Educerie · IB Diploma · Physics
Theme B The particulate nature of matter · B.1 Thermal energy transfers
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Describe solids, liquids and gases with the molecular model | SL, HL | "Outline, in terms of molecules, the differences between a liquid and a gas" (3 marks) |
| Calculate density from ρ = m/V | SL, HL | One step inside a longer question |
| Convert between Celsius and Kelvin, and know a temperature change is the same on both | SL, HL | Paper 1A, and every gas and radiation calculation |
| Relate Kelvin temperature to the average kinetic energy of particles, Ek = (3/2)kBT | SL, HL | "Calculate the average kinetic energy of…" (1 to 2 marks) |
| Define internal energy, and distinguish it from temperature | SL, HL | "Distinguish between internal energy and temperature" (3 marks) |
| State that temperature difference decides the direction of thermal energy transfer | SL, HL | Paper 1A, or one line of an explanation |
| Explain a phase change as a change in potential energy at constant temperature | SL, HL | "Explain why the temperature stays constant while the solid melts" (2 marks) |
| Calculate with Q = mcΔT and Q = mL, including mixtures and phase changes | SL, HL | Paper 2, 3 to 5 marks; Paper 1B from a graph |
| Explain conduction with particle kinetic energy, and calculate with ΔQ/Δt = −kAΔT/Δx | SL, HL | "Calculate the rate of thermal energy transfer through the wall" (2 marks) |
| Describe convection in terms of density differences | SL, HL | "Explain how the heater warms the whole room" (3 marks) |
| Use the Stefan–Boltzmann law L = σAT⁴ for a black body | SL, HL | "Calculate the luminosity of the star" (2 marks) |
| Define apparent brightness and use b = L/4πd² | SL, HL | "Determine the distance to the star" (2 marks) |
| Describe a black-body spectrum and use Wien's law λmaxT = 2.9 × 10⁻³ m K | SL, HL | "Estimate the surface temperature from the spectrum" (2 marks) |
Before you start
You need energy and power from A.3, and the idea that energy is conserved: every calculation in this subtopic is energy in = energy out, written carefully. From A.1 you need gradients of graphs. Light as an electromagnetic wave appears in section 9; you need only know that it carries energy and has a wavelength.
1The idea in one paragraph
Everything is made of particles (atoms or molecules) that are always moving and that pull on each other. That one picture explains a surprising amount. The temperature of a substance tells you the average kinetic energy of its particles. Its internal energy is the total energy of all of them, moving and pulling. Supply energy and either the particles move faster, so the temperature rises (Q = mcΔT), or they pull apart, so the substance changes phase at a steady temperature (Q = mL). Energy travels from a hotter body to a colder one in three ways: conduction, as fast particles jostle slow ones; convection, as warm fluid rises because it is less dense; and thermal radiation, as electromagnetic waves leave the surface. The last one works across empty space, which is how the Sun warms the Earth and how we measure the temperature of a star we will never visit.
2The molecular model, and density
The molecular model describes each phase by how close its particles are, how strongly they are held and how they move. Figure 1 draws the three.
| Solid | Liquid | Gas | |
|---|---|---|---|
| Spacing | close together, in a regular pattern | close together, no regular pattern | far apart, about ten times the spacing in a liquid |
| Forces between particles | strong: each particle held in place | strong enough to keep them together, not in place | very weak, except during collisions |
| Motion | vibrate about fixed positions | vibrate, and move past one another | move fast and randomly in straight lines between collisions |
| So the substance | keeps its shape and volume | keeps its volume, takes the shape of its container | fills any container |
That last row is the point: the everyday behaviour follows from the particle picture. A liquid can be poured because its particles can slide past each other. A gas can be squashed because there is mostly empty space between its particles.
Density ρ is mass per unit volume:
ρ = m/V, unit kg m⁻³
A block of aluminium measures 5.0 cm × 4.0 cm × 2.0 cm and has a mass of 108 g.
Density links straight back to Figure 1. Water at 100 °C has a density of about 960 kg m⁻³; the steam it boils into has a density of about 0.6 kg m⁻³. The same mass takes up roughly 1600 times the volume, and since volume goes as spacing cubed, the particles are about 12 times further apart (∛1600 ≈ 12). That is the "ten times" in the table.
