes less dense than the cooler water above it.
The warmer, less dense water rises, while the cooler, denser water sinks and takes its place.
The sinking water then warms at the bottom in its turn, so a loop forms that carries energy through the whole pot.
Convection needs a liquid or a gas, because the matter itself must be free to move from place to place — in a solid the particles are locked in position.
Worked examples
Worked example 1. A space heater sits on the floor of a cold room. Within minutes, air near the ceiling at the far side of the room is warmer. How did the energy get there?
Step 1
The heater warms the air next to it, and that air expands and becomes less dense than the cooler air around it.
Step 2
The warmer, less dense air rises, and cooler, denser air sinks and flows in to replace it.
Step 3
The moving air itself carries the energy across the room as it circulates.
Step 4
The energy traveled by convection — the warmed air moved in bulk, carrying its energy with it.
Worked example 2. In a lava lamp, blobs of warmed wax climb from the bulb at the base to the top, then drift back down. Why do the blobs rise and then sink?
Step 1
The bulb warms the wax at the base, and the warmed wax becomes less dense than the liquid around it, so it rises.
Step 2
Near the top, the wax cools, becomes denser again, and sinks back toward the bulb.
Step 3
The blobs ride a convection loop — warmer, less dense material rises; cooler, denser material sinks.
You can now state that convection is heat transfer by the bulk movement of a liquid or gas, in which warmer, less dense fluid rises while cooler, denser fluid sinks and takes its place.
Check your understanding
What is convection?
AHeat transfer by the bulk movement of a liquid or gas.correct
BHeat transfer in which energy passes from particle to particle while the particles stay in place.
This option is wrong — you described conduction — in convection the fluid itself moves from place to place, carrying its energy along.
CHeat transfer by waves that can cross empty space.
This option is wrong — you described a transfer that needs no matter — convection is the opposite: moving matter is exactly what carries the energy.
DHeat transfer that happens only inside solids.
This option is wrong — you put convection in the one state it cannot use — a solid's particles are locked in place, so only liquids and gases convect.
Convection is heat transfer by the bulk movement of a liquid or gas. Warmer, less dense fluid rises while cooler, denser fluid sinks and takes its place — the moving fluid carries the energy.
Check your understanding
Why does the air warmed by a candle flame move upward?
AThe warmed air expands and becomes less dense than the cooler air around it, so it rises.correct
BHeat naturally moves upward, and the air follows it.
This option is wrong — you gave heat a built-in direction — heat has none; it is the warm, less dense AIR that rises, carrying energy as it goes.
CThe flame pushes the air up mechanically, like a fan blade.
This option is wrong — you added a mechanical push — no push is needed; the density difference alone makes the warmer air float upward.
DThe warmed air contracts and becomes denser, and denser air always moves up.
This option is wrong — you flipped the density change — warming EXPANDS air, lowering its density, and it is the LESS dense air that rises.
Warming a gas makes it expand, so the same particles fill more space — its density falls. The warmer, less dense air rises, while cooler, denser air sinks and takes its place.
Check your understanding
You open a chest freezer and cold air spills over the rim toward the floor. Why does the cold air sink?
AThe cold air is denser than the warmer room air, so it sinks while the warmer air stays above it.correct
BColdness weighs the air down with an extra ingredient.
This option is wrong — you added a substance called cold — nothing is added; cold air is denser simply because the same particles pack a smaller volume.
CThe freezer's circulation fan pushes the cold air out over the rim and down.
This option is wrong — you added a mechanical push — the spill happens by density alone: denser fluid sinks below less dense fluid.
DCold air rises first and then falls back after mixing.
This option is wrong — you sent the dense air the wrong way first — cooler, denser air sinks directly; it is the warmer, less dense air that rises.
Cooler air is denser than warmer air. Denser fluid sinks below less dense fluid, so the freezer's cold air pours down toward the floor — the same density rule that drives every convection loop.
Lesson 8 of 40 · THM-008
Radiation
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Did You Know?
You have seen visible light as one region of the electromagnetic spectrum. Those waves can also deliver heat.
The idea
Step into sunlight and your face warms, even though nothing hot is touching you.
Heat transfer by electromagnetic waves is called 'radiation'.
Electromagnetic waves can travel through empty space, so radiation needs no matter to carry it.
The Sun's energy crosses about 150 million kilometers of almost completely empty space before it reaches your face — no particles carried it.
Conduction and convection both need particles; radiation is the one heat transfer that works across a vacuum.
Worked examples
Worked example 1. A buffet heat lamp keeps a tray of fries warm from above without touching the food. How does the energy reach the fries?
Step 1
Answer: by radiation — the lamp sends out electromagnetic waves, and the waves deliver energy to whatever they land on.
Worked example 2. You hold your hands out to the SIDE of a bed of glowing coals, level with them, and still feel warmth on your palms. Why must the warmth be arriving by radiation?
Step 1
Answer: your hands touch nothing hot, and the air warmed by the coals rises upward rather than sideways — the energy reaching your palms travels as electromagnetic waves.
You can now state that radiation is heat transfer by electromagnetic waves, which can travel through empty space with no matter carrying them.
Check your understanding
What is radiation, as a way heat travels?
AHeat transfer by electromagnetic waves, which can travel through empty space.correct
BHeat transfer by the bulk movement of a liquid or gas.
This option is wrong — you described convection — radiation needs no moving fluid; the energy rides electromagnetic waves.
CHeat transfer in which energy passes from particle to particle through direct contact.
This option is wrong — you described conduction — radiation crosses gaps, including a vacuum, with no particles passing anything along.
DHeat transfer in which fast particles fly from the hot object to the cold one.
This option is wrong — you sent particles across the gap — no matter travels in radiation; electromagnetic waves carry the energy.
Radiation is heat transfer by electromagnetic waves. Electromagnetic waves can travel through empty space, so radiation needs no matter to carry it.
Check your understanding
Which way of transferring heat still works across a vacuum, where there are no particles at all?
ARadiation, because electromagnetic waves need no matter to travel.correct
BConduction, because a vacuum is an excellent contact surface.
This option is wrong — you gave conduction nothing to work with — conduction needs touching particles, and a vacuum has none.
CConvection, because a vacuum flows more easily than any gas.
This option is wrong — you made emptiness flow — convection needs a liquid or gas to move in bulk, and a vacuum contains no matter to move.
DNo heat transfer works across a vacuum.
This option is wrong — you sealed the vacuum completely — electromagnetic waves cross it easily, which is how the Sun's energy reaches Earth.
Conduction and convection both need particles — touching ones or flowing ones. Radiation is heat transfer by electromagnetic waves, and those waves travel through empty space, so radiation alone crosses a vacuum.
Check your understanding
A ceramic heat lamp warms a reptile in a terrarium from a bulb mounted well above the basking rock, with no fan. Which statement about the energy reaching the reptile directly from the lamp is correct?
AIt arrives as electromagnetic waves — the lamp warms the reptile without touching it and without needing the air to carry the energy down.correct
BIt arrives by air warmed at the bulb sinking down onto the reptile.
This option is wrong — you sent warm air downward — warmed air is less dense and rises; the downward delivery is by electromagnetic waves.
CIt arrives through the glass walls, particle by particle, into the rock.
This option is wrong — you routed the energy through contact — the lamp is not touching the walls or the rock; its waves shine straight down onto the reptile.
DNo energy reaches the reptile directly, because energy cannot cross an air gap without matter carrying it every step.
This option is wrong — you blocked the gap — electromagnetic waves cross air and even empty space, which is exactly how a heat lamp works.
The lamp touches nothing, and warmed air rises rather than sinks — so neither contact nor moving air explains the downward warming. Radiation is heat transfer by electromagnetic waves, and the lamp's waves shine down onto the reptile directly.
Lesson 9 of 40 · THM-009
Classify a heat transfer
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Did You Know?
You have seen the three ways heat travels: conduction, convection, and radiation. A real scene rarely announces which one is at work — but one question sorts almost every case.
The idea
Ask how the energy travels from the hot thing to the thing being warmed.
If the two are in direct contact and energy passes particle to particle where they touch, the transfer is conduction.
If a liquid or gas moves in bulk and carries the energy with it, the transfer is convection.
If the energy crosses a gap as electromagnetic waves — even through empty space — the transfer is radiation.
One campfire shows all three at once.
The fire warms your face from across the fire pit by radiation — the energy crosses the open gap as waves.
One campfire, three transfers: radiation across the gap, conduction along the poker, convection in the rising air.
The metal poker resting in the flames warms along its length to your hand by conduction — energy passes particle to particle through the solid.
The hot air climbing above the flames carries energy upward by convection — the air itself moves in bulk.
Worked examples
Worked example 1. A clothes iron presses a wrinkled shirt, and the fabric under the plate warms at once. Classify the transfer.
Step 1
The hot plate and the fabric are in direct contact, and the energy enters exactly where they touch.
Step 2
Contact plus particle-to-particle transfer is the conduction signature.
Step 3
The transfer is conduction.
Worked example 2. A floor vent blows furnace-warmed air that spreads through a bedroom, warming it. Classify the transfer that warms the room.
Step 1
The warmed air itself travels from the furnace into the room, carrying its energy along.
Step 2
A gas moving in bulk is the convection signature.
Step 3
The transfer is convection.
You can now classify a described heat transfer as conduction, convection, or radiation.
Check your understanding
A hot-water bottle is tucked against cold bed sheets, and the sheets warm where the bottle rests on them. Classify the transfer.
AConductioncorrect
BConvection
This option is wrong — you looked for moving fluid — the water stays in the bottle; the sheets warm at the touching surface, particle to particle.
CRadiation
This option is wrong — you sent the energy across a gap — there is no gap; the warming happens exactly where the bottle and sheets touch.
DNone of the above
This option is wrong — you rejected all three — direct contact with energy passing where the objects touch is conduction.
Ask how the energy travels: the bottle and sheets are in direct contact, and the warming happens where they touch. Energy passing particle to particle through contact is conduction.
Check your understanding
A cat naps on a rug in a shaded corner while a narrow beam of afternoon sunlight falls across its back, far from any heater, and the fur in the beam grows warm. Classify the transfer warming the cat.
ARadiationcorrect
BConduction
This option is wrong — you looked for a hot contact — the rug in the shade is no warmer than the room, and the fur warms only where the beam of sunlight lands on it.
CConvection
This option is wrong — you looked for moving warm air — no warm current flows to the cat; the energy crosses the room as electromagnetic waves.
DNone of the above
This option is wrong — you rejected all three — energy crossing a gap as electromagnetic waves is radiation.
Ask how the energy travels: nothing hot touches the cat, and no warm fluid flows onto it. The Sun's energy crosses the gap as electromagnetic waves — radiation.
Check your understanding
Above a summer parking lot, air warmed at the hot asphalt keeps rising in shimmering columns while cooler air sinks to replace it, spreading the warmth upward. Classify the transfer spreading the warmth through the air.
AConvectioncorrect
BConduction
This option is wrong — you picked the touching step — the asphalt does warm the air it touches, but the SPREADING described is the warmed air itself rising in bulk.
CRadiation
This option is wrong — you picked waves — the described carrier is moving air, matter traveling in bulk, not electromagnetic waves.
DNone of the above
This option is wrong — you rejected all three — a gas moving in bulk and carrying energy along is convection.
Ask how the energy travels: warmed air rises while cooler air sinks — the air itself moves. A liquid or gas moving in bulk and carrying its energy along is convection.
Lesson 10 of 40 · THM-010
System and surroundings
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Did You Know?
A reaction runs in a beaker and the mixture warms. Warms compared with what? Gains energy from where? Before you can track energy anywhere, you have to draw a line around the thing you are studying.
The idea
For a reaction running in a beaker, the reacting chemicals are the part of the world you have chosen to study.
The part of the world chosen for study is called the 'system'.
Everything around the system that can exchange energy with it is called the 'surroundings'.
For the beaker reaction, the surroundings are the solution the chemicals sit in, the beaker itself, and the air around it.
Naming the system first makes energy questions precise: energy the system gives out goes to the surroundings, and energy the system takes in comes from the surroundings.
Worked examples
Worked example 1. A hand warmer packet reacts inside a glove on a ski lift. Identify the system and the surroundings.
Step 1
The thing under study is the reacting mixture sealed in the packet.
Step 2
Everything around it that can exchange energy with it — the packet's wrapper, the glove, and the hand inside — is not the thing under study.
Step 3
System: the reacting chemicals in the packet. Surroundings: the wrapper, the glove, the hand, and the air.
Worked example 2. An antacid tablet fizzes in a glass of water. If you are studying the tablet's reaction, identify the system and the surroundings.
Step 1
The thing under study is the reacting tablet material.
Step 2
The water it fizzes in, the glass, and the air can all exchange energy with the reaction but are not the thing under study.
Step 3
System: the reacting tablet chemicals. Surroundings: the water, the glass, and the air.
You can now identify the system and the surroundings in a described process, where the system is the part of the world chosen for study and the surroundings are everything around it that can exchange energy with it.
Check your understanding
Baking soda and vinegar react in a flask, and you are studying the reaction. What is the system?
AThe reacting baking soda and vinegarcorrect
BThe flask
This option is wrong — you picked the container — the flask can exchange energy with the reaction, which makes it part of the surroundings, not the thing under study.
CThe air in the room
This option is wrong — you picked what is around the setup — the air belongs to the surroundings; the system is the reacting chemicals you chose to study.
DThe flask, the mixture, and the air together
This option is wrong — you bundled everything into the system — the system is only the part chosen for study, and here that is the reacting chemicals.
The system is the part of the world chosen for study. You are studying the reaction, so the reacting baking soda and vinegar are the system — the flask and the air are surroundings.
Check your understanding
A battery powers a toy robot, and you are studying the reaction inside the battery. What are the surroundings?
AThe battery casing, the robot, and the air around themcorrect
BThe reacting chemicals inside the battery
This option is wrong — you named the system — the reacting chemicals are the part under study; the surroundings are everything around them.
COnly the air in the room
This option is wrong — you skipped the nearest neighbors — the casing and the robot can exchange energy with the reaction too, so they belong to the surroundings.
DNothing — a sealed battery has no surroundings
This option is wrong — you let the sealed casing erase the surroundings — the surroundings are whatever is around the system, sealed or not.
The surroundings are everything around the system that can exchange energy with it. The system is the reacting chemicals, so the casing, the robot, and the air all count as surroundings.
Check your understanding
During a reaction, the system gives out energy. Where does that energy go?
AInto the surroundingscorrect
BIt disappears once it leaves the system
This option is wrong — you let the energy vanish at the boundary — energy that leaves the system is received by the surroundings.
CIt stays inside the system in a hidden form
This option is wrong — you kept the energy inside — 'gives out' means it crosses the boundary, and whatever is outside the system is the surroundings.
DIt goes only into the air, never into containers or liquids
This option is wrong — you shrank the surroundings to the air — the solution, the container, and the air are all surroundings, and any of them can receive the energy.
The surroundings are everything around the system that can exchange energy with it. Energy the system gives out goes to the surroundings — the solution, the container, and the air are all on the receiving side of the line.
Lesson 11 of 40 · THM-011
Open, closed, and isolated systems
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Did You Know?
You have seen how to split the world into a system and its surroundings. The next question about any setup is what can cross the line between them.
The idea
Two things can cross a system's boundary: matter and energy.
An 'open system' exchanges both matter and energy with its surroundings.
An open mug of hot tea is an open system — energy leaves through the mug's sides, and matter leaves as vapor drifting off the top.
What can cross the boundary: open — matter and energy; closed — energy only; isolated — neither.
A 'closed system' exchanges energy but not matter.
A sealed, uninsulated bottle of water is a closed system — energy passes through the glass, but no matter gets in or out.
An 'isolated system' exchanges neither matter nor energy.
A sealed vacuum flask comes close — nothing gets in or out and almost no energy leaks — but a perfectly isolated system is an ideal that real containers only approximate.
Worked examples
Worked example 1. A pot of soup simmers on the stove with the lid off. Classify the soup-and-pot setup as an open, closed, or isolated system.
Step 1
Check matter: steam escapes from the uncovered surface, so matter crosses the boundary.
Step 2
Check energy: the stove pours energy in through the base, so energy crosses too.
Step 3
Both matter and energy cross, so the setup is an open system.
Worked example 2. A sealed aluminum can of soda sits in a warm room. Classify the can of soda as an open, closed, or isolated system.
Step 1
Check matter: the can is sealed, so no matter gets in or out.
Step 2
Check energy: the warm room's energy passes through the thin metal, warming the soda.
Step 3
Energy crosses but matter does not, so the can is a closed system.
You can now classify a described setup as an open system, which exchanges both matter and energy with its surroundings, a closed system, which exchanges energy but not matter, or an isolated system, which exchanges neither.
Check your understanding
A zip-sealed plastic pouch of juice sits in a lunchbox on a warm day, and the juice slowly warms. Classify the pouch of juice.
