Skip to main content
← Thermal Physics

Thermal Physics · 21.1

Temperature & Internal Energy

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.

01

Build the model

Connect the measurement to the mechanism.

Temperature determines the direction of spontaneous heat transfer and whether systems are in thermal equilibrium. In the classical ideal-gas model it is proportional to mean translational kinetic energy per molecule. Heat is energy transferred because of a temperature difference.

Internal energy includes microscopic kinetic and potential energies. The thermodynamic temperature scale is not defined by the response of any particular thermometer material.

Simple definition
Temperature describes thermal state; internal energy is the total microscopic kinetic plus potential energy; heat is energy transferred between systems because their temperatures differ.
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 = θ/°C + 273.15

One kelvin and one Celsius degree are equal temperature intervals. Zero kelvin corresponds exactly to −273.15 °C. The offset cancels in a temperature difference: a rise of 1 °C is a rise of 1 K.

Exact offset; 273 is a rounded approximation. No degree symbol on kelvin.

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

An ideal-gas extrapolation, not a real-gas cooling experiment

For a fixed amount and volume of ideal gas, p is proportional to thermodynamic temperature T. Extrapolating this model gives p = 0 at 0 K, exactly −273.15 °C. Real gases deviate and may condense before that limit, so this is not a claim that every actual gas follows a straight line all the way to zero pressure. Thermodynamic temperature is independent of the substance used to measure it.

03

What a thermometer reads

A thermometer reaches thermal equilibrium with the object. It measures a calibrated property, such as liquid-column length, resistance or thermocouple e.m.f., to infer the temperature. It does not directly measure all the object's internal energy; energy can change during a phase transition without a temperature change.

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.

steel 90°Caluminium 20°Cboth settle at 55 °C

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, to the nearest kelvin?

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. Round each answer to the nearest whole unit.
  1. T/K = 25 + 273.15 = 298.15, which rounds to 298 K.
  2. θ/°C = 100 − 273.15 = −173.15, which rounds to −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.
  1. Same substance, so the specific heat cancels: heat lost = heat gained.
  2. 0.100 × (90 − T) = 0.200 × (T − 30).
  3. 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?
  1. 27 °C corresponds to 300.15 K. In this ideal-gas model mean translational KE is proportional to T.
  2. Double the mean translational KE needs 600.30 K.
  3. 600.30 − 273.15 = 327.15 °C, about 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.
  1. Heat lost by steel = heat gained by aluminium: 2.0 × 450 × (100 − T) = 1.0 × 900 × (T − 10).
  2. 900(100 − T) = 900(T − 10) → 100 − T = T − 10 → T = 55 °C.
  3. 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.

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.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

IGCSE

01Fig. 10.1Thermal properties and temperatureIGCSE
An aluminium block with an electric heater and a thermometer in drilled holespower supply48 W immersion heaterthermometeraluminium blockof mass 1.0 kg

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.

Guided questions 5 parts

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.

  1. 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
    1. 48 J
    2. 48 W is a power, the rate at which the heater transfers energy
  2. 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
    1. E = mcΔθ = 1.0 × 900 × 20 = 18 000 J
    2. t = E/P = 18 000 / 48 = 375 s (6.25 minutes)
  3. 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
    1. energy must be conducted through the aluminium from the heater hole across to the thermometer hole
    2. this takes time, so the thermometer lags behind the temperature of the aluminium next to the heater
    3. some of the early energy also warms the heater itself and the metal immediately around it before the block is at a uniform temperature
  4. 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
    1. energy lost in 60 s = mcΔθ = 1.0 × 900 × 0.60 = 540 J
    2. rate of loss = 540/60 = 9.0 W
    3. useful power = 48 − 9.0 = 39 W
    4. 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.

Check answer 3 marks
  1. E = Pt = 48 × 300 = 14 400 J
  2. c = E/(mΔθ) = 14 400 / (0.50 × 12) = 2400 J/(kg °C)
  3. overestimate, because some of the 14 400 J is transferred to the surroundings rather than to the liquid

IB

02Figure 2Thermal energy transfersIB
Insulated container of liquid with an immersed heater and a thermometerpower supply50.0 W0.500 kg of liquidheaterthermometerinsulation

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.

Guided questions 5 parts

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.

