A metal bench and a wooden one sit at the same temperature on a cold morning, yet the metal feels colder. Temperature, heat, and internal energy are three different things — and thermal physics starts by telling them apart.
Temperature measures the average random kinetic energy of molecules. Heat is energy in transit from a hotter object to a colder one. Two objects in contact exchange heat until their temperatures match — thermal equilibrium.
Internal energy is the whole store: the kinetic energy of molecular motion plus the potential energy held in intermolecular bonds. The kelvin scale starts at absolute zero, where that molecular motion is at its minimum.
Simple definition
Temperature measures the average kinetic energy per molecule; internal energy is the total kinetic plus potential energy of all the molecules; heat is the energy that flows between objects at different temperatures.
Example
A sparkler spark at 1000 °C carries far less energy than a bathtub of warm water — the spark has a high temperature but only a few molecules.
Kelvin and CelsiusT(K) = T(°C) + 273
The two scales use the same size of step — kelvin just starts counting at absolute zero, −273 °C, instead of at the freezing point of water.
No degree symbol on kelvin temperatures
Internal energyE(int) = E(K) + E(P)
Heating a substance can speed its molecules up (more KE) or pull them further apart against their bonds (more PE) — internal energy counts both.
Random molecular KE + intermolecular PE
Thermal equilibriumheat flows until T(A) = T(B)
Touching objects trade energy until their temperatures equalise; after that the net flow is zero. This is what a thermometer relies on.
Net flow is always hot → cold
01
Solids, liquids, gases
In a solid the molecules only vibrate about fixed positions. In a liquid some bonds are broken and molecules move about within a fixed volume. In a gas most bonds are broken and the molecules fly freely. Melting and boiling are bond-breaking steps between these arrangements.
02
Why absolute zero exists
Cool a fixed volume of gas and its pressure falls along a straight line. Extrapolate, and every gas — whatever its type — reaches zero pressure at the same temperature: −273 °C. Since pressure cannot go below zero, no temperature can sit below this point.
03
What a thermometer reads
A thermometer left in contact with an object reaches thermal equilibrium with it, and its reading tracks the object’s average molecular kinetic energy. It reads the KE part of internal energy — the PE part hides from it, which matters during melting and boiling.
04
Heating makes things bigger
Almost everything expands when heated. Give the particles more energy and they vibrate or fly about more vigorously, so on average they sit a little further apart and the same mass takes up more volume. How much further depends on what holds them: a gas expands most, because nothing but the container restrains it; a liquid expands far less, its particles still touching; a solid least of all, every atom locked into the lattice and able only to vibrate more widely about the same site. Gases most, liquids next, solids least — and the reason is the strength of the bonds, not the size of the temperature rise.
05
Designing for expansion
Engineers leave room for it rather than fight it. Railway lines are laid with gaps between sections, a bridge deck ends in a toothed expansion joint and sits on rollers, and overhead cables are strung with a deliberate sag that takes up the slack in winter. A bimetallic strip puts the effect to work: two metals that expand by different amounts are bonded together, so heating bends the strip towards the metal that expands less, and that bend can open or close a circuit in a thermostat or a fire alarm.
02
Change one variable at a time
Make the relationship visible.
Steel c = 450, aluminium c = 900 J kg⁻¹ K⁻¹; the aluminium block is fixed at 1.0 kg. Heat flows hot → cold until both blocks share one temperature.
Equilibrium temperature55 °C
Heat transferred31.5 kJ
Directionsteel → aluminium
03
Catch the common trap
Explain before calculating.
What is 27 °C on the kelvin scale?
Choose an answer to test the model.
04
Worked examples
State the rule, substitute, then check units.
EasyConvert 25 °C to kelvin, and 100 K to Celsius.
T(K) = 25 + 273 = 298 K.
T(°C) = 100 − 273 = −173 °C.
Answer298 K; −173 °C
Medium100 g of water at 90 °C is mixed with 200 g of water at 30 °C in an insulated cup. Find the equilibrium temperature.
Same substance, so the specific heat cancels: heat lost = heat gained.
0.100 × (90 − T) = 0.200 × (T − 30).
9 − 0.1T = 0.2T − 6 → 0.3T = 15 → T = 50 °C.
Answer50 °C
HardThe average molecular kinetic energy of a gas is proportional to its kelvin temperature. At what Celsius temperature is the average KE double its value at 27 °C?
27 °C = 300 K.
Double the KE needs double the kelvin temperature: 600 K.
600 − 273 = 327 °C — not 54 °C. Doubling must be done in kelvin.
Answer327 °C (600 K)
ChallengingA 2.0 kg steel block (c = 450 J kg⁻¹ K⁻¹) at 100 °C is clamped to a 1.0 kg aluminium block (c = 900 J kg⁻¹ K⁻¹) at 10 °C, insulated from everything else. Find the equilibrium temperature and the heat that flows.