AA closed systemcorrect
BAn open system
This option is wrong — you let matter cross a sealed boundary — the pouch is sealed, so only energy gets through the plastic.
CAn isolated system
This option is wrong — you blocked the energy — the juice warms, so energy is clearly crossing the pouch's boundary.
DNone of the above
This option is wrong — you rejected all three — energy in, matter blocked is exactly the closed-system pattern.
Run the two checks. Matter: the pouch is sealed, so no matter crosses. Energy: the juice warms, so energy is crossing. Energy but not matter — a closed system.
Check your understanding
A kettle boils with its spout open, sending a plume of steam into the kitchen. Classify the kettle of water.
AAn open systemcorrect
BA closed system
This option is wrong — you blocked the matter — steam is water leaving through the spout, so matter crosses the boundary along with energy.
CAn isolated system
This option is wrong — you sealed everything — both the escaping steam and the energy pouring in from the element cross the boundary.
DNone of the above
This option is wrong — you rejected all three — matter and energy both crossing is exactly the open-system pattern.
Run the two checks. Matter: steam escapes through the spout, so matter crosses. Energy: the heating element pours energy in, so energy crosses. Both cross — an open system.
Check your understanding
Coffee is sealed inside a high-quality vacuum flask and stays hot for many hours. Which classification fits the flask of coffee best?
AApproximately an isolated systemcorrect
BAn open system
This option is wrong — you let matter and energy both cross — the flask is sealed and its vacuum wall lets almost no energy through.
CA closed system
This option is wrong — you let the energy flow freely — a closed system leaks energy readily, but the vacuum wall blocks almost all of it for hours.
DA perfectly isolated system
This option is wrong — you made the ideal real — the coffee does cool eventually, so a little energy leaks; real flasks only approximate isolation.
Run the two checks. Matter: sealed, so none crosses. Energy: almost none crosses for hours — but 'almost' is the key word. Neither matter nor energy (nearly) — the flask approximates an isolated system, an ideal no real container fully reaches.
Lesson 12 of 40 · THM-012
Conservation of energy
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Did You Know?
You have seen energy change form in device after device. Track the total amount through any of those changes and a pattern appears — one with no known exception.
The idea
Energy cannot be created or destroyed.
Energy can only be transferred between objects or transformed from one form to another.
This is the 'law of conservation of energy', and it is also called the 'first law of thermodynamics'.
The electrical energy a toaster draws does not disappear — it leaves as thermal energy and light.
Count every form and every place the energy went — amounts of energy are measured in joules (J), the unit met in earlier science courses — and the total after a change always equals the total before it.
An isolated system exchanges neither matter nor energy with its surroundings, so its total energy stays constant.
Inside a sealed cooler, the cold cans and the warmer sandwiches trade energy with one another, but the total energy inside stays the same.
Worked examples
Worked example 1. A phone's battery runs from full to empty during a movie. Was the battery's stored energy destroyed?
Step 1
Answer: no — it was transformed and transferred, ending up as light and sound from the screen and speaker and as thermal energy in the phone and the air. The total never changed.
Worked example 2. A lamp converts 100 J of electrical energy and gives off 5 J as light. What happened to the other 95 J?
Step 1
Answer: it became thermal energy in the bulb and the air — energy is never destroyed, so all 100 J must be accounted for: 5 J of light plus 95 J of thermal energy.
You can now state the law of conservation of energy, that energy cannot be created or destroyed but only transferred between objects or transformed from one form to another, and that this law is also called the first law of thermodynamics.
Check your understanding
What does the law of conservation of energy state?
AEnergy cannot be created or destroyed — it can only be transferred between objects or transformed from one form to another.correct
BEnergy is slowly used up whenever a device runs, so the total amount of energy shrinks a little every time.
This option is wrong — you let energy be destroyed — a running device transforms energy into other forms; the total never shrinks.
CEnergy can be created inside power stations but destroyed nowhere.
This option is wrong — you let energy be created — power stations only transform energy from one form to another; nothing creates it.
DEnergy always ends up as light, whatever form it starts in.
This option is wrong — you fixed the final form — the law fixes the TOTAL, not the form; energy can end up in any form so long as the total matches.
Energy cannot be created or destroyed — only transferred between objects or transformed from one form to another. This law is also called the first law of thermodynamics.
Check your understanding
A space heater converts 500 J of electrical energy. Which statement must be true?
AThe energy given off — nearly all of it as thermal energy — totals exactly 500 J.correct
BThe energy given off totals less than 500 J, because some energy is used up inside the heater.
This option is wrong — you destroyed the missing energy — 'used up' energy is just energy transformed into another form, and the total stays 500 J.
CThe energy given off totals more than 500 J, because heaters multiply the energy they draw.
This option is wrong — you created energy — no device gives out more energy than goes in; transformation never adds to the total.
DNo total can be given, because thermal energy cannot be counted.
This option is wrong — you dropped the bookkeeping — every joule can be tracked, and the totals before and after must match.
Energy cannot be created or destroyed — only transferred or transformed. All 500 J must therefore leave the heater in some form, and the totals before and after match exactly.
Check your understanding
By which other name is the law of conservation of energy known?
AThe first law of thermodynamicscorrect
BThe zeroth law of thermodynamics
This option is wrong — you picked the wrong number — the conservation law is the FIRST law; the zeroth law is a different statement you will meet next.
CThe second law of thermodynamics
This option is wrong — you picked the wrong number — the conservation law is the FIRST law; the second law is a different statement you will meet soon.
DThe law of conservation of matter
This option is wrong — you swapped the conserved quantity — matter conservation is about atoms; this law is about ENERGY being neither created nor destroyed.
Energy cannot be created or destroyed — only transferred or transformed. That statement carries two names: the law of conservation of energy, and the first law of thermodynamics.
Lesson 13 of 40 · THM-013
Thermal equilibrium and the zeroth law
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Did You Know?
A nurse slips a thermometer under your tongue and tells you to wait. The waiting is not a formality — it is the physics that makes the reading mean anything.
The idea
Leave two objects in contact and heat flows from the hotter one into the colder one.
The hotter object cools and the colder object warms, until the two reach the same temperature.
The condition of sharing one temperature is called 'thermal equilibrium'.
At thermal equilibrium, heat stops flowing between the objects, and their shared temperature holds.
This pattern — objects left in contact reach the same temperature — is the 'zeroth law of thermodynamics'.
A thermometer works by reaching thermal equilibrium with whatever it touches.
A fever thermometer reads your temperature only after the thermometer itself has warmed to your temperature — that is what the waiting is for.
Worked examples
Worked example 1. A can of soda from the refrigerator is left on the kitchen counter all afternoon. What temperature does the soda end up at, and why?
Step 1
Answer: room temperature — heat flows from the warmer room air into the colder can until the two reach thermal equilibrium, and then the flow stops.
Worked example 2. A meat thermometer pushed into a roast takes half a minute before its reading settles. What is happening during that half minute?
Step 1
Answer: heat flows from the hot roast into the cooler thermometer until the two reach thermal equilibrium — only then does the thermometer's reading report the roast's temperature.
You can now state that objects left in contact reach the same temperature, a condition called thermal equilibrium, and that this pattern is the zeroth law of thermodynamics, which is why a thermometer works by reaching thermal equilibrium with what it touches.
Check your understanding
What is thermal equilibrium?
AThe condition in which objects in contact have reached the same temperature and heat has stopped flowing between them.correct
BThe condition in which the hotter object has passed every bit of its energy across to the colder one.
This option is wrong — you emptied the hot object — the flow stops when the TEMPERATURES match, long before the hotter object runs out of energy.
CThe condition in which two objects touch but no energy has moved yet.
This option is wrong — you described the starting point — equilibrium is the ENDPOINT, after heat has flowed and the temperatures have leveled.
DThe condition in which both objects have cooled to the temperature of the air.
This option is wrong — you brought in a third object — equilibrium is between the objects in contact, at a shared temperature that need not match the air's.
Objects left in contact reach the same temperature — that condition is thermal equilibrium. At equilibrium the temperatures match, so heat stops flowing, and the shared temperature holds.
Check your understanding
An 80 °C metal block and a 20 °C metal block are pressed together and left alone. What is the final result?
ABoth blocks end up at one shared temperature between 20 °C and 80 °C.correct
BThe hot block stays at 80 °C and the cold block warms to 80 °C.
This option is wrong — you let the hot block give without cooling — the giver loses energy and cools while the receiver warms; they meet in between.
CThe hot block cools to 20 °C and the cold block stays at 20 °C.
This option is wrong — you let the cold block absorb without warming — the receiver gains energy and warms while the giver cools; they meet in between.
DThe blocks trade temperatures — the hot one ends at 20 °C and the cold one at 80 °C.
This option is wrong — you swapped the temperatures — heat flows one way, hot to cold, until the two SHARE one temperature; it never crosses them over.
Heat flows from the hotter block into the colder one — the hot one cools as the cold one warms. The flow continues until the two reach the same temperature, somewhere between 20 °C and 80 °C — thermal equilibrium.
Check your understanding
A cook stirs an aquarium-style glass thermometer into a pot of caramel and waits for the reading to stop climbing. Why does the settled reading report the caramel's temperature?
AThe thermometer has reached thermal equilibrium with the caramel — the two now share one temperature, and the thermometer reports its own.correct
BThe thermometer measures the caramel from a distance, so no waiting is really needed.
This option is wrong — you skipped the equilibrium step — a contact thermometer must WARM TO the caramel's temperature before its reading means anything.
CThe settled reading shows the average of the thermometer's starting temperature and the caramel's starting temperature.
This option is wrong — you averaged in the thermometer — the tiny thermometer barely changes the caramel; it warms until it matches the caramel's temperature.
DThe reading settles because the caramel has stopped holding energy.
This option is wrong — you emptied the caramel — it holds plenty of thermal energy; the reading settles because the temperature DIFFERENCE between it and the thermometer is gone.
A thermometer works by reaching thermal equilibrium with whatever it touches. Heat flows from the hot caramel into the cooler thermometer until their temperatures match — then the flow stops, the reading holds, and the thermometer's temperature IS the caramel's.
Lesson 14 of 40 · THM-014
Heat flows one way: the second law
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Did You Know?
You have seen objects in contact settle at one shared temperature. The settling always runs in the same direction — and no one has ever seen it run backward on its own.
The idea
On its own, heat always flows from hotter objects to colder ones.
The flow spreads energy out until it is evenly shared.
Heat never flows from a colder object to a hotter one by itself.
This one-way pattern is the 'second law of thermodynamics'.
A hot drink left on a desk cools to room temperature — and it never gathers energy back from the room to reheat itself.
The reverse would not break the energy count, since the energy would all still exist — but it simply never happens on its own.
Worked examples
Worked example 1. A baked potato is left on the counter in a cool kitchen. What happens to it, and can the reverse happen on its own?
Step 1
Answer: heat flows from the hot potato into the cooler air until the potato reaches room temperature — and the reverse, the room's energy re-collecting to reheat the potato, never happens by itself.
Worked example 2. A warm pie cools on a windowsill in cool evening air. In which direction does heat flow, and could the flow ever run the other way on its own?
Step 1
Answer: from the hotter pie into the cooler air, spreading the energy out — and never the other way by itself; that is the second law of thermodynamics.
You can now state that heat on its own always flows from hotter objects to colder ones, spreading energy out until it is evenly shared, and never flows the other way by itself, a pattern called the second law of thermodynamics.
Check your understanding
What does the second law of thermodynamics say about the direction heat flows on its own?
AHeat flows from hotter objects to colder ones, and never from colder to hotter by itself.correct
BHeat flows from colder objects to hotter ones, feeding the hotter object's lead.
This option is wrong — you reversed the one-way street — left alone, energy always passes from the hotter object to the colder one.
CHeat flows equally in both directions at all times, whatever the temperatures.
This option is wrong — you erased the direction — while a temperature difference exists, the flow runs one way: hot to cold.
DHeat does not flow at all unless something pushes it.
This option is wrong — you stalled the flow — a temperature difference is itself enough; heat flows hot-to-cold on its own.
On its own, heat always flows from hotter objects to colder ones, spreading energy out until it is evenly shared. It never flows the other way by itself — that one-way pattern is the second law of thermodynamics.
Check your understanding
A mug of cocoa is forgotten in a cold car overnight. Which statement describes what happens?
AHeat flows from the cocoa into the cold air until the energy is evenly shared, and the cocoa never rewarms on its own.correct
BHeat flows from the cold air into the cocoa, keeping it warm through the night.
This option is wrong — you reversed the direction — the cocoa is the hotter object, so it is the giver, cooling as the night goes on.
CThe cocoa cools first and then gradually rewarms overnight as the cold air gives the energy back to it.
This option is wrong — you let the flow reverse by itself — once the energy has spread out, it never re-collects into the cocoa on its own.
DNothing happens, because the sealed car keeps all heat flows stopped.
This option is wrong — you froze the physics — inside the car, the hot cocoa and cold air are in contact through the mug and the air, and heat flows hot to cold regardless.
On its own, heat always flows from hotter objects to colder ones. The cocoa loses energy to the cold air until the temperatures level out — and the spread-out energy never gathers itself back into the mug.
Check your understanding
A frozen gel pack sits against a warm yogurt cup in a lunch bag, keeping the yogurt cold all morning. Which statement explains the cooling?
AHeat flows from the warmer yogurt into the colder gel pack, so the yogurt loses energy.correct
BColdness flows from the gel pack into the yogurt, chilling it.
This option is wrong — you moved 'cold' as if it were a substance — only energy flows, and it flows out of the warmer yogurt into the colder pack.
CHeat flows from the gel pack into the yogurt, because frozen things push their energy outward.
This option is wrong — you reversed the direction — the pack is the colder object, so it receives energy; it never gives energy to something warmer by itself.
DNo energy moves — the pack simply blocks warmth from existing nearby.
This option is wrong — you replaced a flow with a shield — the yogurt cools because its energy really does drain into the colder pack, hot to cold.
On its own, heat always flows from hotter objects to colder ones. The yogurt is the hotter object, so its energy drains into the frozen pack — the yogurt stays cold because it keeps LOSING energy, not because cold flows in.
Lesson 15 of 40 · THM-015
Absolute zero: the third law
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Have You Ever Wondered?
Wonder this:
Heating has an open road: furnaces run past 3000 °C, and the centers of stars run past millions. Cooling does not. Below a certain temperature, nothing in the universe can go. Where is the floor?
You have seen that temperature measures the average kinetic energy of a sample's particles, so a colder sample's particles move more slowly on average.
The idea
Cooling a sample slows its particles down.
The slowing cannot continue forever — there is a point where particle motion falls to its minimum.
You met that temperature with gases: it is absolute zero — the temperature of least possible particle motion, with no temperature below it.
Absolute zero is −273 °C, which is 0 K (zero kelvin).
The 'third law of thermodynamics' sets absolute zero as a limit: real objects can approach it but never reach it.
Temperature has no ceiling, but it has a floor: absolute zero, −273 °C (0 K).
Laboratory equipment has cooled samples to within a fraction of a degree of −273 °C, but never to −273 °C itself.
Worked examples
Worked example 1. What is the lowest possible temperature, in degrees Celsius and in kelvin?
Step 1
Answer: −273 °C, which is 0 K.
Worked example 2. A research team cools a cloud of sodium atoms to 0.000001 K. Can the team keep removing energy until the cloud sits at 0 K exactly?
Step 1
Answer: No — the third law of thermodynamics sets absolute zero as a limit that real objects can approach but never reach.
You can now state that there is a lowest possible temperature, absolute zero at −273 °C (0 K, zero kelvin), where particle motion falls to its minimum, and that the third law of thermodynamics sets it as a limit that real objects can approach but never reach.
Check your understanding
What temperature is absolute zero?
A−273 °Ccorrect
B0 °C
This option is wrong — you placed the floor at water's freezing point — 0 °C is where water freezes; absolute zero sits far below it, at −273 °C.
C273 °C
This option is wrong — you dropped the minus sign — absolute zero is 273 degrees BELOW zero on the Celsius scale.
D−373 °C
This option is wrong — you mixed in the boiling-point mark of 373 K — absolute zero is −273 °C, which is 0 K.
Absolute zero is the lowest possible temperature. It sits at −273 °C, which is 0 K.
Check your understanding
What happens to the motion of a sample's particles as the sample is cooled toward absolute zero?
AIt falls toward its minimum.correct
BIt speeds up as the particles pack closer together.
This option is wrong — you flipped the direction — cooling means the particles' average kinetic energy is falling, so their motion slows.
CIt stays the same, because particle motion does not depend on temperature.
This option is wrong — you cut the tie between temperature and motion — temperature measures the particles' average kinetic energy, so cooling means slowing.
DIt already reached its minimum at 0 °C.
This option is wrong — you placed the floor at water's freezing point — particles in ice at 0 °C still move plenty; the minimum lies at −273 °C.
Cooling a sample slows its particles down. At absolute zero, −273 °C, particle motion falls to its minimum.