3Temperature, and what it measures
Two scales are in use. The Celsius scale (°C) is set by water: 0 °C at melting, 100 °C at boiling. The Kelvin scale (K) starts at absolute zero, the lowest possible temperature, where the particles have the least possible energy. The steps are the same size on both scales, so:
T / K = θ / °C + 273, and a temperature change is the same number on both scales: a rise of 60 °C is a rise of 60 K
So 20 °C is 293 K, and heating water from 20 °C to 80 °C raises its temperature by 60 K. In every formula that uses a temperature itself (Ek = (3/2)kBT, L = σAT⁴, Wien's law) you must use kelvin. In formulas that use a temperature change (Q = mcΔT, conduction) either scale gives the same number.
Kelvin temperature has a direct physical meaning. It measures the average kinetic energy of the particles' random motion:
Ek = (3/2)kBT, where kB = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant
At room temperature, 300 K:
A tiny number, because it belongs to one particle. The equation says that the average kinetic energy is proportional to the kelvin temperature. Double the kelvin temperature and you double it. From 27 °C (300 K), doubling means going to 600 K, which is 327 °C, not 54 °C. Double the Celsius number and you have done nothing meaningful.
4Internal energy, and which way energy flows
The internal energy of a system is the energy its particles hold between them. Add up the kinetic energy of every particle's random motion, add the potential energy stored in the forces that particles exert on one another, and the sum is the internal energy.
Internal energy = total random kinetic energy + total intermolecular potential energy
Note the two words "total". Internal energy is a sum over every particle, so it depends on how many particles there are. Temperature depends on the average kinetic energy, so it does not. A swimming pool at 25 °C has vastly more internal energy than a cup of tea at 80 °C, because it has vastly more molecules, but its temperature is lower.
That distinction settles which way energy goes. When two bodies are in thermal contact, the resultant thermal energy transfer is always from the body at the higher temperature to the body at the lower temperature, whatever their internal energies. Put the cup of tea on the edge of the pool and energy flows from the tea to the pool, although the pool has far more internal energy. The flow stops when the temperatures are equal. That state is thermal equilibrium.
The word heat (symbol Q) means energy transferred because of a temperature difference. A body does not "contain heat"; it contains internal energy, and heat is what moves.
5Specific heat capacity
Supply energy to a substance that is not changing phase and its temperature rises. How much it rises depends on how much there is and on what it is made of.
Q = mcΔT, where c is the specific heat capacity: the energy needed to raise the temperature of 1 kg of the substance by 1 K. Unit: J kg⁻¹ K⁻¹
Water has an unusually large c, about 4180 J kg⁻¹ K⁻¹; copper's is about 385 J kg⁻¹ K⁻¹. So the same energy warms a kilogram of copper about eleven times as much as a kilogram of water.
One body. A kettle holds 1.2 kg of water at 18 °C. How much energy brings it to 100 °C, and how long does a 2.2 kW kettle take, if no energy is lost?
Two bodies reaching equilibrium. A 0.50 kg copper block at 100 °C is dropped into 0.20 kg of water at 20 °C in an insulated cup. The energy lost by the copper equals the energy gained by the water, and both end at the same temperature T.
Write each ΔT as (higher − lower) so both sides are positive. Sense-check: the final temperature must lie between the two starting temperatures, and nearer the water's, because the water has the larger mc.
Measuring c. Figure 2 is the standard method. An electrical heater and a thermometer sit in holes in a metal block of known mass, wrapped in insulation. An ammeter and voltmeter give the power P = VI; a stopwatch gives t; so the energy supplied is VIt. The temperature is recorded over time, and if the heater power is steady, the gradient of a temperature–time graph gives ΔT/Δt, so c = P ÷ (m × gradient).
The main weakness is energy lost to the surroundings. Some of the heater's energy leaves the block instead of warming it, so the temperature rises more slowly than it should, the gradient is too small, and the calculated c comes out too large. Insulation, a drop of oil in the thermometer hole for good thermal contact, and starting a little below room temperature and finishing the same amount above it all reduce the error.
6Phase changes and specific latent heat
A phase change is a change in particle behaviour, from solid to liquid or liquid to gas and back, that happens with a change in energy at constant temperature. The terms you must use:
- melting (solid → liquid) and freezing (liquid → solid), at the melting point
- boiling (liquid → gas, throughout the liquid, at the boiling point) and condensing (gas → liquid)
- evaporation: liquid → gas from the surface only, at any temperature below the boiling point. The fastest molecules escape, so the average kinetic energy of those left behind falls, which is why evaporation cools. Sweat works this way.