Heat lost by steel = heat gained by aluminium: 2.0 × 450 × (100 − T) = 1.0 × 900 × (T − 10).
900(100 − T) = 900(T − 10) → 100 − T = T − 10 → T = 55 °C.
Q = 900 × (100 − 55) = 40 500 J ≈ 41 kJ flows from steel to aluminium — and the same 40 500 J arrives, confirming the bookkeeping.
Answer55 °C; about 41 kJ transferred
Exam diagrams for this topic3 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.Open subsectionHide subsection
See it. Read it. Work it.
These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.
01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.
IGCSE
01Fig. 10.1Thermal properties and temperatureIGCSEOpen diagramClose diagram
Figure comment
Fig. 10.1The apparatus is seen from the side. A rectangular aluminium block of mass 1.0 kg stands on the bench. A 48 W electric immersion heater is pushed down into a narrow hole drilled near one side of the block, and a thermometer is pushed down into a second narrow hole near the other side, its bulb reaching well inside the block. Two leads run from the top of the heater across to a power supply standing beside the apparatus.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. Read the 48 W as a rate, not an amount: the drawing fixes the energy delivered each second, and the two holes are drilled apart so the thermometer reads the block, not the heater.
aState State the energy transferred to the aluminium block by the heater shown in Fig. 10.1 in one second, and state the quantity the label 48 W measures.
recall2 marks
Check answer · 2 marks
48 J
48 W is a power, the rate at which the heater transfers energy
bCalculate The specific heat capacity of aluminium is 900 J/(kg °C). Calculate the time for which the heater in Fig. 10.1 must run to raise the temperature of the block by 20 °C, assuming no thermal energy escapes.
routine2 marks
Check answer · 2 marks
E = mcΔθ = 1.0 × 900 × 20 = 18 000 J
t = E/P = 18 000 / 48 = 375 s (6.25 minutes)
cExplain The heater and the thermometer occupy holes at opposite sides of the block. Explain why the thermometer reading rises more slowly in the first minute after switching on than it does later in the run.
demanding3 marks
Check answer · 3 marks
energy must be conducted through the aluminium from the heater hole across to the thermometer hole
this takes time, so the thermometer lags behind the temperature of the aluminium next to the heater
some of the early energy also warms the heater itself and the metal immediately around it before the block is at a uniform temperature
dDetermine The student switches the heater off at the end of the run and finds that the block cools by 0.60 °C in the next minute. Determine the rate at which the block was losing thermal energy at the end of the run, and hence determine the percentage of the heater's 48 W that was actually raising the temperature of the block at that moment.
top of the paper4 marks
Show a hint
The block cools at the same rate whether or not the heater is on, so the cooling run measures the loss that was happening during the heating run.
Check answer · 4 marks
energy lost in 60 s = mcΔθ = 1.0 × 900 × 0.60 = 540 J
rate of loss = 540/60 = 9.0 W
useful power = 48 − 9.0 = 39 W
percentage = 39/48 × 100 = 81%
Transfer challenge
The same 48 W heater is used instead to warm 0.50 kg of a liquid in a beaker, and the temperature of the liquid rises by 12 °C in 5.0 minutes. Calculate a value for the specific heat capacity of the liquid, and state whether it is an overestimate or an underestimate of the true value.
overestimate, because some of the 14 400 J is transferred to the surroundings rather than to the liquid
IB
02Figure 2Thermal energy transfersIBOpen diagramClose diagram
Figure comment
Figure 2Section through the apparatus. A container holding the liquid, labelled 0.500 kg of liquid, is surrounded on its sides and base by hatched insulation. A coiled heating element is immersed near the bottom of the liquid, and its two leads run up out of the open top of the container to a box labelled power supply, 50.0 W. A thermometer stands in the liquid with its bulb well below the surface and its stem projecting above the container.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The thermometer bulb sits in the liquid and not against the coil, and 50.0 W is labelled on the supply, so it is the electrical input rather than the power reaching the liquid.
aState State what the 50.0 W on the power supply measures, and state one feature of the drawing that means less than 50.0 W warms the liquid.
recall2 marks
Check answer · 2 marks
the rate at which electrical energy is supplied to the heating element
the container is open at the top, so energy escapes there by evaporation and convection (or: the leads, the heater and the container itself absorb energy)
bDetermine The liquid has specific heat capacity 2.4 × 10³ J kg⁻¹ K⁻¹. Determine how long the heater must run to raise the temperature of the liquid shown by 1.00 K, assuming all the electrical energy reaches it.
cDetermine The open top is the one surface the hatched insulation does not cover, and 0.42 g of liquid evaporates from it each minute. The specific latent heat of vaporisation is 8.5 × 10⁵ J kg⁻¹. Determine the power this carries away and the resulting error in c.