Check your understanding
Deep space sits at about 3 K. Can any object be colder than 0 K?
ANo — 0 K is the lowest possible temperature.correct
BYes — temperatures below 0 K occur naturally in the emptiest parts of space.
This option is wrong — you let nature pass the floor — even the coldest parts of space sit a few kelvin ABOVE absolute zero, and nothing sits below it.
CYes — laboratory freezers routinely cool samples below 0 K.
This option is wrong — you treated the floor as an equipment problem — laboratories can approach 0 K, but no equipment can reach or pass it.
DNo — but ordinary home freezers already run at 0 K.
This option is wrong — you placed everyday cold at the floor — a home freezer runs near −18 °C, which is 255 K, far above absolute zero.
Absolute zero, 0 K, is the lowest possible temperature. Real objects can approach it but never reach it — and nothing sits below it.
Lesson 16 of 40 · THM-016
Match the observation to the law
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You have seen the four laws of thermodynamics one at a time: conservation of energy, thermal equilibrium, one-way heat flow, and the absolute-zero limit. Everyday observations usually illustrate exactly one of them — this lesson is the matching routine.
The idea
Match an observation to a law by asking what the observation is about.
An observation about energy changing form or place while the total stays the same illustrates the first law.
An observation about touching objects ending up at the same temperature illustrates the zeroth law.
An observation about heat flowing only from hot to cold, never the reverse on its own, illustrates the second law.
An observation about a lowest temperature that can be approached but never reached illustrates the third law.
The zeroth and second laws sit close together: reaching the SAME temperature points to the zeroth law, while the one-way DIRECTION of the flow points to the second.
Matching an observation to a law
The observation is about…
Law
touching objects ending up at the same temperature
zeroth law
energy changing form or place while the total stays the same
first law
heat flowing only from hot to cold, never the reverse on its own
second law
a lowest temperature approached but never reached
third law
Ask what the observation is about, then read off the law.
Ice cream melting in a warm room is heat flowing one way, from the warm air into the cold ice cream — the second law.
A wall thermostat reads the room correctly only because the sensor inside it has reached the room's temperature — the zeroth law.
A flashlight's chemical energy leaves as light and a little warmth, with nothing lost — the first law.
A physics team cooling atoms ever closer to −273 °C without ever arriving — the third law.
Worked examples
Worked example 1. A blacksmith's glowing horseshoe dropped into a water barrel always cools the horseshoe and warms the water — barrel water never re-heats a cold horseshoe by itself. Which law does this illustrate?
Step 1
Ask what the observation is about: the one-way direction of heat flow — hot to cold, never the reverse on its own.
Step 2
The second law of thermodynamics.
Worked example 2. A cheesemaker's temperature probe must sit in the milk vat for a minute before its display settles. Which law does the probe's working depend on?
Step 1
Ask what the observation is about: the probe and the milk reaching the same temperature.
Step 2
The zeroth law of thermodynamics.
Worked example 3. An electric fan draws electrical energy and delivers moving air, a little sound, and a little warmth — measured together, the outputs equal the energy drawn. Which law does the measurement illustrate?
Step 1
Ask what the observation is about: energy changing form while the total stays the same.
Step 2
The first law of thermodynamics.
You can now identify which of the four laws of thermodynamics an everyday observation illustrates.
Check your understanding
Left in a warm kitchen, a tray of ice cubes always melts — the kitchen air never freezes the tray harder while warming itself. Which law of thermodynamics does this illustrate?
AThe second law of thermodynamicscorrect
BThe zeroth law of thermodynamics
This option is wrong — you matched the scene to temperature-matching — the observation stresses which way the energy moves (warm air to cold ice, never the reverse), and one-way direction is the second law.
CThe first law of thermodynamics
This option is wrong — you looked for an energy total — nothing here is about the total staying the same; the observation is about the one-way direction of heat flow.
DThe third law of thermodynamics
This option is wrong — you matched melting ice to a temperature floor — nothing here approaches absolute zero; the observation is about heat's one-way direction.
Ask what the observation is about: heat moving from the warm air into the cold ice, and never the reverse on its own. One-way direction of heat flow is the second law of thermodynamics.
Check your understanding
A candy thermometer clipped into hot syrup needs a moment before its reading settles at the syrup's temperature. Which law does the thermometer's working depend on?
AThe zeroth law of thermodynamicscorrect
BThe second law of thermodynamics
This option is wrong — you matched the scene to heat's direction — the thermometer works because it and the syrup END UP at the same temperature, and that shared final temperature is the zeroth law.
CThe first law of thermodynamics
This option is wrong — you looked for an energy total — the thermometer's working rests on reaching the syrup's temperature, which is thermal equilibrium, the zeroth law.
DThe third law of thermodynamics
This option is wrong — you matched a thermometer to the absolute-zero limit — nothing here approaches −273 °C; the reading settles because probe and syrup reach one temperature.
Ask what the observation is about: the thermometer and the syrup reaching the same temperature. Touching objects ending up at the same temperature is thermal equilibrium — the zeroth law of thermodynamics.
Check your understanding
A battery-powered toy car's stored chemical energy leaves as motion, sound, and warmth — careful measurement finds the outputs together equal the energy the battery supplied. Which law does the measurement illustrate?
AThe first law of thermodynamicscorrect
BThe second law of thermodynamics
This option is wrong — you matched the warmth to heat's direction — the point of the measurement is that the energy TOTAL is unchanged, which is conservation of energy, the first law.
CThe zeroth law of thermodynamics
This option is wrong — you matched the scene to temperature-matching — no two objects are settling at a shared temperature here; the observation is about the unchanged energy total.
DThe third law of thermodynamics
This option is wrong — you matched a toy car to the absolute-zero limit — nothing here approaches −273 °C; the observation is about the unchanged energy total.
Ask what the observation is about: chemical energy changing into motion, sound, and warmth while the total stays the same. An unchanged energy total through a transformation is the first law of thermodynamics.
Summary video — Energy, heat, and the laws of thermodynamics
You have seen how to pick out a process's system and its surroundings, and you have seen reactions that warm the beaker around them. This lesson names that pattern precisely.
The idea
In some processes, the system releases energy to the surroundings.
A process that releases energy to the surroundings is called 'exothermic'.
The released energy has to go somewhere: it enters the surroundings, so the surroundings warm up.
A burning candle is an exothermic process: the burning wax is the system, and it releases energy that warms the air around the flame.
Exothermic: the system releases energy to the surroundings, so the surroundings warm up.
Worked examples
Worked example 1. A charcoal grill burns, and the air above it shimmers with warmth. In this exothermic process, name the system and state which way the energy moves.
Step 1
Answer: the burning charcoal is the system, and it releases energy to the surroundings — the air above the grill, which warms up.
Worked example 2. When cement powder is mixed with water, the mixture slowly warms itself, the bucket, and the air. Is the process exothermic, and what tells you?
Step 1
Answer: yes — the surroundings (the bucket and the air) warm up, so the system released energy to them.
You can now state that a process is exothermic when the system releases energy to the surroundings, so the surroundings warm up.
Check your understanding
In an exothermic process, which way does energy move?
AFrom the system to the surroundings.correct
BFrom the surroundings to the system.
This option is wrong — you flipped the direction — an exothermic process RELEASES energy, so the energy leaves the system and enters the surroundings.
CIt stays inside the system.
This option is wrong — you kept the energy at home — 'releases' means the energy leaves the system and enters the surroundings.
DIt disappears as the process runs.
This option is wrong — you let energy vanish — energy cannot be destroyed; the released energy enters the surroundings, which is why they warm up.
A process that releases energy to the surroundings is exothermic. The released energy leaves the system, enters the surroundings, and warms them.
Check your understanding
A pocket hand warmer is snapped, and the paste inside reacts. The reacting paste is the system. What happens to the glove and hand around it?
AThey warm up as they receive energy from the system.correct
BThey cool down as they supply energy to the system.
This option is wrong — you ran the energy the wrong way — a hand warmer's reacting paste releases energy, and the surroundings that receive it warm up.
CTheir temperature stays the same, because the energy stays inside the paste.
This option is wrong — you kept the released energy inside the system — released energy leaves the system and enters the surroundings.
DThey warm up by creating their own energy.
This option is wrong — you created energy from nothing — the warmth arrives FROM the reacting paste; energy is transferred, never created.
The reacting paste is the system, and it releases energy. The glove and hand are surroundings: they receive that energy and warm up.
Check your understanding
A self-heating coffee can holds a sealed compartment where calcium oxide reacts with water; the coffee around the compartment heats to drinking temperature. Which statement describes the energy flow?
AThe reacting compartment releases energy to the coffee around it.correct
BThe coffee supplies energy to the reacting compartment.
This option is wrong — you flipped the direction — the coffee WARMS, so it must be receiving energy, not supplying it.
CEnergy is created inside the compartment and added to the world's total.
This option is wrong — you created energy from nothing — the reaction releases energy it transfers to the coffee; no energy is created.
DThe coffee warms because the reaction absorbs energy from it.
This option is wrong — you paired warming with absorbing — a system that absorbs energy COOLS its surroundings; warming surroundings mean the system released energy.
The reacting compartment is the system; the coffee is part of the surroundings. The coffee warms, so energy is entering it — released by the reacting system. That is the exothermic pattern: system releases, surroundings warm.
Lesson 18 of 40 · THM-018
Endothermic processes
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You have seen exothermic processes: the system releases energy, and the surroundings warm up. Other processes run the flow in reverse.
The idea
In some processes, the system absorbs energy from the surroundings.
A process that absorbs energy from the surroundings is called 'endothermic'.
The absorbed energy has to come from somewhere: it leaves the surroundings, so the surroundings cool down.
An endothermic process needs its energy fed in the whole time it runs — and the energy has to come from somewhere: if the surroundings stop supplying it, the process draws energy from the reacting mixture itself, and the mixture grows colder and colder.
Endothermic: the system absorbs energy from the surroundings, so the surroundings cool down.
A chemical cold pack is an endothermic process: the reacting contents are the system, and they absorb energy from your skin, so the skin they touch cools.
Worked examples
Worked example 1. A cake batter bakes only while the oven keeps supplying energy — take it out early and the baking stops. Is baking the batter endothermic, and what tells you?
Step 1
Answer: yes — the batter (the system) absorbs energy from the hot oven air the whole time it bakes; a process that absorbs energy from its surroundings is endothermic.
Worked example 2. When certain salts dissolve in a glass of water, the glass turns noticeably colder. Which way did the energy move?
Step 1
Answer: from the surroundings into the system — the water and glass cooled because the dissolving salt absorbed energy from them.
You can now state that a process is endothermic when the system absorbs energy from the surroundings, so the surroundings cool down.
Check your understanding
In an endothermic process, which way does energy move?
AFrom the surroundings into the system.correct
BFrom the system into the surroundings.
This option is wrong — you flipped the direction — an endothermic process ABSORBS energy, so the energy leaves the surroundings and enters the system.
CIt stays inside the surroundings.
This option is wrong — you froze the flow — 'absorbs' means energy leaves the surroundings and enters the system, which is why the surroundings cool.
DNew energy appears inside the system from nowhere.
This option is wrong — you created energy from nothing — the system's gained energy is taken FROM the surroundings; energy is transferred, never created.
A process that absorbs energy from the surroundings is endothermic. The absorbed energy leaves the surroundings, which is why they cool down.
Check your understanding
Frost forms on the outside of a beaker while a reaction runs inside it. What does the beaker's chilled surface tell you about the reaction?
AThe reaction is absorbing energy from the beaker and the air around it.correct
BThe reaction is releasing energy to the beaker and the air around it.
This option is wrong — you paired cooling with releasing — released energy would WARM the beaker; a chilled beaker means energy is being drawn out of it.
CThe reaction has stopped exchanging energy with anything.
This option is wrong — you read frost as no-exchange — the beaker got cold because energy is actively leaving it for the system.
DThe reaction is passing coldness out into the beaker.
This option is wrong — you treated cold as a substance that flows — cold is not a thing that moves; the beaker chills because ENERGY moves out of it into the reacting system.
The beaker and nearby air are surroundings, and they cooled. Cooling surroundings mean the system is absorbing energy from them. That is the endothermic pattern: system absorbs, surroundings cool.
Check your understanding
A reaction that absorbs energy is running in a flask standing in warm water. The flask is moved onto an insulating pad, where nothing supplies it energy. What happens?
AThe reaction starts drawing energy from the mixture itself, so the mixture gets colder and colder.correct
BThe reaction keeps running with no other change, because reactions make their own energy.
This option is wrong — you created energy from nothing — the absorbed energy has to come from somewhere; with no outside supply, the reaction drains the mixture's own energy, and the mixture cools.
CThe reaction speeds up, because the insulation traps energy inside the flask.
This option is wrong — you treated insulation as an energy source — insulation only blocks transfer; it adds nothing, and the reaction still has to take its energy from somewhere.
DThe reaction stops instantly, because no energy at all is left for it.
This option is wrong — you cut off every source at once — one source is still there: the mixture's own energy; the reaction can draw on that, cooling the mixture as it does.
The absorbed energy has to come from somewhere. On the insulating pad the surroundings supply nothing, so the reaction draws energy from the mixture itself. That is why the mixture gets colder and colder while the reaction runs.
Lesson 19 of 40 · THM-019
Breaking bonds costs energy
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Did You Know?
You have seen why atoms bond: the atoms end up with lower energy bonded together than they have when separate. That energy valley has a price of admission — in reverse.
The idea
Bonded atoms sit at lower energy than the same atoms separated.
Pulling them apart pushes them back up to the higher-energy separated state.
So breaking a chemical bond always requires an energy input.
Breaking a bond is an uphill change: energy must be supplied to lift the atoms to the separated state.
No bond breaks for free — the energy must be supplied from outside.
Splitting water molecules into hydrogen and oxygen needs a steady supply of electrical energy; cut the electricity, and the splitting stops.
Worked examples
Worked example 1. High in the atmosphere, an ozone molecule splits when one of its oxygen–oxygen bonds breaks. What had to be supplied for the bond to break, and where did it come from?
Step 1
Answer: an energy input — supplied by sunlight striking the molecule.
Worked example 2. In industry, sodium is made by running an electric current through molten salt without pause. Why can the process not continue if the current is switched off?
Step 1
Answer: breaking the bonds between the ions requires a steady energy input, and without the electrical supply no more bonds can be broken.
You can now state that breaking a chemical bond always requires an energy input.
Check your understanding
What does breaking a chemical bond always require?
AAn input of energy.correct
BA release of energy.
This option is wrong — you flipped the direction — bonded atoms sit at LOWER energy, so pulling them apart is an uphill change that must be paid for with energy in.
CNo energy change at all.
This option is wrong — you made bond breaking free — the bonded and separated states sit at different energies, and climbing from lower to higher always costs energy.
DA drop to lower energy.
This option is wrong — you sent the atoms downhill — the separated atoms sit HIGHER in energy than the bonded ones, so breaking a bond is the uphill direction.
Bonded atoms sit at lower energy than the same atoms separated. Breaking the bond pushes them uphill to the separated state, so it always requires an energy input.
Check your understanding
Hydrogen gas is made of H₂ molecules. What must happen for an H₂ molecule to split into two separate hydrogen atoms?
AEnergy must be supplied to break the bond.correct
BEnergy must be released as the bond breaks.
This option is wrong — you flipped the direction — breaking a bond never gives energy out; it takes energy in.
CNothing — the atoms drift apart on their own with no energy involved.
This option is wrong — you made bond breaking free — the bonded molecule rests at lower energy, and it stays bonded until outside energy pushes the atoms apart.
DThe separated atoms must first drop to lower energy than the molecule.
This option is wrong — you inverted the energy levels — the separated atoms sit HIGHER in energy than the bonded molecule, which is exactly why separating them costs energy.
The bonded H₂ molecule sits at lower energy than the two separate atoms. Splitting it is an uphill change, so energy must be supplied from outside.
Check your understanding
A student says: 'Breaking bonds releases energy.' What is wrong with the claim?
ABreaking a bond never releases energy — it always requires an energy input.correct
BNothing — breaking bonds does release energy.
This option is wrong — you endorsed the flip — bonded atoms sit at lower energy, so breaking a bond is an uphill change that absorbs energy, never releases it.
CBreaking a bond releases energy only in fuels.
This option is wrong — you kept the flip alive for one class of substances — bond breaking costs energy in every substance, fuels included.
DBreaking a bond neither absorbs nor releases energy.
This option is wrong — you made bond breaking energy-neutral — the separated state sits higher in energy, and reaching it always costs an input.
Bonded atoms sit at lower energy than separated atoms. Breaking a bond pushes them uphill, so it always requires an energy input — in every substance.
Lesson 20 of 40 · THM-020
Forming bonds releases energy
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Did You Know?
You have seen that breaking a chemical bond always requires an energy input. Bond forming runs the same energy hill in the other direction.
The idea
Separated atoms sit at higher energy than the same atoms bonded together.