Figure 3 heats ice from −18 °C to steam at a steady power. The temperature rises, stops, rises, stops, rises.
On the sloping parts, energy goes into random kinetic energy: the particles move faster and the temperature climbs. On the flat parts, the temperature does not change, so the average kinetic energy does not change either. The energy is going into intermolecular potential energy, doing work against the forces that hold the particles together, breaking the bonds that keep them in place (melting) or together (boiling). The internal energy rises all the way along the graph; only on the slopes does the temperature follow it.
The energy needed per kilogram for a phase change is the specific latent heat L:
Q = mL, where L is the energy needed to change the phase of 1 kg of the substance at constant temperature. Unit: J kg⁻¹
There are two for each substance: the specific latent heat of fusion Lf (solid ↔ liquid) and of vaporization Lv (liquid ↔ gas). For water, Lf = 3.34 × 10⁵ J kg⁻¹ and Lv = 2.26 × 10⁶ J kg⁻¹. Lv is almost seven times Lf, which is why the boiling plateau in Figure 3 is so long: melting only loosens the particles, boiling must separate them completely.
Take 0.25 kg of ice from a freezer at −18 °C to water at 20 °C. Use cice = 2100 J kg⁻¹ K⁻¹ and cwater = 4180 J kg⁻¹ K⁻¹. There are three stages, and each gets its own line.
Almost three-quarters of the energy goes into melting, during which the thermometer does not move at all. That is why ice cools a drink so well.
7Conduction
Conduction is the transfer of energy through a material by collisions between its particles, with no bulk movement of the material. At the hot end, particles have a larger average kinetic energy. They collide with their slower neighbours and pass on some of that energy, which pass it on in turn. The resultant flow is from high kinetic energy to low, which is from high temperature to low. In metals there is a second, faster route: free electrons move through the whole metal and carry kinetic energy quickly from the hot region to the cold, which is why metals are much better conductors than non-metals.
The rate of energy transfer through a slab (Figure 4) depends on three things you can see in the equation:
ΔQ/Δt = −kA ΔT/Δx
- A is the cross-sectional area the energy flows through: twice the area, twice the flow.
- ΔT/Δx is the temperature gradient: the temperature change across the slab divided by its thickness. A thin slab with a large temperature difference has a steep gradient and a fast flow.
- k is the thermal conductivity of the material, in W m⁻¹ K⁻¹: large for metals (copper about 400), small for insulators (glass about 0.8, still air about 0.025).
The minus sign says energy flows down the gradient: temperature falls in the direction of flow, so ΔT/Δx is negative (Figure 4b), and the minus makes the rate positive. In a calculation, use the size of the temperature difference and a positive answer.
A single pane of glass 1.5 m² in area and 4.0 mm thick has its inner surface at 12 °C and its outer surface at 4 °C. Take k = 0.80 W m⁻¹ K⁻¹.
Replace the glass with a layer of still air 12 mm thick, with the same 8 K across it: 0.025 × 1.5 × 8 ÷ 0.012 = 25 W, about a hundredth. This is why double glazing traps a layer of gas, and why feathers, fur and foam insulate: they hold air still.
Conduction also explains why a metal bench feels colder than a wooden one on the same cold morning, although both are at the same temperature. Metal conducts energy away from your skin much faster, so your skin cools faster, and your nerves report the rate of energy loss, not the temperature of the bench.
8Convection
Convection is the transfer of energy by the bulk movement of a fluid (a liquid or a gas), driven by differences in density. It cannot happen in a solid, whose particles cannot move from place to place.
Follow the air in Figure 5. Air next to the radiator is warmed. Its particles move faster and spread out, so the warm air expands and its density falls. Less dense fluid floats upwards through denser fluid, just as a cork rises through water, so the warm air rises to the ceiling and spreads across it. Away from the radiator it transfers energy to the cooler walls and room, contracts, becomes denser and sinks. Cooler air is drawn along the floor to replace the air that rose. The loop that forms is a convection current, and it carries energy around the whole room.
The same explanation, in the same order (heated, expands, less dense, rises; cools, denser, sinks), answers every convection question: a pan of soup, a sea breeze, the circulation of the atmosphere, and the churning outer layers of the Sun. Marks come from the word density. "Heat rises" scores nothing: it is the warm fluid that rises, and it rises because it is less dense.