  1. 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
    1. the rate at which electrical energy is supplied to the heating element
    2. 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)
  2. 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.

    routine3 marks

    Check answer 3 marks
    1. E = mcΔT = 0.500 × 2.4 × 10³ × 1.00 = 1.20 × 10³ J
    2. t = E / P = 1.20 × 10³ / 50.0
    3. t = 24.0 s
  3. 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
    1. mass evaporating per second = 4.2 × 10⁻⁴ / 60 = 7.0 × 10⁻⁶ kg s⁻¹
    2. power carried away = 7.0 × 10⁻⁶ × 8.5 × 10⁵ = 6.0 W
    3. 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
    4. c comes out about 14% too high
  4. 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
    1. 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
    2. 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
    3. the corrected value of c is smaller than the uncorrected one, because part of the 50.0 W was never warming the liquid
    4. 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
  1. ΔT = 38 − 15 = 23 K
  2. for a steady flow the power is (m/t)cΔT, not mcΔT for a fixed mass
  3. P = 0.075 × 4.2 × 10³ × 23 = 7.2 × 10³ W
  4. 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 lawsIB
Sealed rigid container holding nitrogen gasnitrogen gasV = 5.0 × 10⁻³ m³p = 2.4 × 10⁵ PaT = 290 Ksealed rigid container

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.

Guided questions 5 parts

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.

  1. 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
    1. the volume, 5.0 × 10⁻³ m³, because the container is rigid
    2. the amount of gas, because the container is sealed
  2. 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
    1. V and n are fixed, so p/T is constant: p₁/T₁ = p₂/T₂
    2. T₂ = 290 × (3.0 × 10⁵ / 2.4 × 10⁵) = 362.5 K
    3. T₂ = 363 K, that is about 89 °C
  3. 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
    1. average kinetic energy = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J
    2. mass of one molecule = 28.0 × 10⁻³ / 6.02 × 10²³ = 4.65 × 10⁻²⁶ kg
    3. ½mv²_rms = 6.0 × 10⁻²¹, so v²_rms = 2.58 × 10⁵ m² s⁻²
    4. v_rms = 5.1 × 10² m s⁻¹
  4. 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
    1. p = nRT/V with n halved and V and T unchanged, so the pressure halves to 1.2 × 10⁵ Pa
    2. the average kinetic energy of a molecule is unchanged, because it depends only on the temperature
    3. 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
    4. 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
  1. n is constant, so p₁V₁/T₁ = p₂V₂/T₂
  2. V₂ = 12 × (1.0 × 10⁵ / 2.6 × 10⁴) × (220/290)
  3. V₂ = 35 m³
  4. 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

Cambridge 9702 · focused teaching and practice

Complete the objective

Open a topic for its explanation, worked example and original practice. The AS and additional A-Level requirements remain separate. Practical preparation does not replace supervised laboratory work.

Additional A-Level · 14.2.1

Choose and calibrate a thermometric property

Teaching, example & practice

General browsing: select Cambridge AS or A2 and your examination year for a recommended route.

A thermometer measures a physical property that changes reproducibly with temperature. Calibration associates property readings with independently known temperatures. Leave the sensor in thermal contact long enough to reach equilibrium; otherwise it may report its own changing temperature rather than the object's temperature.

Different properties need different controls. A thermocouple responds to the temperatures of both junctions, so its reference junction must be known. A gas-volume thermometer needs a fixed amount of gas and constant pressure. Do not assume every property is exactly linear across a large range: use a calibration curve when necessary.

For an assumed linear calibration: θ/°C = 100(X − X₀)/(X₁₀₀ − X₀)

Device/propertyChange used in its calibrated rangeEssential control
Liquid thermometer: densityFor fixed mass, falling density gives rising volume and liquid-column length.Use the specified liquid and calibrated range; allow expansion relative to the glass.
Gas thermometer: volumeFixed amount of dilute gas expands as temperature rises.Keep pressure constant; do not let gas leak.
Metal resistance thermometerMetal resistance generally increases with temperature.Use a small measuring current to limit self-heating.
Thermocouple: e.m.f.Junction-temperature difference changes the measured e.m.f.Keep the reference-junction temperature known and use its calibration.

Worked example

A metal-resistance thermometer reads 100.0 Ω at 0.0 °C and 138.5 Ω at 100.0 °C. Assuming linear behaviour, what temperature corresponds to 119.25 Ω?