When atoms come together and a bond forms, they drop to the lower-energy bonded state.
The energy difference does not vanish — it leaves the atoms and enters whatever is around them.
Forming a bond is a downhill change: the atoms drop to the bonded state, and the difference is released.
So forming a chemical bond always releases energy.
When hydrogen and oxygen combine into water molecules, energy is released as the new bonds form.
Worked examples
Worked example 1. High in the atmosphere, two separate oxygen atoms meet and bond into an O₂ molecule. What happens to the pair's energy as the bond forms?
Step 1
Answer: the atoms drop to the lower-energy bonded state, and the difference is released as energy.
Worked example 2. Two chlorine atoms join into a Cl₂ molecule. Does forming the new bond absorb energy, release energy, or neither?
Step 1
Answer: it releases energy — forming a chemical bond always releases energy.
You can now state that forming a chemical bond always releases energy.
Check your understanding
What happens to energy when a chemical bond forms?
AEnergy is released.correct
BEnergy is absorbed.
This option is wrong — you flipped the direction — forming a bond is the downhill change to lower energy, and the difference comes OUT.
CNothing happens to energy.
This option is wrong — you made bond forming energy-neutral — the bonded state sits lower in energy, and the difference must go somewhere: it is released.
DEnergy is stored in the bond, waiting to be released when the bond breaks.
This option is wrong — you stored the energy for later — the energy leaves the atoms as the bond forms; breaking the bond later will COST energy, not pay any out.
Separated atoms sit at higher energy than bonded atoms. Forming the bond drops them to the lower level, and the difference is released as energy.
Check your understanding
Two hydrogen atoms approach and bond into an H₂ molecule. Which statement compares the molecule's energy with the energy of the two separate atoms?
AThe molecule has lower energy — the difference was released as the bond formed.correct
BThe molecule has higher energy — the difference was absorbed as the bond formed.
This option is wrong — you flipped both halves — bonding is the drop to LOWER energy, and the difference is released, not absorbed.
CThe molecule has the same energy as the two separate atoms.
This option is wrong — you made bonding energy-neutral — if the energies were equal, nothing would hold the atoms together.
DThe molecule has lower energy — the difference was absorbed into the bond.
This option is wrong — you got the levels right but parked the energy inside the bond — the difference LEAVES the atoms as the bond forms.
Bonded atoms sit at lower energy than the same atoms separated. The drop happens as the bond forms, and the energy difference is released to the outside.
Check your understanding
Iron and oxygen atoms bond as rust forms on a nail. What does the forming of each new bond do?
AIt releases energy.correct
BIt absorbs energy.
This option is wrong — you flipped the direction — forming a bond always releases energy, in rust as in everything else.
CIt neither absorbs nor releases energy.
This option is wrong — you made bond forming energy-neutral — the atoms drop to lower energy, and the difference is released.
DIt releases energy only if the nail is heated first.
This option is wrong — you added a heating condition — forming a bond releases energy every time, warm nail or cold.
Forming a chemical bond always releases energy. Each new iron–oxygen bond drops the atoms to a lower-energy state, and the difference is released.
Lesson 21 of 40 · THM-021
Why reactions release or absorb energy
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Have You Ever Wondered?
Wonder this:
Burning methane breaks strong bonds, and breaking bonds costs energy. Yet a lit gas stove pours heat into the kitchen. If the first step of burning swallows energy, where does the stove's heat come from?
You have seen the two halves separately: breaking bonds requires energy, and forming bonds releases energy. Every reaction does both.
The idea
Every chemical reaction breaks bonds in the reactants and forms new bonds in the products.
Breaking the reactants' bonds requires energy in.
Forming the products' bonds releases energy out.
The overall direction comes from comparing the two amounts.
A reaction is exothermic when forming the products' new bonds releases more energy than breaking the reactants' bonds requires.
Compare energy in (breaking) with energy out (forming): the larger side sets the reaction's direction.
A reaction is endothermic when breaking the reactants' bonds requires more energy than forming the products' bonds releases.
Burning methane is exothermic because its new bonds release more energy than its old bonds cost to break — the surplus is the stove's heat.
Worked examples
Worked example 1. When hydrogen burns in oxygen, forming the water molecules' new bonds releases more energy than breaking the hydrogen and oxygen bonds requires. Classify the reaction.
Step 1
Compare the two amounts: more energy is released by bond forming than is required by bond breaking.
Step 2
More out than in leaves a surplus for the surroundings.
Step 3
The reaction is exothermic.
Worked example 2. When nitrogen and oxygen react to form nitrogen monoxide, breaking the strong N₂ and O₂ bonds requires more energy than forming the new bonds releases. Classify the reaction.
Step 1
Compare the two amounts: more energy is required by bond breaking than is released by bond forming.
Step 2
More in than out leaves a shortfall that the surroundings must supply.
Step 3
The reaction is endothermic.
You can now explain that a reaction is exothermic when forming the products' new bonds releases more energy than breaking the reactants' bonds requires, and endothermic when bond breaking requires more energy than bond forming releases.
Check your understanding
In a reaction, forming the products' bonds releases more energy than breaking the reactants' bonds requires. Classify the reaction.
AExothermiccorrect
BEndothermic
This option is wrong — you flipped the comparison — more energy out (forming) than in (breaking) leaves a surplus for the surroundings, which is the exothermic pattern.
CNeither — the two amounts always cancel exactly
This option is wrong — you assumed the books always balance — the stem says forming releases MORE, and that surplus is released to the surroundings.
DIt cannot be classified without measuring a temperature
This option is wrong — you demanded a thermometer — the bond-energy comparison alone settles the direction: more out than in means exothermic.
Compare the two amounts: energy out from forming the products' bonds is larger than energy in to break the reactants' bonds. A reaction is exothermic when forming the products' new bonds releases more energy than breaking the reactants' bonds requires.
Check your understanding
Propane burns exothermically in a camping heater. Which statement about its bond energies must be true?
AForming the products' bonds releases more energy than breaking the reactants' bonds requires.correct
BBreaking the reactants' bonds requires more energy than forming the products' bonds releases.
This option is wrong — you wrote the endothermic condition — a shortfall would pull energy IN from the surroundings, and a heater pours energy out.
CBreaking the reactants' bonds releases the energy, and forming the products' bonds absorbs it.
This option is wrong — you swapped the two roles — breaking always REQUIRES energy and forming always RELEASES it, whatever the reaction.
DNo bonds break when propane burns — only new bonds form.
This option is wrong — you skipped the first half of every reaction — the propane and oxygen bonds must break before the products' bonds can form.
Every reaction both breaks bonds (energy in) and forms bonds (energy out). The heater releases a surplus, so the forming side must release more than the breaking side requires.
Check your understanding
Plants make glucose and oxygen only while absorbing light energy — photosynthesis is endothermic. Which bond-energy comparison explains why the process needs that steady supply?
ABreaking the reactants' bonds requires more energy than forming the products' bonds releases.correct
BForming the products' bonds releases more energy than breaking the reactants' bonds requires.
This option is wrong — you wrote the exothermic condition — a surplus would flow OUT, and photosynthesis takes energy in.
CBreaking the reactants' bonds releases energy, and forming the products' bonds requires it.
This option is wrong — you swapped the two roles — breaking always requires energy and forming always releases it; only the AMOUNTS differ between reactions.
DThe light supplies energy because no bonds form during photosynthesis.
This option is wrong — you removed the forming half — new bonds do form in glucose and oxygen; the light covers the SHORTFALL between breaking's cost and forming's release.
A reaction is endothermic when bond breaking requires more energy than bond forming releases. The gap between the two amounts is the shortfall the absorbed light keeps paying.
Lesson 22 of 40 · THM-022
Read an energy diagram
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Did You Know?
You have seen exothermic and endothermic named in words and explained by bond energies. Chemists also show a reaction's energy change as a picture that can be read at a glance.
The idea
An 'energy diagram' is a graph of energy, read on the vertical axis, across the course of a reaction.
The energy of the reactants is drawn as a flat level on the left.
Four energy diagrams. Products below reactants: energy released (exothermic). Products above reactants: energy absorbed (endothermic).
The energy of the products is drawn as a flat level on the right.
A smooth curve joins the two levels.
When the product level sits LOWER than the reactant level, the reaction released energy — it is exothermic.
When the product level sits HIGHER than the reactant level, the reaction absorbed energy — it is endothermic.
The vertical gap between the two levels shows how much energy was released or absorbed.
In the figure, reaction 1's products sit below its reactants — exothermic — and reaction 2's products sit above its reactants — endothermic.
Worked examples
Worked example 1. Panel C of the figure shows the energy diagram for reaction 3. Classify reaction 3 as exothermic or endothermic.
Step 1
Read the two levels: reaction 3's product level sits lower than its reactant level.
Step 2
A product level below the reactant level means energy was released.
Step 3
Reaction 3 is exothermic.
Worked example 2. Panel D of the figure shows the energy diagram for reaction 4. Classify reaction 4 as exothermic or endothermic.
Step 1
Read the two levels: reaction 4's product level sits higher than its reactant level.
Step 2
A product level above the reactant level means energy was absorbed.
Step 3
Reaction 4 is endothermic.
You can now classify a reaction as exothermic or endothermic from its energy diagram, a graph that shows the energy of the reactants and of the products as horizontal levels: exothermic when the product level sits lower than the reactant level, endothermic when it sits higher.
Check your understanding
The energy diagram for a reaction is shown. Classify the reaction.
AExothermiccorrect
BEndothermic
This option is wrong — you flipped the reading — a product level BELOW the reactant level means the reaction released energy, which is exothermic.
CNeither — the diagram shows no energy change
This option is wrong — you missed the gap — the two levels sit at different heights, and that vertical difference IS the energy change.
DIt cannot be classified from an energy diagram
This option is wrong — you assumed the picture carries no direction — the relative heights of the two levels are exactly what settle exothermic against endothermic.
Read the two flat levels: the products sit lower than the reactants. When the product level sits lower than the reactant level, the reaction released energy — it is exothermic.
Check your understanding
The energy diagram for a different reaction is shown. Classify the reaction.
AEndothermiccorrect
BExothermic
This option is wrong — you flipped the reading — a product level ABOVE the reactant level means the reaction absorbed energy, which is endothermic.
CNeither — the diagram shows no energy change
This option is wrong — you missed the gap — the products sit clearly above the reactants, and that vertical difference is the absorbed energy.
DIt cannot be classified from an energy diagram
This option is wrong — you assumed the picture carries no direction — the relative heights of the two levels settle the classification.
Read the two flat levels: the products sit higher than the reactants. When the product level sits higher than the reactant level, the reaction absorbed energy — it is endothermic.
Check your understanding
Diagrams A and B show the energy diagrams of two reactions. Which diagram shows an exothermic reaction?
ADiagram Bcorrect
BDiagram A
This option is wrong — you flipped the reading — diagram A's products sit above its reactants, which is the endothermic pattern; exothermic needs the products BELOW.
CBoth diagrams
This option is wrong — you classified any level change as exothermic — only a product level below the reactant level shows released energy.
DNeither diagram
This option is wrong — you found no release in either — diagram B's products sit below its reactants, which is exactly the released-energy pattern.
Compare each diagram's two levels. Diagram B's products sit below its reactants — energy released, exothermic. Diagram A's products sit above its reactants — energy absorbed, endothermic.
Lesson 23 of 40 · THM-023
Sketch an energy diagram
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Did You Know?
You have seen how to read an energy diagram. Drawing one is the reverse skill: from a description of a process, you place the levels yourself.
The idea
Draw a vertical axis and label it 'energy'.
Draw a flat level on the left and label it 'reactants'.
The drawing target for an exothermic process: labeled axis, labeled flat levels, smooth curve, labeled gap.
Decide the products' height from the process: exothermic means the products end up LOWER in energy, endothermic means they end up HIGHER.
Draw the flat products level on the right at that height, and label it 'products'.
Join the two levels with one smooth curve.
Mark the vertical gap between the levels, and label it 'energy released' for an exothermic process or 'energy absorbed' for an endothermic one.
Worked examples
Worked example 1. Sketch the energy diagram for the exothermic burning of magnesium.
Step 1
Draw the energy axis and the flat reactants level on the left.
Step 2
The process is exothermic, so the products end up lower in energy.
Step 3
Draw the flat products level on the right, below the reactants level, and join the levels with one smooth curve.
Step 4
Mark the vertical gap and label it 'energy released'.
Step 5
A diagram with the products level below the reactants level, both labeled, joined by a smooth curve, with the gap labeled 'energy released'.
Worked example 2. Sketch the energy diagram for the endothermic decomposition of limestone in a kiln.
Step 1
Draw the energy axis and the flat reactants level on the left.
Step 2
The process is endothermic, so the products end up higher in energy.
Step 3
Draw the flat products level on the right, above the reactants level, and join the levels with one smooth curve.
Step 4
Mark the vertical gap and label it 'energy absorbed'.
Step 5
A diagram with the products level above the reactants level, both labeled, joined by a smooth curve, with the gap labeled 'energy absorbed'.
You can now sketch an energy diagram for a described exothermic or endothermic process, placing the reactant and product energy levels at the correct relative heights and labeling the energy difference as released or absorbed.
Check your understanding
Panels A to D each show a student's energy diagram for an ENDOTHERMIC reaction. Which panel is drawn correctly?
APanel Acorrect
BPanel B
This option is wrong — you drew the exothermic shape — an endothermic reaction's products end up HIGHER in energy than its reactants.
CPanel C
This option is wrong — you drew no energy change — an endothermic reaction absorbs energy, so its two levels must sit at different heights with the gap marked.
DPanel D
This option is wrong — you placed the levels correctly but labeled the gap as released — a rise from reactants to products is energy ABSORBED.
Endothermic: the products end up higher in energy than the reactants. Panel A places the products level above the reactants level and labels the gap 'energy absorbed' — both parts must be right. A correct diagram also keeps the levels flat and labeled, with one smooth joining curve.
Your turn
A hand warmer's iron powder reacts with oxygen in an exothermic process. On paper, sketch the energy diagram for this process: label the axis, both levels, and the energy gap. Then select Continue to compare your sketch with the model answer.
Model answer. A diagram with a vertical axis labeled 'energy'. A flat level labeled 'reactants' sits on the left, and a flat level labeled 'products' sits on the right, clearly LOWER than the reactants level. One smooth curve joins the levels. The vertical gap between the levels is marked and labeled 'energy released'.
The vertical axis is labeled 'energy'.
Both levels are flat, with 'reactants' on the left and 'products' on the right, each labeled.
The products level sits LOWER than the reactants level.
One smooth curve joins the two levels.
The vertical gap is marked and labeled 'energy released'.
Your turn
Citric acid reacting with baking soda in water is an endothermic process. On paper, sketch the energy diagram for this process: label the axis, both levels, and the energy gap. Then select Continue to compare your sketch with the model answer.
Model answer. A diagram with a vertical axis labeled 'energy'. A flat level labeled 'reactants' sits on the left, and a flat level labeled 'products' sits on the right, clearly HIGHER than the reactants level. One smooth curve joins the levels. The vertical gap between the levels is marked and labeled 'energy absorbed'.
The vertical axis is labeled 'energy'.
Both levels are flat, with 'reactants' on the left and 'products' on the right, each labeled.
The products level sits HIGHER than the reactants level.
One smooth curve joins the two levels.
The vertical gap is marked and labeled 'energy absorbed'.
Lesson 24 of 40 · THM-024
Thermochemical equations
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Did You Know?
You have seen a reaction's energy change drawn as an energy diagram. A chemical equation can carry the same information in writing — you have seen that an equation's reactants sit left of the arrow and its products sit right.
The idea
A 'thermochemical equation' is a chemical equation that writes the energy change as if it were a reactant or a product.
Energy written on the product side means the reaction gives energy out — exothermic.
Energy written on the reactant side means the reaction takes energy in — endothermic.
2H₂ + O₂ → 2H₂O + energy reads: hydrogen and oxygen react, and energy comes out with the products — exothermic.
The energy term sits on one side only, and that side settles the classification.
Worked examples
Worked example 1. Classify the reaction shown by the thermochemical equation C + O₂ → CO₂ + energy.
Step 1
Find the energy term: it is written on the product side.
Step 2
Energy written as a product means energy given out — the reaction is exothermic.
Worked example 2. Classify the reaction shown by the thermochemical equation CaCO₃ + energy → CaO + CO₂.
Step 1
Find the energy term: it is written on the reactant side.
Step 2
Energy written as a reactant means energy taken in — the reaction is endothermic.
You can now classify a reaction as exothermic or endothermic from a thermochemical equation, a chemical equation that writes the energy change as a reactant or as a product: energy written as a product means exothermic, and energy written as a reactant means endothermic.
Check your understanding
Classify the reaction shown by the thermochemical equation CH₄ + 2O₂ → CO₂ + 2H₂O + energy.
AExothermiccorrect
BEndothermic
This option is wrong — you flipped the rule — energy written on the PRODUCT side means the reaction gives energy out, which is exothermic.