9Thermal radiation, black bodies and stars
Thermal radiation is energy transferred by electromagnetic waves emitted from the surface of a body because of its temperature. Every body above absolute zero emits it. Unlike conduction and convection it needs no material, so it crosses a vacuum. At everyday temperatures the waves are mostly infrared; at a few thousand kelvin a large part is visible light.
A black body is an idealised object that absorbs all the radiation that falls on it and emits the most radiation possible for its temperature. Stars, and the Sun, are close to black bodies. The total power a body radiates is its luminosity L, in watts, and for a black body it is given by the Stefan–Boltzmann law:
L = σAT⁴, where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, A is the surface area and T is the absolute temperature
The fourth power makes temperature dominate. Double the temperature and the power rises 2⁴ = 16 times. For a sphere of radius R, A = 4πR².
The spectrum. A black body does not emit equally at all wavelengths. Figure 6 plots the intensity it emits against wavelength at three temperatures. Each curve rises to a peak and falls away, and as the temperature rises the whole curve grows and its peak moves to shorter wavelength.
The peak position is given by Wien's displacement law:
λmaxT = 2.9 × 10⁻³ m K
That is how the surface temperature of a star is measured: record its spectrum, find the peak, divide. A red star has a peak at long wavelength and is cool; a blue-white star peaks in the ultraviolet and is hot. The area under each curve is the total power per square metre, which is why the 6000 K curve encloses so much more than the 4000 K curve: it is the T⁴ of Stefan–Boltzmann, shown as a picture.
Apparent brightness. From Earth we cannot measure a star's luminosity directly. We measure the power arriving per square metre of our detector, the apparent brightness b, in W m⁻². The star's power spreads out over a sphere centred on it, and at distance d that sphere has area 4πd² (Figure 7).
b = L / 4πd²
Double the distance and the same power covers four times the area, so the brightness falls to a quarter. That is an inverse-square law, and it appears again for gravitational and electric fields.
The Sun, worked through. The Sun's surface temperature is about 5800 K and its radius is 6.96 × 10⁸ m. The Earth is 1.50 × 10¹¹ m away.
That last number is close to the measured value at the top of the atmosphere, about 1.36 kW m⁻², which is a good test of treating the Sun as a black body.
A star, worked backwards. Luminosities are often given as multiples of the Sun's, L⊙ = 3.83 × 10²⁶ W. An (invented) star has its spectrum peak at 290 nm and a luminosity of 25 L⊙. Its apparent brightness at Earth is 2.0 × 10⁻⁹ W m⁻². Find its temperature, radius and distance.
Three measurements made on Earth (a peak wavelength, a brightness and a luminosity found by other means) have given the size and distance of an object no one can reach. That is the guide's question in action: observing one quantity lets you determine others. Radiation is also how solar panels and the Earth's climate are understood, which is where B.2 picks up.
10Where marks are lost
- Using Celsius in a formula that needs kelvin. Ek = (3/2)kBT, L = σAT⁴ and λmaxT all need kelvin. Only temperature differences can stay in Celsius.
- Confusing temperature with internal energy. Temperature depends on the average kinetic energy of a particle; internal energy is the total kinetic plus potential energy of all of them. A large cold body can have more internal energy than a small hot one.
- Saying energy flows from more internal energy to less. It flows from higher temperature to lower temperature.
- Saying the kinetic energy rises during melting. At constant temperature the average kinetic energy is constant; the energy supplied increases the intermolecular potential energy.
- Leaving out a stage in a phase-change problem. Ice at −18 °C to water at 20 °C is three stages: warm the ice, melt it, warm the water. Write each as its own line.
- "Heat rises." Warm fluid rises, because it has expanded and its density is lower. Without density the convection mark is lost.
- Forgetting that area in L = σAT⁴ is the surface area 4πR², and that d in b = L/4πd² is distance to the observer. Using R where d belongs (or πR² where 4πR² belongs) is the commonest star error.
- Arguing that a larger measured c means a better experiment. Energy lost to the surroundings makes the measured c too large. Say which way the error goes and why.
11Draw it right
- On a heating curve (Figure 3), label the axes with quantity and unit, make the phase-change sections exactly horizontal, and label each section with the phase and whether Ek or Ep is increasing.