  1. Fraction of calibrated interval = (119.25 − 100.0)/(138.5 − 100.0) = 0.500.
  2. θ = 0.500 × 100.0 °C = 50.0 °C.
  3. This interpolation relies on the stated linear approximation; two calibration points alone do not prove linearity.

50.0 °C under the linear calibration assumption.

Scope and sources

Original GioPhysics teaching and practice, checked against the syllabus. This section does not certify unrelated videos, downloads or the rest of a shared lesson.

Additional A-Level · 14.2.2

Why thermodynamic temperature is substance-independent

Teaching, example & practice

General browsing: select Cambridge AS or A2 and your examination year for a recommended route.

A scale made by dividing a particular liquid's expansion into equal steps is tied to that liquid's behaviour. Another material can agree at the calibration points yet give a different intermediate reading if its response is nonlinear. This is a limitation of that thermometer's calibration, not evidence that the object has two temperatures.

Thermodynamic temperature is a universal temperature scale: it is not defined by the expansion, density, resistance or e.m.f. of a chosen substance. Thermometers made from different materials are instruments for measuring that same temperature. Each needs calibration to it.

A dilute-gas thermometer illustrates the idea. As gases approach ideal-gas behaviour, the same pressure-temperature or volume-temperature ratio is approached independently of gas species. This is a useful model, not a claim that real gases remain gaseous down to zero kelvin. In the modern SI, the kelvin is defined through a fixed value of the universal Boltzmann constant; the properties of one selected thermometer material do not define it.

Worked example

Two thermometers agree at 0 °C and 100 °C but read 49 °C and 51 °C in the same well-stirred bath at equilibrium. Does this mean the bath has two thermodynamic temperatures?

  1. At equilibrium both sensors are sampling the same bath temperature.
  2. Their different material responses or calibration errors can produce different indicated readings.
  3. Compare each with a suitable reference over the interval; do not average the two values and call that the definition of temperature.

No. There is one thermodynamic temperature; the thermometer indications need calibration or correction.

Scope and sources

Original GioPhysics teaching and practice, checked against the syllabus. This section does not certify unrelated videos, downloads or the rest of a shared lesson.

  • Cambridge syllabus · printed page 27 — 2025–2027: verified numbered demand on the printed page; teaching and questions are newly authored.
  • Cambridge syllabus · printed page 27 — 2028–2030: checked directly against the downloaded official PDF; the targeted demand is unchanged.
  • www.bipm.org — BIPM primary source for the modern fixed-Boltzmann-constant definition; included as context, not an additional Cambridge memorisation demand.
Additional A-Level · 14.2.3

Exact Kelvin–Celsius conversion and temperature intervals

Teaching, example & practice

General browsing: select Cambridge AS or A2 and your examination year for a recommended route.

Kelvin and Celsius use equal-sized intervals but different zeros. The exact numerical offset is 273.15: 0 °C is 273.15 K, and 0 K corresponds to −273.15 °C. Kelvin is written K, not °K.

Subtracting two converted temperatures cancels the offset. A rise of 12.0 °C is a rise of 12.0 K; never add 273.15 to a temperature change. Ratios such as 'twice the temperature' in ideal-gas equations must use absolute temperatures in kelvin, not Celsius readings.

The rounded offset 273 is an approximation. Use the exact relationship first, retain sensible guard digits, then round the final answer to the precision justified by the data or the question.

T/K = θ/°C + 273.15

θ/°C = T/K − 273.15

ΔT/K = Δθ/°C

Worked example

A sample warms from −10.0 °C to 25.0 °C. Convert both temperatures and calculate the increase.

  1. T₁ = −10.0 + 273.15 = 263.15 K, or 263.2 K to 0.1 K.
  2. T₂ = 25.0 + 273.15 = 298.15 K, or 298.2 K to 0.1 K.
  3. ΔT = 298.15 − 263.15 = 35.0 K, identical in size to 35.0 °C.

263.2 K to 298.2 K; increase 35.0 K.

Scope and sources

Original GioPhysics teaching and practice, checked against the syllabus. This section does not certify unrelated videos, downloads or the rest of a shared lesson.

Extra challenging questions

Take the ideas further with course-specific extension questions, supplied-data graphs and worked solutions.