CNeither — the energy term cancels out
This option is wrong — you treated the energy term as bookkeeping — it sits on one side only, and that side carries the classification.
DIt cannot be classified from the equation
This option is wrong — you assumed the equation carries no energy information — a thermochemical equation shows the direction by which side the energy term sits on.
Find the energy term: it sits on the product side. Energy written as a product means energy given out — the reaction is exothermic.
Check your understanding
Classify the reaction shown by the thermochemical equation 2HgO + energy → 2Hg + O₂.
AEndothermiccorrect
BExothermic
This option is wrong — you flipped the rule — energy written on the REACTANT side means the reaction takes energy in, which is endothermic.
CNeither — the energy term cancels out
This option is wrong — you treated the energy term as bookkeeping — it sits on one side only, and here that side is the reactant side.
DIt cannot be classified from the equation
This option is wrong — you assumed the equation carries no energy information — energy written with the reactants means the reaction absorbs it.
Find the energy term: it sits on the reactant side. Energy written as a reactant means energy taken in — the reaction is endothermic.
Check your understanding
Which of these thermochemical equations shows an endothermic reaction?
AN₂ + O₂ + energy → 2NOcorrect
B2Mg + O₂ → 2MgO + energy
This option is wrong — you read energy-anywhere as endothermic — here the energy term sits on the PRODUCT side, so the reaction gives energy out.
CS + O₂ → SO₂ + energy
This option is wrong — you matched the wrong side — energy with the products means exothermic; endothermic needs the energy term with the reactants.
D2K + 2H₂O → 2KOH + H₂ + energy
This option is wrong — you may have judged by the violent-looking chemistry — potassium in water is dramatic AND exothermic; the equation's energy term sits with the products.
Endothermic means energy taken in — the energy term must sit on the reactant side. Only N₂ + O₂ + energy → 2NO writes energy with the reactants. The other three write energy with the products, so they are exothermic.
Summary video — Exothermic and endothermic processes and how to show them
An electric kettle is one of the hungriest machines in a kitchen — it works flat out for two whole minutes to heat one pot's worth of water. Warming water swallows a surprising amount of energy. How much, exactly?
You have seen that heat is energy transferred from hotter objects to colder ones. To put numbers on heating, you first need a unit for energy itself.
The idea
Energy is measured in 'joules' (J) — the SI unit of energy.
One joule is small: lifting an apple from the floor to a table takes about one joule.
Different materials need different amounts of energy to warm up.
The 'specific heat capacity' of a material is the energy needed to raise the temperature of one gram of it by one degree Celsius.
Its unit is joules per gram per degree Celsius, written J/(g·°C).
Water's specific heat capacity is 4.18 J/(g·°C): raising 1 g of water by 1 °C takes 4.18 J.
That per-gram, per-degree price is why the kettle works so hard — every gram of water charges 4.18 J for every degree.
Water's specific heat capacity, 4.18 J/(g·°C): 4.18 J raises 1 g of water by exactly 1 °C.
Worked examples
Worked example 1. Copper's specific heat capacity is 0.385 J/(g·°C). What does that value mean?
Step 1
Answer: raising the temperature of 1 g of copper by 1 °C takes 0.385 J.
Worked example 2. What unit is energy measured in, and what is its symbol?
Step 1
Answer: the joule, symbol J.
You can now state what the specific heat capacity of a material means: the energy needed to raise the temperature of one gram of the material by one degree Celsius.
Check your understanding
Ethanol's specific heat capacity is 2.44 J/(g·°C). What does this value tell you?
ARaising 1 g of ethanol by 1 °C takes 2.44 J.correct
BRaising 1 g of ethanol by 2.44 °C takes 1 J.
This option is wrong — you swapped the roles of the numbers — the value is joules per gram per degree: 2.44 J buys one degree for one gram.
CEvery gram of ethanol contains 2.44 J of energy.
This option is wrong — you read the value as stored energy content — specific heat capacity is the energy needed to WARM the material, not energy sitting inside it.
DRaising any mass of ethanol by 1 °C takes 2.44 J.
This option is wrong — you dropped the per-gram part — the value prices one degree for one GRAM; more grams cost more.
Specific heat capacity is the energy needed to raise 1 g of the material by 1 °C. For ethanol that price is 2.44 J per gram per degree.
Check your understanding
What is the specific heat capacity of water?
A4.18 J/(g·°C)correct
B1.00 J/(g·°C)
This option is wrong — you recalled the one-per-gram figure from the calorie unit — in JOULES, water's specific heat capacity is 4.18 J/(g·°C).
C0.418 J/(g·°C)
This option is wrong — you slipped the decimal point one place down — water's value is 4.18 J/(g·°C).
D41.8 J/(g·°C)
This option is wrong — you slipped the decimal point one place up — water's value is 4.18 J/(g·°C).
Water's specific heat capacity is 4.18 J/(g·°C). That means 4.18 J raises 1 g of water by 1 °C.
Check your understanding
Which unit does specific heat capacity carry?
AJ/(g·°C)correct
BJ
This option is wrong — you kept only the energy part — specific heat capacity is energy PER gram PER degree, so the gram and the degree belong in the unit.
CJ/g
This option is wrong — you dropped the per-degree part — the value prices one degree of warming for one gram, so °C belongs in the unit.
D°C/g
This option is wrong — you lost the energy part — specific heat capacity is measured in joules per gram per degree Celsius.
Specific heat capacity is the energy (J) to warm one gram (g) by one degree (°C). All three pieces appear in the unit: J/(g·°C).
Lesson 26 of 40 · THM-026
Compare materials by specific heat
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Have You Ever Wondered?
Wonder this:
On a summer afternoon the dry sand scorches your bare feet while the sea, under the very same sun all day, stays cool. Same sunshine, wildly different temperatures. Why?
You have seen what a specific heat capacity value means for one gram of a material. A table of those values lets you compare whole materials.
The idea
Equal masses of two materials can absorb the same energy and end up at very different temperatures.
The material with the lower specific heat capacity needs less energy per degree, so the same energy pushes its temperature further.
For equal masses absorbing the same energy, the material with the LOWER specific heat capacity shows the LARGER temperature change.
In the table, sand's specific heat capacity (0.84 J/(g·°C)) is far lower than water's (4.18 J/(g·°C)).
So a gram of beach sand warms about five times more than a gram of seawater absorbing the same sunshine — scorching sand, cool sea.
Specific heat capacities
Material
Specific heat capacity (J/(g·°C))
water
4.18
ethanol
2.44
olive oil
1.97
aluminum
0.897
sand
0.84
iron
0.449
copper
0.385
lead
0.128
Real specific heat capacities. Lower value: the same energy warms the material further.
Worked examples
Worked example 1. Equal masses of iron (specific heat capacity 0.449 J/(g·°C)) and ethanol (2.44 J/(g·°C)) each absorb the same energy. Which sample warms more?
Step 1
Compare the specific heat capacities: iron's 0.449 J/(g·°C) is lower than ethanol's 2.44 J/(g·°C).
Step 2
For equal masses absorbing the same energy, the material with the lower specific heat capacity shows the larger temperature change.
Step 3
The iron warms more.
Worked example 2. Equal masses of copper (0.385 J/(g·°C)) and lead (0.128 J/(g·°C)) each absorb the same energy. Which sample warms more?
Step 1
Compare the specific heat capacities: lead's 0.128 J/(g·°C) is lower than copper's 0.385 J/(g·°C).
Step 2
For equal masses absorbing the same energy, the material with the lower specific heat capacity shows the larger temperature change.
Step 3
The lead warms more.
You can now predict from a table of specific heat capacities which of two equal-mass samples warms more when each absorbs the same energy, where the material with the lower specific heat capacity shows the larger temperature change.
Check your understanding
Using the table, equal masses of aluminum and silver (specific heat capacity 0.235 J/(g·°C)) each absorb the same energy. Which sample's temperature rises more?
Specific heat capacities
Material
Specific heat capacity (J/(g·°C))
aluminum
0.897
silver
0.235
AThe silver sample.correct
BThe aluminum sample.
This option is wrong — you picked the higher specific heat capacity — a higher value means MORE energy needed per degree, so the same energy warms it less.
CBoth rise by the same amount.
This option is wrong — you matched same energy to same rise — equal energy only gives equal rises when the specific heat capacities match, and these differ.
DIt cannot be predicted from the table.
This option is wrong — you set the table aside — with mass and energy held equal, the specific heat capacities alone settle which sample warms more.
Read the table: silver's 0.235 J/(g·°C) is lower than aluminum's 0.897 J/(g·°C). For equal masses absorbing the same energy, the material with the lower specific heat capacity shows the larger temperature change. So the silver warms more.
Check your understanding
Using the table, equal masses of water and olive oil each absorb the same energy in a kitchen experiment. Which sample's temperature rises more?
Specific heat capacities
Material
Specific heat capacity (J/(g·°C))
water
4.18
olive oil
1.97
AThe olive oil.correct
BThe water.
This option is wrong — you picked the higher specific heat capacity — water's high value means each degree costs more energy, so the same energy warms it less.
CBoth rise by the same amount.
This option is wrong — you matched same energy to same rise — the two materials charge different amounts per degree, so the same energy buys different rises.
DIt cannot be predicted from the table.
This option is wrong — you set the table aside — with mass and energy held equal, the lower specific heat capacity marks the sample that warms more.
Read the table: olive oil's 1.97 J/(g·°C) is lower than water's 4.18 J/(g·°C). For equal masses absorbing the same energy, the lower specific heat capacity shows the larger temperature change. So the olive oil warms more — one reason oil heats so fast in a pan.
Check your understanding
Using the table, equal masses of ethanol and iron each absorb the same energy. Which sample ends up with the smaller temperature change?
Specific heat capacities
Material
Specific heat capacity (J/(g·°C))
ethanol
2.44
iron
0.449
AThe ethanol sample.correct
BThe iron sample.
This option is wrong — you gave the smaller change to the lower specific heat capacity — the LOWER value warms MORE; the higher value, ethanol's, warms less.
CBoth changes are the same size.
This option is wrong — you matched same energy to same rise — different specific heat capacities turn the same energy into different rises.
DIt cannot be predicted from the table.
This option is wrong — you set the table aside — with mass and energy held equal, the specific heat capacities settle it: the higher value changes less.
The question asks for the SMALLER change, so look for the higher specific heat capacity. Ethanol's 2.44 J/(g·°C) is higher than iron's 0.449 J/(g·°C), so each of ethanol's degrees costs more energy. The same energy therefore pushes ethanol's temperature less far.
Lesson 27 of 40 · THM-027
More mass, more energy
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Did You Know?
You have seen the per-gram meaning of specific heat capacity: warming 1 g of water by 1 °C takes 4.18 J. Real samples hold many grams.
The idea
Take two samples of the same material, warmed through the same temperature change.
Each gram needs its own share of energy, so twice the grams need twice the energy.
Warming a larger mass of the same material through the same temperature change takes proportionally more energy.
Warming 200 g of water from 20 °C to 30 °C takes twice the energy that warming 100 g of water between the same temperatures does.
Three times the mass, three times the energy; ten times the mass, ten times the energy.
Doubling the mass doubles the energy the same warming needs.
Worked examples
Worked example 1. Warming 150 g of ethanol from 15 °C to 25 °C takes a certain amount of energy. How does the energy needed to warm 450 g of ethanol from 15 °C to 25 °C compare?
Step 1
The material is the same, and the temperature change is the same.
Step 2
Only the mass changed: 450 g is three times 150 g.
Step 3
Three times the mass needs three times the energy.
Step 4
Warming 450 g takes three times the energy.
Worked example 2. Warming 100 g of water from 20 °C to 30 °C takes 4180 J. How much energy does warming 500 g of water from 20 °C to 30 °C take?
Step 1
The material is the same, and the temperature change is the same.
Step 2
Only the mass changed: 500 g is five times 100 g.
Step 3
Five times the mass needs five times the energy: 5 × 4180 J = 20900 J.
Step 4
Warming 500 g takes 20900 J.
You can now predict that warming a larger mass of the same material through the same temperature change takes proportionally more energy, with the material and the temperature change held constant.
Check your understanding
A camper warms 300 g of water for tea, and later warms 600 g of water for washing up. Both warmings run from 10 °C to 90 °C. How does the energy needed for the 600 g compare with the energy for the 300 g?
ATwice as much.correct
BThe same, because the temperature change is the same.
This option is wrong — you priced the warming by its temperature change alone — every gram needs its own share, so doubling the grams doubles the energy.
CHalf as much.
This option is wrong — you inverted the proportion — more mass needs MORE energy, not less.
DFour times as much.
This option is wrong — you squared the factor — doubling the mass doubles the energy, it does not multiply it by four.
Same material, same temperature change — only the mass differs. 600 g is twice 300 g, and twice the mass needs twice the energy.
Check your understanding
Warming 100 g of ethanol by 10 °C takes 2440 J. How much energy does warming 300 g of ethanol by 10 °C take? Give your answer in joules.
Answer: 7320J(tolerance ±1)
Same material, same temperature change — only the mass differs. 300 g is three times 100 g. Three times the mass needs three times the energy: 3 × 2440 J = 7320 J.
Check your understanding
A hot tub holds 400 times the mass of water that a kettle holds. Both are warmed through the same temperature change. How does the hot tub's energy need compare with the kettle's?
A400 times as much.correct
BThe same, because the temperature change is the same.
This option is wrong — you priced the warming by its temperature change alone — each of the hot tub's extra grams needs its own share of energy.
C400 times less.
This option is wrong — you inverted the proportion — the larger mass needs proportionally MORE energy.
DMore, but the factor cannot be found without the specific heat capacity.
This option is wrong — you reached for the specific heat capacity — both containers hold the same material, so its value scales both equally, and the factor comes from mass alone.
Same material (water), same temperature change — only the mass differs. 400 times the mass needs 400 times the energy.
Lesson 28 of 40 · THM-028
Bigger temperature change, more energy
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Did You Know?
You have seen that doubling a sample's mass doubles the energy a warming needs. The size of the temperature change works the same way.
The idea
Take one sample — the same material and the same mass — and warm it through different temperature changes.
Each degree of warming needs its own share of energy, so twice the degrees need twice the energy.
Warming the same sample through a larger temperature change takes proportionally more energy.
Warming 100 g of water by 20 °C takes twice the energy that warming it by 10 °C does.
What counts is the CHANGE in temperature: from 20 °C to 30 °C is a 10 °C change, and from 20 °C to 40 °C is a 20 °C change.
Doubling the temperature change doubles the energy the warming needs.
Worked examples
Worked example 1. A 200 g block of aluminum is warmed from 25 °C to 35 °C. How does the energy needed to warm the same block from 25 °C to 55 °C compare?
Step 1
The sample is the same — same material, same mass.
Step 2
Compare the changes: 35 − 25 = 10 °C, and 55 − 25 = 30 °C — three times the change.
Step 3
Three times the temperature change needs three times the energy.
Step 4
Warming the block to 55 °C takes three times the energy.
Worked example 2. Warming 50 g of water from 20 °C to 30 °C takes 2090 J. How much energy does warming the same 50 g of water from 20 °C to 60 °C take?
Step 1
The sample is the same — same material, same mass.
Step 2
Compare the changes: 30 − 20 = 10 °C, and 60 − 20 = 40 °C — four times the change.
Step 3
Four times the temperature change needs four times the energy: 4 × 2090 J = 8360 J.
Step 4
Warming to 60 °C takes 8360 J.
You can now predict that warming the same sample through a larger temperature change takes proportionally more energy, with the mass and the material held constant.
Check your understanding
A 300 g pan of milk is warmed by 15 °C one day and by 45 °C the next. How does the energy for the 45 °C warming compare with the energy for the 15 °C warming?
AThree times as much.correct
BThe same, because it is the same pan of milk.
This option is wrong — you priced the warming by the sample alone — each degree needs its own share of energy, and the second warming buys three times the degrees.
CNine times as much.
This option is wrong — you squared the factor — tripling the temperature change triples the energy, nothing more.
DThirty times as much.
This option is wrong — you used the difference between the two changes, 45 − 15 = 30, as the factor — the factor is their RATIO, 45 ÷ 15 = 3.
Same material, same mass — only the temperature change differs. 45 °C is three times 15 °C, and three times the change needs three times the energy.
Check your understanding
Warming 100 g of water from 20 °C to 30 °C takes 4180 J. How much energy does warming the same 100 g of water from 20 °C to 50 °C take? Give your answer in joules.
Answer: 12540J(tolerance ±1)
Work with the CHANGES, not the final temperatures. First change: 30 − 20 = 10 °C. Second change: 50 − 20 = 30 °C — three times as large. Three times the change needs three times the energy: 3 × 4180 J = 12540 J.
Check your understanding
The same 250 g of water is warmed twice: warming 1 runs from 10 °C to 20 °C, and warming 2 runs from 60 °C to 70 °C. How do their energy needs compare?