- For a cooling curve, the same shape runs downhill, with the plateau at the freezing point.
- In a particle diagram (Figure 1), keep particles the same size in all three phases; only spacing, order and motion change.
- For conduction, show the energy flow arrow from the hot face to the cold face, and label A, Δx and the two temperatures.
- For convection, draw a closed loop: rising above the heat source, sinking where the fluid cools, and returning along the bottom. Label "less dense, rises" and "denser, sinks".
- On a black-body spectrum, put wavelength on the x axis and intensity on the y axis, start every curve at zero at short wavelength, make the hotter curve higher everywhere with its peak further left, and mark λmax.
- For an experiment graph (Paper 1B), draw the best-fit line through the points, and take the gradient from a large triangle whose corners are on the line.
12Try it
Marks in brackets. Use c(water) = 4180 J kg⁻¹ K⁻¹ and Lf(water) = 3.34 × 10⁵ J kg⁻¹ where needed. Answers and marker's notes follow.
Q1. Distinguish between the temperature of a body and its internal energy. 3 marks
Q2. Ice cubes of total mass 0.040 kg at 0 °C are added to 0.30 kg of a drink at 25 °C. The drink has the same specific heat capacity as water. Assuming no energy is exchanged with the surroundings, calculate the final temperature once all the ice has melted. 4 marks
Q3. (Data-based, Paper 1B style.) A student heats a 1.00 kg aluminium block with a 50.0 W heater, using the apparatus in Figure 2, and records the temperature every minute. The results are plotted in Figure 8. 6 marks
(a) Determine the rate of temperature rise of the block. 2 marks
(b) Calculate the specific heat capacity of aluminium from these results. 2 marks
(c) The accepted value is 900 J kg⁻¹ K⁻¹. Explain why the experimental value is larger, and suggest one improvement to the method. 2 marks
Q4. A cool box is made of polystyrene 25 mm thick, with a total wall area of 0.60 m². The thermal conductivity of polystyrene is 0.033 W m⁻¹ K⁻¹. The outside surface is at 30 °C and the inside surface is at 0 °C, kept there by melting ice. 4 marks
(a) Calculate the rate of thermal energy transfer into the box. 2 marks
(b) Calculate how long 1.0 kg of ice at 0 °C takes to melt completely. 2 marks
Q5. The spectrum of a star peaks at 725 nm. Its luminosity is 0.20 L⊙, where L⊙ = 3.83 × 10²⁶ W is the luminosity of the Sun. The Sun's surface temperature is 5800 K. 6 marks
(a) Calculate the surface temperature of the star. 1 mark
(b) Determine the radius of the star as a multiple of the Sun's radius. 3 marks
(c) The apparent brightness of the star at Earth is 1.0 × 10⁻¹⁰ W m⁻². Calculate its distance from Earth. 2 marks
Q6. An electric heater is placed near the floor of a cold room. Explain how the air in the whole room becomes warm. 3 marks
13In one breath
Solids have close, ordered particles that vibrate in place; liquids have close, disordered particles that slide past each other; gases have widely spaced particles moving fast and randomly; density is ρ = m/V. Kelvin temperature is Celsius plus 273, a temperature change is the same on both scales, and the kelvin temperature is proportional to the average kinetic energy of the particles, Ek = (3/2)kBT. Internal energy is the total random kinetic energy plus the total intermolecular potential energy, and energy flows from higher temperature to lower until the temperatures are equal. Heating without a phase change raises the kinetic energy and the temperature: Q = mcΔT. A phase change happens at constant temperature, raising or lowering the potential energy: Q = mL. Conduction passes kinetic energy from particle to particle (and, in metals, by free electrons) at a rate ΔQ/Δt = −kAΔT/Δx; convection moves warm, less dense fluid upwards and cool, denser fluid down; thermal radiation is electromagnetic waves from a surface. A black body radiates L = σAT⁴, its spectrum peaks at λmax = 2.9 × 10⁻³ m K ÷ T, and its brightness at distance d is b = L/4πd², which is how a star's temperature, size and distance are measured from Earth.
Answers
Q1. Temperature is a measure of the average random kinetic energy of the particles of the body; for a given body it does not depend on how many particles there are. Internal energy is the total random kinetic energy of all the particles plus the total intermolecular potential energy between them, so it depends on the amount of substance and includes potential energy. Two bodies at the same temperature can have very different internal energies. 1 for temperature as a measure of the average kinetic energy of particles, 1 for internal energy as the total kinetic plus potential energy, 1 for a clear contrast (average versus total, or potential energy included only in internal energy). "Temperature is how hot something is" scores 0.