demanding4 marks
Check answer · 4 marks
mass evaporating per second = 4.2 × 10⁻⁴ / 60 = 7.0 × 10⁻⁶ kg s⁻¹
power carried away = 7.0 × 10⁻⁶ × 8.5 × 10⁵ = 6.0 W
only 50.0 − 6.0 = 44.0 W actually warms the liquid, so a value calculated from 50.0 W is too large by a factor 50.0/44.0 = 1.14
c comes out about 14% too high
dSuggest The student switches the supply off and keeps reading the thermometer as the liquid cools. Suggest how the cooling readings can be used to correct her value of c, and outline the assumption the correction rests on.
top of the paper4 marks
Check answer · 4 marks
after switch-off nothing but the losses is acting, so the rate of fall measured at a given temperature is a direct measure of the loss at that temperature
at that same temperature the heating run gives 50.0 = mc × (rate of rise) + mc × (rate of fall on cooling), so adding the two gradients yields c without needing the loss in watts first
the corrected value of c is smaller than the uncorrected one, because part of the 50.0 W was never warming the liquid
assumes the rate of loss depends only on the excess temperature over the surroundings, so it is the same whether the heater is on or off
Transfer challenge
An electric shower raises water from 15 °C to 38 °C as it flows through at 0.075 kg s⁻¹. Determine the minimum electrical power the shower must draw. Take c for water as 4.2 × 10³ J kg⁻¹ K⁻¹.
Check answer · 4 marks
ΔT = 38 − 15 = 23 K
for a steady flow the power is (m/t)cΔT, not mcΔT for a fixed mass
P = 0.075 × 4.2 × 10³ × 23 = 7.2 × 10³ W
this is a minimum because any energy lost to the shower body and the surroundings must be supplied on top of it
03Figure 3Thermal energy transfers · Gas lawsIBOpen diagramClose diagram
Figure comment
Figure 3A rectangular container with thick walls and a stopper in its top is labelled sealed rigid container. Inside, ten molecules are drawn as dots, each carrying a short arrow of its own direction and length so that the molecules are moving randomly, several of them towards the walls. To the right of the container stand the labels nitrogen gas, V = 5.0 × 10⁻³ m³, p = 2.4 × 10⁵ Pa and T = 290 K.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Reading cue. The arrows differ in length as well as direction, so the drawing shows a spread of speeds; the rigid walls fix V, which is why warming the gas raises p rather than expanding it.
aState State the two quantities shown by or implied by the drawing that cannot change when the container is warmed, and give a reason for each.
recall2 marks
Check answer · 2 marks
the volume, 5.0 × 10⁻³ m³, because the container is rigid
the amount of gas, because the container is sealed
bDetermine Determine the temperature to which the gas must be warmed for the pressure to reach 3.0 × 10⁵ Pa.
routine3 marks
Check answer · 3 marks
V and n are fixed, so p/T is constant: p₁/T₁ = p₂/T₂
T₂ = 290 × (3.0 × 10⁵ / 2.4 × 10⁵) = 362.5 K
T₂ = 363 K, that is about 89 °C
cDetermine The arrows represent a spread of molecular speeds. Determine the root mean square speed of a nitrogen molecule at the temperature shown. The molar mass of N₂ is 28.0 g mol⁻¹.
demanding4 marks
Show a hint
The mass of one molecule is the molar mass divided by the Avogadro constant, not the molar mass itself.
Check answer · 4 marks
average kinetic energy = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J
mass of one molecule = 28.0 × 10⁻³ / 6.02 × 10²³ = 4.65 × 10⁻²⁶ kg
½mv²_rms = 6.0 × 10⁻²¹, so v²_rms = 2.58 × 10⁵ m² s⁻²
v_rms = 5.1 × 10² m s⁻¹
dExplain A valve is opened briefly and exactly half the nitrogen escapes, the temperature being held at 290 K. Explain how the pressure, the average kinetic energy of a molecule, and the rate of collisions on a given wall each change.
top of the paper4 marks
Check answer · 4 marks
p = nRT/V with n halved and V and T unchanged, so the pressure halves to 1.2 × 10⁵ Pa
the average kinetic energy of a molecule is unchanged, because it depends only on the temperature
each remaining molecule moves just as fast and hits just as hard, but there are half as many, so the rate of collisions on a given wall halves
the pressure falls because the collisions are fewer, not because they are gentler
Transfer challenge
A weather balloon holds 12 m³ of helium at 1.0 × 10⁵ Pa and 290 K at ground level, in an envelope free to expand. Determine its volume at an altitude where the pressure is 2.6 × 10⁴ Pa and the temperature is 220 K, assuming no gas escapes.
Check answer · 4 marks
n is constant, so p₁V₁/T₁ = p₂V₂/T₂
V₂ = 12 × (1.0 × 10⁵ / 2.6 × 10⁴) × (220/290)
V₂ = 35 m³
unlike the rigid container, the flexible envelope lets V change: the pressure drop expands it nearly fourfold and the cooling claws back about a quarter of that