AThe same — both are 10 °C changes.correct
BWarming 2 needs more, because the water is hotter.
This option is wrong — you judged by how hot the water is — the energy follows the SIZE of the change, and both changes are 10 °C.
CWarming 1 needs more, because cold water is harder to warm.
This option is wrong — you gave cold water a premium — a degree of warming costs the same wherever it starts; both 10 °C changes cost the same.
DWarming 2 needs six times as much, because 60 is six times 10.
This option is wrong — you compared the starting temperatures — the energy follows the temperature CHANGE, and the two changes are equal.
Find each change: 20 − 10 = 10 °C, and 70 − 60 = 10 °C. Equal changes on the same sample need equal energy — the starting temperature does not set the price.
Lesson 29 of 40 · THM-029
The equation q = mcΔT
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Did You Know?
You have seen, one lesson at a time, that the energy a warming sample needs depends on its material, on its mass, and on its temperature change. One equation carries all three at once.
The equation
The heat equation combines the three effects you have seen into one statement: q = mcΔT.
q is the energy transferred as heat, measured in joules (J).
q
energy transferred as heat (J)
m
mass of the sample (g)
c
specific heat capacity of the material (J/(g·°C))
ΔT
temperature change, final temperature minus initial temperature (°C)
m is the mass of the sample, in grams (g).
c is the specific heat capacity of the material, in J/(g·°C).
ΔT is the temperature change in °C: the final temperature minus the initial temperature.
For water warming from 20 °C to 30 °C, ΔT = 30 − 20 = 10 °C.
The equation says what the three lessons showed: more mass, a bigger temperature change, or a higher specific heat capacity each means more energy transferred.
Worked examples
Worked example 1. In the heat equation q = mcΔT, what does c stand for, and in what unit is it measured?
Step 1
c is the specific heat capacity of the material — the energy that warms one gram of it by one degree Celsius.
Step 2
c stands for the specific heat capacity, in J/(g·°C).
Worked example 2. A pan of water warms from 18.0 °C to 43.0 °C. What is ΔT?
Step 1
Write down the values in the question
final temperature = 43.0 °C
initial temperature = 18.0 °C
Step 2
Write down the equation
ΔT = final temperature − initial temperature
Step 3
Substitute in the values, and calculate
ΔT = 43.0 − 18.0
ΔT = 25.0 °C
You can now state the heat equation q = mcΔT and identify each symbol: q is the energy transferred as heat in joules (J), m is the mass in grams (g), c is the specific heat capacity in J/(g·°C), and ΔT is the temperature change in °C, final temperature minus initial temperature.
Check your understanding
Write the equation that gives the energy transferred as heat from a sample's mass, its specific heat capacity, and its temperature change. Use the symbols q, m, c, and ΔT.
Accepted answer: q = mcΔT
The heat equation multiplies three measured quantities. q = mcΔT q is in joules (J), m in grams (g), c in J/(g·°C), and ΔT in °C.
Check your understanding
In the heat equation q = mcΔT, what does ΔT stand for?
AThe temperature change: the final temperature minus the initial temperature.correct
BThe final temperature of the sample when the heating ends.
This option is wrong — you took ΔT to be a single thermometer reading — it is the CHANGE in temperature, final minus initial.
CThe initial temperature minus the final temperature.
This option is wrong — you flipped the subtraction — ΔT is the final temperature minus the initial temperature.
DThe sample's temperature converted from degrees Celsius to kelvin.
This option is wrong — you swapped in a unit conversion — ΔT is a change in temperature, measured in °C in this equation.
ΔT is the temperature change in °C: the final temperature minus the initial temperature. A single reading is not a change — two readings are always needed. For water warming from 20 °C to 30 °C, ΔT = 30 − 20 = 10 °C.
Check your understanding
A beaker of water warms from 21.5 °C to 34.0 °C. What is ΔT, in °C?
Answer: 12.5°C(tolerance ±0.05)
Write down the values in the question: final temperature = 34.0 °C initial temperature = 21.5 °C Write down the equation: ΔT = final temperature − initial temperature Substitute in the values, and calculate: ΔT = 34.0 − 21.5 ΔT = 12.5 °C
Lesson 30 of 40 · THM-030
Calculate q
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Did You Know?
You have seen the heat equation q = mcΔT and what each symbol means. Now it produces numbers: given the mass, the material, and the temperature change, the equation gives the energy transferred.
The equation
Write down the values in the question: the mass, the specific heat capacity, and the temperatures.
If the question gives an initial and a final temperature, calculate ΔT first: for a sample that warmed, the final temperature minus the initial temperature.
Write down the equation: q = mcΔT.
Substitute in the values, and calculate — the result is the energy transferred, in joules (J).
If the sample warmed, it absorbed that energy; if it cooled, it released that energy.
For a sample that cooled, use the size of the temperature drop as ΔT — the higher reading minus the lower one — so the energy stays a positive amount, with the word 'released' carrying the direction.
Water's specific heat capacity is high, so expect big numbers for water: warming a glass of it by a few degrees takes thousands of joules.
Worked examples
Worked example 1. How much energy does it take to warm 100.0 g of water from 22.0 °C to 42.0 °C? (c of water = 4.18 J/(g·°C))
Step 1
Write down the values in the question
m = 100.0 g
c = 4.18 J/(g·°C)
ΔT = 42.0 − 22.0 = 20.0 °C
Step 2
Write down the equation
q = mcΔT
Step 3
Substitute in the values, and calculate
q = 100.0 × 4.18 × 20.0
q = 8360 J
Worked example 2. A 250.0 g copper pipe warms from 15.0 °C to 35.0 °C in the sun. How much energy does the copper absorb? (c of copper = 0.385 J/(g·°C))
Step 1
Write down the values in the question
m = 250.0 g
c = 0.385 J/(g·°C)
ΔT = 35.0 − 15.0 = 20.0 °C
Step 2
Write down the equation
q = mcΔT
Step 3
Substitute in the values, and calculate
q = 250.0 × 0.385 × 20.0
q = 1925 J
You can now calculate the heat absorbed or released by a sample from its mass, specific heat capacity, and temperature change using q = mcΔT.
Check your understanding
How much energy does it take to warm 200.0 g of water from 20.0 °C to 25.0 °C, in J? (c of water = 4.18 J/(g·°C))
Answer: 4180J(tolerance ±0.5)
Write down the values in the question: m = 200.0 g c = 4.18 J/(g·°C) ΔT = 25.0 − 20.0 = 5.0 °C Write down the equation: q = mcΔT Substitute in the values, and calculate: q = 200.0 × 4.18 × 5.0 q = 4180 J
Check your understanding
A 50.0 g aluminum spoon warms from 18.0 °C to 38.0 °C in hot dishwater. How much energy does the aluminum absorb, in J? (c of aluminum = 0.897 J/(g·°C))
Answer: 897J(tolerance ±0.5)
Write down the values in the question: m = 50.0 g c = 0.897 J/(g·°C) ΔT = 38.0 − 18.0 = 20.0 °C Write down the equation: q = mcΔT Substitute in the values, and calculate: q = 50.0 × 0.897 × 20.0 q = 897 J
Check your understanding
A 400.0 g bottle of ethanol cools from 22.0 °C to 19.5 °C in a refrigerator. How much energy does the ethanol release, in J? (c of ethanol = 2.44 J/(g·°C))
Answer: 2440J(tolerance ±0.5)
Write down the values in the question: m = 400.0 g c = 2.44 J/(g·°C) ΔT = 22.0 − 19.5 = 2.5 °C (a drop of 2.5 °C) Write down the equation: q = mcΔT Substitute in the values, and calculate: q = 400.0 × 2.44 × 2.5 q = 2440 J — released, because the ethanol cooled
Lesson 31 of 40 · THM-031
Rearrange q = mcΔT
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Did You Know?
You have seen q = mcΔT give the energy when everything else is known. Real problems often run the other way: the energy is known, and the mass, the specific heat capacity, or the temperature change is not.
The equation
q = mcΔT still holds when the unknown sits on the right-hand side.
Make the unknown the subject before substituting.
To make c the subject, divide both sides by m × ΔT, giving c = q ÷ (m × ΔT).
To make m the subject, divide both sides by c × ΔT, giving m = q ÷ (c × ΔT).
To make ΔT the subject, divide both sides by m × c, giving ΔT = q ÷ (m × c).
It is one equation rearranged three ways, not three new equations to memorize: the known energy is always divided by the two known right-hand quantities.
Worked examples
Worked example 1. A 200.0 g metal block absorbs 1540 J of energy and warms from 25.0 °C to 45.0 °C. What is the metal's specific heat capacity?
Step 1
Write down the values in the question
q = 1540 J
m = 200.0 g
ΔT = 45.0 − 25.0 = 20.0 °C
Step 2
Write down the equation
q = mcΔT
Step 3
Make the unknown the subject
c = q ÷ (m × ΔT)
Step 4
Substitute in the values, and calculate
c = 1540 ÷ (200.0 × 20.0)
c = 0.385 J/(g·°C)
Worked example 2. How many grams of water can 8360 J of energy warm by 10.0 °C? (c of water = 4.18 J/(g·°C))
Step 1
Write down the values in the question
q = 8360 J
c = 4.18 J/(g·°C)
ΔT = 10.0 °C
Step 2
Write down the equation
q = mcΔT
Step 3
Make the unknown the subject
m = q ÷ (c × ΔT)
Step 4
Substitute in the values, and calculate
m = 8360 ÷ (4.18 × 10.0)
m = 200.0 g
You can now calculate an unknown mass, specific heat capacity, or temperature change by rearranging q = mcΔT to make the unknown the subject.
Check your understanding
A 100.0 g metal bar absorbs 470 J of energy and warms by 20.0 °C. What is the metal's specific heat capacity, in J/(g·°C)?
Answer: 0.235J/(g·°C)(tolerance ±0.0005)
Write down the values in the question: q = 470 J m = 100.0 g ΔT = 20.0 °C Write down the equation: q = mcΔT Make c the subject: c = q ÷ (m × ΔT) Substitute in the values, and calculate: c = 470 ÷ (100.0 × 20.0) c = 0.235 J/(g·°C)
Check your understanding
How many grams of water can 1045 J of energy warm by 2.0 °C? (c of water = 4.18 J/(g·°C))
Answer: 125g(tolerance ±0.05)
Write down the values in the question: q = 1045 J c = 4.18 J/(g·°C) ΔT = 2.0 °C Write down the equation: q = mcΔT Make m the subject: m = q ÷ (c × ΔT) Substitute in the values, and calculate: m = 1045 ÷ (4.18 × 2.0) m = 125 g
Check your understanding
A 250.0 g glass of water absorbs 5225 J of energy from the sun. By how many degrees Celsius does the water warm? (c of water = 4.18 J/(g·°C))
Answer: 5.0°C(tolerance ±0.05)
Write down the values in the question: q = 5225 J m = 250.0 g c = 4.18 J/(g·°C) Write down the equation: q = mcΔT Make ΔT the subject: ΔT = q ÷ (m × c) Substitute in the values, and calculate: ΔT = 5225 ÷ (250.0 × 4.18) ΔT = 5.0 °C
Lesson 32 of 40 · THM-032
The coffee-cup calorimeter
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Have You Ever Wondered?
Wonder this:
A dissolving solid releases energy into the water around it — but energy is invisible. No instrument shows joules flowing. How can a foam cup and a thermometer measure something you cannot see?
You have seen that heat flows from hotter to colder and that an isolated system exchanges neither matter nor energy with its surroundings. Chemists build a cheap benchtop setup that puts both ideas to work.
The idea
The trick is to trap the invisible energy in something whose temperature you can read.
The setup is an insulated foam cup with a lid, holding a measured mass of water, with a thermometer through the lid — a 'coffee-cup calorimeter'.
The insulated cup and lid keep energy from escaping, so the setup approximates an isolated system.
The insulated cup traps the energy, the water carries it, and the thermometer reads the result.
The measured mass of water absorbs or supplies the energy being measured.
The thermometer records the water's temperature change.
A process running in the water leaves its mark as a temperature change you can read off the thermometer.
Worked examples
Worked example 1. Which part of a coffee-cup calorimeter keeps the exchanged energy inside the setup?
Step 1
The insulated foam cup and its lid — they stop energy escaping, so the setup approximates an isolated system.
Step 2
The insulated cup (with its lid).
Worked example 2. What job does the measured mass of water do in a coffee-cup calorimeter?
Step 1
The water absorbs or supplies the energy being measured.
Step 2
It absorbs or supplies the energy being measured.
You can now identify the parts of a coffee-cup calorimeter and each part's job: an insulated cup that keeps energy from escaping, so the setup approximates an isolated system, a measured mass of water that absorbs or supplies the energy being measured, and a thermometer that records the water's temperature change.
Check your understanding
A reaction running in a coffee-cup calorimeter releases energy. Which part of the setup absorbs that energy so it can be measured?
AThe measured mass of watercorrect
BThe insulated foam cup
This option is wrong — you gave the insulation the absorbing job — the cup's job is to keep energy from escaping, not to take it in.
CThe thermometer
This option is wrong — you gave the reader the absorbing job — the thermometer records the water's temperature change; the water holds the energy.
DThe air above the lid
This option is wrong — you sent the energy outside the setup — the insulation exists precisely to stop energy reaching the air.
The measured mass of water absorbs or supplies the energy being measured. The insulated cup and lid keep that energy from escaping. The thermometer then records the water's temperature change.
Check your understanding
Which part of a coffee-cup calorimeter has the job of keeping energy from escaping the setup?
AThe insulated foam cup and its lidcorrect
BThe measured mass of water
This option is wrong — you gave the water the insulating job — the water's job is to absorb or supply the energy being measured.
CThe thermometer
This option is wrong — you gave the thermometer the insulating job — its job is to record the water's temperature change.
DNone of the parts — energy is meant to escape freely
This option is wrong — you dropped the isolation requirement — if energy escapes, the water's temperature change misses part of the story.
The insulated foam cup and its lid keep energy from escaping. That is what lets the setup approximate an isolated system. Every joule the process exchanges then stays where the water can register it.
Check your understanding
Because its insulated cup keeps energy from escaping, a coffee-cup calorimeter approximates which kind of system?
AAn isolated systemcorrect
BAn open system
This option is wrong — you chose the system that exchanges both matter and energy — the insulation is there to stop the energy exchange.
CA closed system
This option is wrong — you chose the system that still exchanges energy — the insulated cup blocks energy exchange too, which is what makes the setup approximately isolated.
DNone of the above
This option is wrong — you treated the setup as unclassifiable — a setup exchanging neither matter nor energy is the definition of an isolated system.
An isolated system exchanges neither matter nor energy with its surroundings. The lidded, insulated cup blocks both exchanges as far as a foam cup can. So the calorimeter approximates an isolated system — close enough for the measurement to work.
Lesson 33 of 40 · THM-033
Heat lost equals heat gained
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Did You Know?
You have seen that energy cannot be created or destroyed, that heat flows from hotter to colder, and that an insulated calorimeter approximates an isolated system. Put together, they force a simple accounting rule.
The idea
Drop a hot object into cooler water inside an insulated container, and energy flows from the hot object into the water.
Energy cannot be created or destroyed — only transferred or transformed.
Inside an isolated setup, the transferred energy has nowhere else to go.
So the heat lost by the hotter object equals the heat gained by the colder object.
The balance holds only when no energy escapes to anything else.
If the container leaks energy to the surroundings, the colder object gains less than the hotter object lost.
Worked examples
Worked example 1. A hot copper block loses 3000 J inside an insulated container of water. How much heat does the water gain?
Step 1
The setup is insulated, so no energy escapes to anything else.
Step 2
The heat lost by the hotter object equals the heat gained by the colder object.
Step 3
The water gains 3000 J.
Worked example 2. The same hot copper block loses 3000 J while cooling in an uninsulated ceramic mug of water. Does the water gain 3000 J?
Step 1
The mug is not insulated, so some energy escapes to the mug and the air.
Step 2
The balance holds only when no energy escapes to anything else.
Step 3
No — the water gains less than 3000 J, because part of the energy leaks away.
You can now state that when two objects exchange energy inside an isolated setup, the heat lost by the hotter object equals the heat gained by the colder object, a balance that holds only when no energy escapes to anything else.
Check your understanding
A hot horseshoe dropped into the water of an insulated tub loses 1250 J as it cools. How much heat does the water gain, in J?
Answer: 1250J(tolerance ±0.5)
The tub is insulated, so no energy escapes to anything else. The heat lost by the hotter object equals the heat gained by the colder object. The horseshoe lost 1250 J, so the water gained 1250 J.
Check your understanding
Under what condition does the heat lost by a hotter object exactly equal the heat gained by the colder object it touches?
AWhen no energy escapes to anything else.correct
BWhen the two objects have equal masses.
This option is wrong — you attached the balance to the masses — the balance is about where the energy goes, and it holds for any masses as long as none escapes.