Q2. Energy to melt the ice = 0.040 × 3.34 × 10⁵ = 13 360 J. Energy to warm the melted ice from 0 °C to T = 0.040 × 4180 × T = 167.2T. Energy lost by the drink = 0.30 × 4180 × (25 − T) = 31 350 − 1254T. Setting gained = lost: 13 360 + 167.2T = 31 350 − 1254T, so 1421.2T = 17 990 and T = 12.7 °C. M1 for mLf for the ice, M1 for warming the melted ice from 0 °C, M1 for equating energy gained and lost, A1 for 12.7 °C. Leaving out the warming of the melted ice gives 14.3 °C and loses the second M1 and the A1.
Q3. (a) A line of best fit through the points runs from 20.0 °C at t = 0 to about 35.6 °C at t = 300 s. Rate = (35.6 − 20.0) ÷ 300 = 0.052 K s⁻¹ (accept 0.050 to 0.054). (b) Energy supplied per second equals the rate of energy gain of the block: P = mc × (ΔT/Δt), so c = 50.0 ÷ (1.00 × 0.052) = 960 J kg⁻¹ K⁻¹ (accept 930 to 1000 from their gradient). (c) Some of the heater's energy is transferred to the surroundings instead of to the block, so the temperature rises more slowly than it would with no losses; the gradient is too small and the calculated c too large. Improvement: add more insulation around the block (or a lid over the top, or oil in the thermometer and heater holes for better thermal contact). (a) M1 for a gradient from the line using a large interval, A1 for the value with unit. (b) M1 for c = P ÷ (m × gradient), A1 for the answer consistent with (a). (c) R1 for energy lost to the surroundings making the gradient smaller so c larger, A1 for one sensible improvement. "Repeat the experiment" alone does not score the improvement mark, because repeats do not remove a systematic error.
Q4. (a) ΔQ/Δt = kA × ΔT/Δx = 0.033 × 0.60 × 30 ÷ 0.025 = 23.8 W (about 24 W). (b) Energy needed = mLf = 1.0 × 3.34 × 10⁵ = 3.34 × 10⁵ J. Time = 3.34 × 10⁵ ÷ 23.8 = 1.4 × 10⁴ s, about 3.9 hours. (a) M1 for substitution with Δx in metres, A1 for 23.8 W. Leaving Δx as 25 gives 0.024 W and scores M0. (b) M1 for mLf ÷ rate, A1 for 1.4 × 10⁴ s (error carried forward from (a)).
Q5. (a) T = 2.9 × 10⁻³ ÷ (725 × 10⁻⁹) = 4.0 × 10³ K. (b) L = σ4πR²T⁴ for both stars, so L/L⊙ = (R/R⊙)² × (T/T⊙)⁴. Then (R/R⊙)² = 0.20 ÷ (4000/5800)⁴ = 0.20 ÷ 0.226 = 0.884, and R = 0.94 R⊙. (c) d = √(L ÷ 4πb) = √(0.20 × 3.83 × 10²⁶ ÷ (4π × 1.0 × 10⁻¹⁰)) = 2.5 × 10¹⁷ m. (a) A1. (b) M1 for using L ∝ R²T⁴ (or for finding R in metres and dividing by the Sun's radius), M1 for the fourth power of the temperature ratio, A1 for 0.94. (c) M1 for rearranging b = L/4πd², A1 for 2.5 × 10¹⁷ m. Using the Sun's luminosity instead of 0.20 L⊙ gives 5.5 × 10¹⁷ m and loses the A1.
Q6. The heater warms the air next to it. That air expands, so its density decreases, and being less dense than the surrounding air it rises. Cooler, denser air from elsewhere in the room moves in along the floor to take its place and is warmed in turn. The warm air spreads across the ceiling, transfers energy to the cooler parts of the room, becomes denser and sinks. The circulation (a convection current) carries energy throughout the room. 1 for warmed air expanding and becoming less dense, 1 for less dense air rising and denser air sinking to replace it, 1 for a continuous circulation carrying energy around the room. "Heat rises" with no mention of density scores at most 1.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section B.1 Thermal energy transfers. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.