CWhen the two objects are made of the same material.
This option is wrong — you attached the balance to the materials — conservation of energy does not care what the objects are made of.
DWhen the two objects start at the same temperature.
This option is wrong — you removed the transfer entirely — objects at the same temperature exchange no net heat at all.
The balance comes from conservation of energy inside an isolated setup. The heat lost by the hotter object equals the heat gained by the colder object only when no energy escapes to anything else. Masses, materials, and starting temperatures set how big the transfer is — not whether the balance holds.
Check your understanding
A hot stone cools in a bucket of water that has no lid and no insulation. The stone loses 2000 J. Which statement about the water is correct?
AThe water gains less than 2000 J, because some energy escapes to the bucket and the air.correct
BThe water gains exactly 2000 J, because heat lost always equals heat gained.
This option is wrong — you dropped the validity condition — the balance holds only when no energy escapes to anything else, and this bucket leaks.
CThe water gains more than 2000 J, because the air adds extra energy.
This option is wrong — you ran the leak backwards — room-temperature air takes energy from the warmed water; it does not top it up.
DThe water gains no energy, because open setups cannot transfer heat.
This option is wrong — you turned a leak into a wall — the stone still heats the water; the leak only means part of the energy goes elsewhere.
Heat lost equals heat gained only when no energy escapes to anything else. An open, uninsulated bucket leaks energy to the bucket itself and the air. So the water gains less than the 2000 J the stone lost.
Lesson 34 of 40 · THM-034
Measure a process's heat
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Did You Know?
You have seen the coffee-cup calorimeter's parts and that, inside it, heat lost equals heat gained. Those two ideas turn a thermometer reading into a measurement of a process's energy.
The equation
A process running in a calorimeter — a solid dissolving, a reaction going — exchanges its energy with the water around it.
Whatever the process releases, the water gains; whatever the process absorbs, the water supplies.
The water warmed 5.0 °C, so the dissolving released energy — q = 200.0 × 4.18 × 5.0 = 4180 J.
So measure the water: q = mcΔT, with the water's mass, water's specific heat capacity of 4.18 J/(g·°C), and the water's temperature change.
That q is the heat the process released or absorbed.
Counting only the water is an approximation — the cup and the thermometer soak up a little energy too — but the foam cup takes very little, and this is the standard school-lab practice.
The water warmed — the process released the energy; the process is exothermic.
The water cooled — the process absorbed the energy; the process is endothermic.
Worked examples
Worked example 1. A spoonful of solid dissolves in 200.0 g of water in a coffee-cup calorimeter, and the water warms from 21.0 °C to 26.0 °C. How much heat did the dissolving release?
Step 1
Write down the values in the question
m = 200.0 g (the water)
c = 4.18 J/(g·°C)
ΔT = 26.0 − 21.0 = 5.0 °C
Step 2
Write down the equation
q = mcΔT
Step 3
Substitute in the values, and calculate
q = 200.0 × 4.18 × 5.0
q = 4180 J — released, because the water warmed
Worked example 2. A different solid dissolves in 150.0 g of calorimeter water, and the water cools from 22.0 °C to 18.0 °C. How much heat did the dissolving absorb?
Step 1
Write down the values in the question
m = 150.0 g (the water)
c = 4.18 J/(g·°C)
ΔT = 22.0 − 18.0 = 4.0 °C (a drop of 4.0 °C)
Step 2
Write down the equation
q = mcΔT
Step 3
Substitute in the values, and calculate
q = 150.0 × 4.18 × 4.0
q = 2508 J — absorbed, because the water cooled
You can now calculate the heat a process releases or absorbs from the temperature change of the calorimeter water around it, using q = mcΔT on the water and reading the direction from whether the water warmed or cooled.
Check your understanding
4.0 g of a solid dissolves in 100.0 g of water in a coffee-cup calorimeter, and the water warms from 20.0 °C to 23.0 °C. How much heat did the dissolving release, in J? (c of water = 4.18 J/(g·°C))
Answer: 1254J(tolerance ±0.5)
Write down the values in the question — for the water: m = 100.0 g c = 4.18 J/(g·°C) ΔT = 23.0 − 20.0 = 3.0 °C Write down the equation: q = mcΔT Substitute in the values, and calculate: q = 100.0 × 4.18 × 3.0 q = 1254 J — released, because the water warmed
Check your understanding
A powder stirred into 250.0 g of calorimeter water makes the water cool from 21.5 °C to 19.5 °C. How much heat did the dissolving absorb, in J? (c of water = 4.18 J/(g·°C))
Answer: 2090J(tolerance ±0.5)
Write down the values in the question — for the water: m = 250.0 g c = 4.18 J/(g·°C) ΔT = 21.5 − 19.5 = 2.0 °C (a drop of 2.0 °C) Write down the equation: q = mcΔT Substitute in the values, and calculate: q = 250.0 × 4.18 × 2.0 q = 2090 J — absorbed, because the water cooled
Check your understanding
A small reaction vial rests in 50.0 g of calorimeter water. As the reaction runs, the water warms from 18.0 °C to 26.0 °C. How much heat did the reaction release, in J? (c of water = 4.18 J/(g·°C))
Answer: 1672J(tolerance ±0.5)
Write down the values in the question — for the water: m = 50.0 g c = 4.18 J/(g·°C) ΔT = 26.0 − 18.0 = 8.0 °C Write down the equation: q = mcΔT Substitute in the values, and calculate: q = 50.0 × 4.18 × 8.0 q = 1672 J — released, because the water warmed
Lesson 35 of 40 · THM-035
Solve a two-substance calorimetry problem
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Did You Know?
You have seen the calorimeter measure one substance at a time. The full routine handles two: a hot sample dropped into cooler water, both ending at one shared temperature.
The equation
A hot sample dropped into cooler calorimeter water exchanges energy until both reach one shared final temperature.
The heat lost by the hot sample equals the heat gained by the water.
The block cools 50.0 °C while the water warms 20.0 °C — one exchange, two temperature changes, one shared final temperature.
Write that balance with q = mcΔT on each side: mass × c × ΔT of the sample = mass × c × ΔT of the water.
Each ΔT is that substance's own temperature change — the sample drops to the shared temperature, and the water rises to it.
Work in two steps: first calculate the water's heat, then make the unknown the subject and divide.
When hot water is poured into cool water, the same balance applies with c = 4.18 J/(g·°C) on both sides.
Worked examples
Worked example 1. A 100.0 g metal block at 90.0 °C is dropped into 25.0 g of water at 20.0 °C in a coffee-cup calorimeter. Both settle at 40.0 °C. What is the metal's specific heat capacity? (c of water = 4.18 J/(g·°C))
Step 1
Write down the values in the question
water: m = 25.0 g, c = 4.18 J/(g·°C), ΔT = 40.0 − 20.0 = 20.0 °C
metal: m = 100.0 g, c = unknown, ΔT = 90.0 − 40.0 = 50.0 °C
Calculate the water's heat first:
q = mcΔT = 25.0 × 4.18 × 20.0 = 2090 J
The heat lost by the metal equals the heat gained by the water, so the metal lost 2090 J.
Step 2
Write down the equation
mass × c × ΔT of the metal = heat gained by the water
Step 3
Make the unknown the subject
c = 2090 ÷ (m × ΔT)
Step 4
Substitute in the values, and calculate
c = 2090 ÷ (100.0 × 50.0)
c = 0.418 J/(g·°C)
Worked example 2. 100.0 g of hot water at 90.0 °C is poured into 300.0 g of cool water at 10.0 °C in a calorimeter. What shared temperature do they reach? (c of water = 4.18 J/(g·°C))
Step 1
Write down the values in the question
hot water: m = 100.0 g, ΔT = 90.0 − T
cool water: m = 300.0 g, ΔT = T − 10.0
Both substances are water, so c is the same on both sides and divides out.
Substitute into the balance, then solve for T step by step:
100.0 × (90.0 − T) = 300.0 × (T − 10.0)
9000 − 100T = 300T − 3000
12,000 = 400T
Step 2
Write down the equation
mass × c × ΔT of the hot water = mass × c × ΔT of the cool water
Step 3
Substitute in the values, and calculate
T = 12,000 ÷ 400
T = 30.0 °C
You can now calculate an unknown final temperature or specific heat capacity when a hot sample and cooler water reach one shared temperature, by setting the heat lost by the hot sample equal to the heat gained by the water.
Check your understanding
A 100.0 g metal cylinder at 53.0 °C is dropped into 200.0 g of water at 20.0 °C in a coffee-cup calorimeter. Both settle at 23.0 °C. What is the metal's specific heat capacity, in J/(g·°C)? (c of water = 4.18 J/(g·°C))
Answer: 0.836J/(g·°C)(tolerance ±0.0005)
Write down the values in the question: water: m = 200.0 g, c = 4.18 J/(g·°C), ΔT = 23.0 − 20.0 = 3.0 °C metal: m = 100.0 g, ΔT = 53.0 − 23.0 = 30.0 °C Calculate the water's heat first: q = 200.0 × 4.18 × 3.0 = 2508 J Write down the balance — heat lost by the metal = heat gained by the water: 100.0 × c × 30.0 = 2508 Make c the subject: c = 2508 ÷ (100.0 × 30.0) c = 0.836 J/(g·°C)
Check your understanding
50.0 g of hot water at 90.0 °C is poured into 200.0 g of cool water at 15.0 °C in a calorimeter. What shared final temperature do they reach, in °C? (c of water = 4.18 J/(g·°C))
Answer: 30.0°C(tolerance ±0.05)
Write down the values in the question: hot water: m = 50.0 g, ΔT = 90.0 − T cool water: m = 200.0 g, ΔT = T − 15.0 Write down the balance — heat lost by the hot water = heat gained by the cool water (c divides out): 50.0 × (90.0 − T) = 200.0 × (T − 15.0) Solve step by step: 4500 − 50T = 200T − 3000 7500 = 250T T = 30.0 °C
Check your understanding
A 400.0 g metal disc at 95.0 °C is lowered into 200.0 g of water at 18.0 °C. Both settle at 25.0 °C. What is the metal's specific heat capacity, in J/(g·°C)? (c of water = 4.18 J/(g·°C))
Answer: 0.209J/(g·°C)(tolerance ±0.0005)
Write down the values in the question: water: m = 200.0 g, c = 4.18 J/(g·°C), ΔT = 25.0 − 18.0 = 7.0 °C metal: m = 400.0 g, ΔT = 95.0 − 25.0 = 70.0 °C Calculate the water's heat first: q = 200.0 × 4.18 × 7.0 = 5852 J Write down the balance — heat lost by the metal = heat gained by the water: 400.0 × c × 70.0 = 5852 Make c the subject: c = 5852 ÷ (400.0 × 70.0) c = 0.209 J/(g·°C)
Summary video — Specific heat capacity and calorimetry
Hold an ice cube in your fist and your hand quickly aches with cold. The ice is not pushing cold into you — something is being taken from you. What?
You have seen processes classified as energy-absorbing or energy-releasing, and you have seen the phase changes named: melting, freezing, and the liquid-to-gas and gas-to-liquid changes. Every phase change belongs to one energy direction.
The idea
Every phase change moves energy in or out of the sample — the melting ice absorbs energy from your hand, which is why the hand aches.
Melting absorbs energy.
'Vaporization' — the liquid-to-gas change you have seen called evaporation, and which includes boiling — absorbs energy.
Freezing releases energy.
Condensation releases energy.
The pattern: changes running in the solid → liquid → gas direction absorb energy, and changes running in the gas → liquid → solid direction release energy.
A change and its reverse always have opposite energy directions: melting absorbs exactly where freezing releases.
Worked example 1. On a cold morning, water vapor from the air condenses on a car windshield. Does the condensation absorb or release energy?
Step 1
Condensation runs from gas to liquid.
Step 2
Changes running in the gas → liquid → solid direction release energy.
Step 3
The condensation releases energy.
Worked example 2. A chocolate bar melts in your pocket. Does the melting absorb or release energy?
Step 1
Melting runs from solid to liquid.
Step 2
Changes running in the solid → liquid → gas direction absorb energy.
Step 3
The melting absorbs energy — supplied by your body's warmth.
You can now classify each phase change by its energy direction: melting and vaporization absorb energy, while freezing and condensation release energy.
Check your understanding
Which pairing of a phase change with its energy direction is correct?
ACondensation — releases energycorrect
BMelting — releases energy
This option is wrong — you flipped melting — it runs toward gas on the map (solid → liquid), so it absorbs energy.
CVaporization — releases energy
This option is wrong — you flipped vaporization — liquid → gas is the absorbing direction; think of the energy a stove must keep supplying.
DFreezing — absorbs energy
This option is wrong — you flipped freezing — liquid → solid runs toward solid, the releasing direction.
Changes running solid → liquid → gas absorb energy. Changes running gas → liquid → solid release energy. Condensation is gas → liquid, so it releases energy — the correct pairing.
Check your understanding
A puddle dries up in the sun as its water turns to vapor. What is the energy direction of this vaporization?
AIt absorbs energy from the surroundings.correct
BIt releases energy to the surroundings.
This option is wrong — you flipped the direction — liquid → gas is the absorbing direction, which is why sunshine is needed to dry the puddle.
CIt moves no energy, because the water only changes form.
This option is wrong — you treated a phase change as energy-free — every phase change moves energy in or out.
DIt absorbs energy at first and then releases the same energy back.
This option is wrong — you bounced the energy back — the absorbed energy leaves with the vapor; it does not return to the puddle.
Vaporization runs from liquid to gas. Changes running solid → liquid → gas absorb energy. The puddle's water absorbs energy from the sun and the warm air as it becomes vapor.
Check your understanding
Molten candle wax drips onto a table and hardens. What is the energy direction of this freezing?
AIt releases energy to the surroundings.correct
BIt absorbs energy from the surroundings.
This option is wrong — you flipped the direction — liquid → solid runs toward solid, the releasing direction.
CIt moves no energy, because the wax stays wax.
This option is wrong — you treated a phase change as energy-free — the substance is unchanged, but energy still moves out as the liquid becomes solid.
DIt absorbs energy, because only cold things can harden.
This option is wrong — you tied hardening to taking in cold — hardening happens because the wax RELEASES energy, cooling as it solidifies.
Hardening liquid wax is freezing: liquid → solid. Changes running gas → liquid → solid release energy. The wax releases energy to the table and the air as it solidifies.
Lesson 37 of 40 · THM-037
Read a heating or cooling curve
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Wonder this:
Put a pot of ice on a steady burner and watch the thermometer. It climbs, then sticks at 0 °C for minutes — while the burner keeps adding energy the whole time. A single graph captures that entire story, pauses and all.
You have seen that melting and vaporization absorb energy. A 'heating curve' shows what the thermometer does while that energy goes in.
The idea
A heating curve graphs a sample's temperature against the energy added to it.
Sloped segments are a single state warming up — solid, liquid, or gas gaining temperature as energy goes in.
Panel A: water's heating curve, with its plateaus named. Panels B and C: ethanol heated and lead cooled, plateaus unnamed — read what is happening, and read each plateau's height off the temperature axis.
Flat plateaus are phase changes: the first plateau is melting, and the second is vaporization (boiling).
The height of a plateau reads off the phase-change temperature: the melting plateau sits at the melting temperature, the boiling plateau at the boiling temperature.
Between the plateaus you can name the state: below the first plateau solid, between the plateaus liquid, above the second plateau gas — and on a plateau, two states mixed.
In the figure, panel A shows all of this on water's curve: the first plateau is melting at 0 °C, and the second is boiling at 100 °C.
A 'cooling curve' is the same graph traversed with energy removed: segments slope downward as a single state cools, and the plateaus are freezing and condensation at those same temperatures.
Worked examples
Worked example 1. Panel B of the figure shows the heating curve of ethanol. What is happening along its upper plateau, and at what temperature?
Step 1
A plateau is a phase change.
Step 2
The upper plateau — the second one reached as energy goes in — is vaporization.
Step 3
Read the plateau's height off the temperature axis.
Step 4
The liquid ethanol is boiling — turning to gas — and the plateau's height shows it boils at 78 °C.
Worked example 2. Panel C of the figure shows a cooling curve for lead. What is happening along its flat segment, and what does that segment's height tell you?
Step 1
On a cooling curve, a plateau is a phase change with energy being removed.
Step 2
Coming down from the liquid, the change on the plateau is freezing.
Step 3
The plateau's height reads off the phase-change temperature.
Step 4
The lead is freezing, and the plateau's height of 327 °C is the temperature at which lead freezes (and melts).
You can now identify the sloped segments of a heating or cooling curve, a graph of temperature against energy added or removed, as a single state warming or cooling and the flat plateaus as phase changes, reading the melting and boiling temperatures from the plateau heights.
Check your understanding
The figure shows the heating curve of acetic acid. What is happening along the flat segment at 118 °C?
AThe liquid is boiling — turning to gas.correct
BThe liquid is warming toward its boiling temperature.
This option is wrong — you read a plateau as a slope — along a flat segment the temperature is NOT climbing; a phase change is happening.
CThe solid is melting.
This option is wrong — you picked the wrong plateau — melting is the FIRST plateau, at 17 °C on this curve; the second plateau is boiling.
DThe gas is warming.
This option is wrong — you placed the gas too early — gas only exists above the second plateau, after the boiling finishes.
Flat plateaus are phase changes. The second plateau, at the higher temperature, is vaporization — boiling. Its height, 118 °C, is acetic acid's boiling temperature.
Check your understanding
The figure shows the heating curve of naphthalene, the classic mothball solid. Read naphthalene's melting temperature from the curve, in °C.
Answer: 80°C(tolerance ±4.0)
Find the first flat plateau on the curve. The first plateau is melting, and its height is the melting temperature. The plateau sits at 80 °C — naphthalene melts at 80 °C.
Check your understanding
The figure shows the heating curve of benzene. What state is the benzene in along the sloped segment between the two plateaus?
ALiquidcorrect
BSolid
This option is wrong — you placed the solid too late — solid benzene exists only below the first plateau at 5.5 °C, where melting happens.
CGas
This option is wrong — you placed the gas too early — gas appears only above the second plateau at 80 °C, after boiling.
DA mixture of solid and liquid
This option is wrong — you put a plateau's contents on a slope — two states coexist ON a plateau; a sloped segment is one single state warming.
Name the state by position: below the first plateau solid, between the plateaus liquid, above the second plateau gas. The marked segment lies between the 5.5 °C and 80 °C plateaus. So the benzene there is entirely liquid, warming as energy goes in.
Lesson 38 of 40 · THM-038
Why the temperature holds during a phase change
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You have seen a heating curve's flat plateaus — the figure shows water's boiling plateau again. The burner keeps adding energy along that whole flat stretch, yet the thermometer refuses to move. Where is the energy going?
The idea
Along a plateau, energy keeps flowing into the sample while the temperature holds still.
Along the plateau the added energy pulls particles away from their neighbors instead of speeding them up — so the thermometer holds at 100 °C.
Temperature measures the average kinetic energy of the particles — how fast they move on average.
The temperature stays constant along the plateau because during a phase change the added energy goes into pulling particles away from their neighbors and rearranging them, not into speeding them up.
Particles that are not speeding up cannot read hotter on a thermometer.
Only when the phase change is complete does added energy go back into speeding the particles up — and the temperature climbs again.
The same holds in reverse on a cooling curve: while a liquid freezes, particles settling back beside their neighbors release energy, and that release replaces what is being removed, so the temperature holds until the freezing finishes.
Worked examples
Worked example 1. Liquid nitrogen in an open flask boils away at −196 °C, and the liquid stays at −196 °C until the last of it is gone. Why does its temperature not rise while energy pours in from the room?
Step 1
The nitrogen is in the middle of a phase change — liquid turning to gas.
Step 2
The temperature stays constant because during a phase change the added energy goes into pulling particles away from their neighbors and rearranging them, not into speeding them up.
Step 3
The room's energy is spent separating nitrogen particles, not speeding them up — so the liquid holds at −196 °C until the boiling finishes.
Worked example 2. Molten candle wax cooling in a dish holds a steady temperature while it hardens, then cools further once fully solid. Why does the temperature hold during the hardening?
Step 1
Hardening is freezing — a phase change with energy being removed.
Step 2
As particles settle back beside their neighbors, they release energy.
Step 3
That released energy replaces what the cool air removes, so the particles do not slow down yet.
Step 4
The energy released by particles locking into place offsets the energy being removed, so the temperature holds until the wax is fully solid.
You can now explain why the temperature stays constant along a heating-curve plateau, because during a phase change the added energy goes into pulling particles away from their neighbors and rearranging them, not into speeding them up.
Check your understanding
A flask of ethanol boils steadily at 78 °C on a hot plate. Energy flows in the whole time, yet the thermometer stays at 78 °C until the liquid is gone. Why?
ABecause during the phase change the added energy goes into pulling particles away from their neighbors and rearranging them, not into speeding them up.correct
BBecause the ethanol stops absorbing energy from the hot plate while it boils, and only starts again once it is all gas.
This option is wrong — you cut off the energy flow — the boiling liquid keeps absorbing energy the whole time; the energy is just spent on separation, not speed.
CBecause a thermometer cannot give accurate readings while a phase change is happening in the liquid around its bulb.
This option is wrong — you blamed the instrument — the thermometer reads perfectly; the particles genuinely are not moving faster on average.
DBecause the energy added during the phase change is destroyed the moment the liquid ethanol turns into gas and escapes from the flask.
This option is wrong — you destroyed energy — energy cannot be created or destroyed; it is carried away in the separated gas particles.
Temperature measures the average kinetic energy of the particles — their average speed. The temperature stays constant because during a phase change the added energy goes into pulling particles away from their neighbors and rearranging them, not into speeding them up. Only once the last liquid becomes gas does added energy speed the particles up again.
Check your understanding
During a melting plateau, where does the energy added to the sample go?
AInto pulling particles away from their neighbors and rearranging them.correct
BInto making the particles move faster on average.
This option is wrong — you sent the energy into speed — faster particles would mean a rising temperature, and the plateau shows the temperature holding.
CInto raising the temperature of the newly formed liquid only.
This option is wrong — you split the sample into two temperatures — solid and liquid share one temperature on the plateau, and it is not rising.
DIt escapes to the surroundings as fast as it arrives.
This option is wrong — you turned absorption into a leak — the sample keeps the energy; it is stored in the pulled-apart arrangement of the particles.
During a phase change the added energy goes into pulling particles away from their neighbors and rearranging them, not into speeding them up. Speed unchanged means temperature unchanged. The energy is not lost — it is stored in the new, looser arrangement of the particles.
Check your understanding
A crucible of naphthalene sits exactly at the end of its melting plateau — the last crystal has just melted — and energy keeps flowing in. What happens to the temperature next?
AIt starts rising, because the added energy now goes into speeding the particles up.correct
BIt keeps holding steady, because plateaus continue as long as energy flows in.
This option is wrong — you let the plateau outlive the phase change — the plateau lasts only while two states coexist; with the solid gone, the pause is over.
CIt falls briefly, because melting used up the sample's energy.
This option is wrong — you made melting a debt to repay — the absorbed energy is stored in the particle arrangement, not owed back.
DIt jumps straight to the boiling temperature.
This option is wrong — you skipped the liquid-warming climb — the liquid warms degree by degree up the sloped segment before the boiling plateau.
The plateau's pause lasts only while particles are still being pulled away from their neighbors. Once the phase change is complete, added energy goes back into speeding the particles up. So the temperature climbs the next sloped segment toward the boiling plateau.
Lesson 39 of 40 · THM-039
Sketch a heating curve
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You have seen how to read a heating curve and why its plateaus sit flat. Now you draw the curve yourself, from just two numbers.
The idea
To sketch a substance's heating curve you need its melting temperature and its boiling temperature.
Label the axes first: temperature (°C) on the vertical axis, energy added on the horizontal axis.
Start at the given starting temperature and draw a rising sloped segment: the solid warming.
The five-piece target, drawn for water from −20 °C to 120 °C: slope, plateau at 0 °C, slope, plateau at 100 °C, slope.
At the melting temperature, draw a flat plateau: melting.
From the end of that plateau, draw a rising sloped segment up to the boiling temperature: the liquid warming.
At the boiling temperature, draw a second flat plateau: boiling.
After it, draw a final rising sloped segment: the gas warming.
Each plateau must sit exactly flat at its phase-change temperature — that height is what a reader will take off your curve.
If a phase-change temperature lies outside the range you are asked to draw, its plateau simply does not appear.
Worked examples
Worked example 1. Sketch the heating curve for acetic acid from 0 °C to 130 °C. Acetic acid melts at 17 °C and boils at 118 °C.
Step 1
Axes: temperature (°C) vertical, energy added horizontal; start at 0 °C.
Step 2
Rising slope from 0 °C to 17 °C — the solid warming.
Step 3
Flat plateau at 17 °C — melting.
Step 4
Rising slope from 17 °C to 118 °C — the liquid warming.
Step 5
Flat plateau at 118 °C — boiling.
Step 6
Rising slope above 118 °C to 130 °C — the gas warming.
Step 7
Five pieces in order: slope, flat plateau at 17 °C, slope, flat plateau at 118 °C, slope.
Worked example 2. Sketch the heating curve for tin from 100 °C to 400 °C. Tin melts at 232 °C and boils at 2602 °C.
Step 1
Axes as before; start at 100 °C.
Step 2
Rising slope from 100 °C to 232 °C — the solid warming.
Step 3
Flat plateau at 232 °C — melting.
Step 4
Rising slope above 232 °C to 400 °C — the liquid warming.
Step 5
Tin's boiling temperature, 2602 °C, lies far above the drawn range, so no second plateau appears.
Step 6
Three pieces only: slope, flat plateau at 232 °C, slope — the boiling plateau is off the top of this graph.
You can now sketch a heating curve for a substance from its melting and boiling temperatures, drawing sloped segments for single-state warming and flat plateaus at the two phase-change temperatures.
Check your understanding
A pure substance melts at 40 °C and boils at 190 °C. Panels A to D each show a curve drawn for it from 20 °C to 220 °C. Which panel shows a correct heating curve?
APanel Acorrect
BPanel B
This option is wrong — you put the boiling plateau first — the FIRST plateau reached from below must sit at the melting temperature, the lower of the two.
CPanel C
This option is wrong — you drew no plateaus — each phase change holds the temperature, so the curve must flatten at 40 °C and at 190 °C.
DPanel D
This option is wrong — you let the plateaus drift upward — the temperature holds constant during a phase change, so each plateau must be exactly flat.
A correct heating curve has five pieces: slope, flat plateau at the melting temperature, slope, flat plateau at the boiling temperature, slope. Here that means flat plateaus at exactly 40 °C and 190 °C, in that order from below. Only Panel A has both plateaus flat, at the right heights, in the right order.
Your turn
On paper, sketch the heating curve for naphthalene from 20 °C to 250 °C. Naphthalene melts at 80 °C and boils at 218 °C. Label both axes and mark the temperature of each plateau, then select Continue to compare your sketch with the model answer.
Model answer. Axes labeled temperature (°C) vertical and energy added horizontal. A rising slope from 20 °C to 80 °C, a flat plateau at 80 °C labeled melting, a rising slope from 80 °C to 218 °C, a flat plateau at 218 °C labeled boiling, and a rising slope from 218 °C to 250 °C.
The vertical axis is labeled temperature (°C) and the horizontal axis energy added.
The first plateau is exactly flat and sits at 80 °C, the melting temperature.
The second plateau is exactly flat and sits at 218 °C, the boiling temperature.
The three segments before, between, and after the plateaus all slope upward.
The pieces run in order: slope, plateau, slope, plateau, slope.
Your turn
On paper, sketch the heating curve for ethanol from −130 °C to 100 °C. Ethanol melts at −114 °C and boils at 78 °C. Label both axes and mark the temperature of each plateau, then select Continue to compare your sketch with the model answer.
Model answer. Axes labeled temperature (°C) vertical and energy added horizontal. A rising slope from −130 °C to −114 °C, a flat plateau at −114 °C labeled melting, a rising slope from −114 °C to 78 °C, a flat plateau at 78 °C labeled boiling, and a rising slope from 78 °C to 100 °C.
The vertical axis is labeled temperature (°C) and the horizontal axis energy added.
The first plateau is exactly flat and sits at −114 °C, the melting temperature.
The second plateau is exactly flat and sits at 78 °C, the boiling temperature.
The three segments before, between, and after the plateaus all slope upward.
Negative temperatures are drawn below 0 °C on the vertical axis, with the curve still rising left to right.
Lesson 40 of 40 · THM-040
Evaluate a heating curve
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You have seen how to sketch a heating curve from a substance's data. The last skill runs the other way: given someone else's curve, decide whether it tells the truth.
The idea
To evaluate a heating curve, check it against the substance's melting and boiling data, one feature at a time.
Check 1 — plateau heights: the first plateau must sit at the melting temperature, and the second at the boiling temperature.
Check 2 — plateau shape: each plateau must be exactly flat, because the temperature holds during a phase change.
Check 3 — the segments: every piece between and beyond the plateaus must slope upward, because a single state warms as energy is added.
A curve that fails any one check misrepresents the substance; a curve that passes all three is correct.
The figure shows all three failures next to a correct curve for the same substance.
Check heights, check flatness, check slopes — each flawed panel fails exactly one check.
Worked examples
Worked example 1. Panel B of the figure shows a student's heating curve for naphthalene, which melts at 80 °C and boils at 218 °C. The curve's first plateau sits at 100 °C. Evaluate the curve.
Step 1
Check 1 — plateau heights: the first plateau must sit at the melting temperature, 80 °C.
Step 2
Panel B's first plateau sits at 100 °C instead — the student borrowed water's boiling temperature.
Step 3
One failed check is enough.
Step 4
Flawed — the melting plateau is drawn at 100 °C, but naphthalene melts at 80 °C.
Worked example 2. Panel C of the figure shows another heating curve for naphthalene whose plateaus sit at 80 °C and 218 °C but drift gently upward instead of lying flat. Evaluate the curve.
Step 1
Check 1 passes: the plateaus sit at 80 °C and 218 °C.
Step 2
Check 2 — plateau shape: each plateau must be exactly flat, because the temperature holds during a phase change.
Step 3
Panel C's plateaus drift upward, so check 2 fails.
Step 4
Flawed — the plateaus must be exactly flat, and these drift upward.
You can now evaluate whether a supplied heating curve correctly represents a substance's melting and boiling data, checking that each plateau sits flat at the correct phase-change temperature and that single-state segments slope.
Check your understanding
Benzene melts at 5.5 °C and boils at 80 °C. The figure shows a student's heating curve for benzene, with flat plateaus at 5.5 °C and 100 °C. Which verdict is correct?
AFlawed — the second plateau sits at 100 °C, but benzene boils at 80 °C.correct
BFlawed — the first plateau sits at 5.5 °C, but melting plateaus belong at 0 °C.
This option is wrong — you defaulted to water's melting temperature — each substance's plateaus sit at ITS OWN phase-change temperatures, and 5.5 °C is correct for benzene.
CCorrect — both plateaus are flat, and flat plateaus are all the checks require.
This option is wrong — you stopped at the flatness check — the heights must also match the substance's data, and 100 °C does not match benzene's 80 °C boiling temperature.
DFlawed — a heating curve for benzene should have no plateaus at all.
This option is wrong — you removed the phase changes — benzene melts and boils inside this range, so the curve must flatten twice.
Run the checks in order: heights, flatness, slopes. First plateau: 5.5 °C matches benzene's melting temperature — passes. Second plateau: 100 °C is water's boiling temperature, not benzene's 80 °C — check 1 fails, so the curve is flawed.
Check your understanding
Tin melts at 232 °C. The figure shows a heating curve drawn for tin from 100 °C to 400 °C as one straight rising line with no flat segment. Which verdict is correct?
AFlawed — a flat plateau must appear at 232 °C, where tin melts.correct
BCorrect — the curve rises steadily, and rising is all a heating curve must do.
This option is wrong — you skipped the plateau check — melting happens at 232 °C inside this range, and the temperature must hold flat there.
CFlawed — the curve needs plateaus at 0 °C and 100 °C.
This option is wrong — you defaulted to water's temperatures — plateaus belong at TIN's phase-change temperatures, and only its 232 °C melting sits in this range.
DFlawed — the line should slope downward, not upward.
This option is wrong — you flipped the graph — added energy warms the sample, so a heating curve's segments rise; the missing plateau is the real flaw.
Check the substance's data against the drawn range: tin melts at 232 °C, inside 100–400 °C. So the curve must flatten at 232 °C — the temperature holds during a phase change. A straight line with no plateau misses the melting entirely: flawed.
Check your understanding
Acetic acid melts at 17 °C and boils at 118 °C. The figure shows a heating curve drawn for it from 5 °C to 135 °C. Run the three checks — heights, flatness, slopes. Which verdict is correct?
ACorrect — flat plateaus at 17 °C and 118 °C, rising segments everywhere else.correct
BFlawed — the first plateau should sit at 0 °C.
This option is wrong — you defaulted to water's melting temperature — acetic acid's own melting temperature is 17 °C, and that is where its plateau belongs.
CFlawed — the plateaus should slope gently upward instead of lying flat.
This option is wrong — you bent the plateaus — the temperature holds during a phase change, so plateaus are exactly flat.
DFlawed — a correct curve needs a third plateau between 17 °C and 118 °C.
This option is wrong — you added a phase change that does not exist — between melting and boiling there is only liquid warming, one rising segment.
Check 1 — heights: 17 °C and 118 °C match acetic acid's data. Passes. Check 2 — flatness: both plateaus lie exactly flat. Passes. Check 3 — slopes: every other segment rises. Passes — the curve is correct.