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This lesson covers the fundamental concepts of thermal energy and internal energy as required by AQA A-Level Physics specification section 3.6. Understanding internal energy is the foundation upon which all thermal physics is built. You must be able to define internal energy precisely, explain what happens to it during heating and during changes of state, and apply the first law of thermodynamics.
Spec mapping (AQA 7408 §3.6.2.1 — Thermal energy transfer): This lesson addresses the definition of internal energy as the sum of randomly distributed kinetic and potential energies of molecules within a system, the distinction between heat and internal energy, the behaviour of internal energy during changes of state, and the qualitative application of the first law of thermodynamics. The Specification Reference 3.6.2.1 also expects students to interpret what happens at the molecular level when a system is heated, cooled, or undergoes a phase change. (Refer to the official AQA specification document for exact wording.)
Synoptic links: This lesson underpins the entire thermal physics topic. The molecular picture of internal energy connects directly to specific heat capacity (§3.6.2.2), where the energy supplied raises the kinetic component without state change, and to specific latent heat (§3.6.2.2), where energy raises the potential component without temperature change. The first-law framework reappears in molecular kinetic theory (§3.6.2.3), where for an ideal gas U is purely translational kinetic energy and the link 〈½m〈c²〉 = (3/2)kT directly equates microscopic motion to internal energy. Synoptic exam routes also reach back to GCSE energy stores (chemical, kinetic, elastic) and forward to AS particle physics wherever Boltzmann-type statistics appear. Examiners frequently combine §3.6.2.1 with §3.6.1 (temperature scales) on six-mark items requiring candidates to distinguish heat, temperature and internal energy in a single coherent answer.
Key Definition: The internal energy of a system is the sum of the randomly distributed kinetic energies and potential energies of all its molecules.
Internal energy has two components:
Exam Tip: When defining internal energy, you must mention both kinetic and potential energy AND state that the energies are randomly distributed. Stating just "the total energy of all the molecules" is insufficient for full marks.
Temperature is a measure of the average kinetic energy of the molecules in a substance. When you heat a substance (without causing a change of state), you increase the average kinetic energy of the molecules, which means:
During a change of state (melting, boiling, freezing, condensing), the temperature remains constant even though energy is being supplied (or removed). This is one of the most important and frequently tested ideas in thermal physics.
When a solid melts:
When a liquid boils:
Exam Tip: A very common exam question asks what happens to internal energy during a change of state. The model answer must state three things: (1) temperature remains constant, (2) kinetic energy does not change, and (3) only the potential energy component of internal energy increases. Many candidates lose marks by failing to mention that kinetic energy stays the same.
The first law of thermodynamics is a statement of conservation of energy applied to thermal processes:
ΔU = Q − W
where:
| Quantity | Positive means... | Negative means... |
|---|---|---|
| Q | Energy transferred INTO the system (heating) | Energy transferred OUT of the system (cooling) |
| W | Work done BY the system (expansion) | Work done ON the system (compression) |
| ΔU | Internal energy increases | Internal energy decreases |
Case 1: Heating at constant volume If a gas is heated in a rigid container, it cannot expand, so W = 0. Therefore ΔU = Q — all the energy supplied goes into increasing the internal energy.
Case 2: Heating at constant pressure If a gas is heated at constant pressure, it expands and does work on its surroundings. The work done is W = pΔV. Therefore ΔU = Q − pΔV — some of the energy supplied goes into work, and the rest increases the internal energy.
Case 3: Adiabatic expansion If a gas expands without any heat transfer (Q = 0), then ΔU = −W. The gas does work at the expense of its own internal energy, so it cools down. This is why gas escaping from a pressurised container feels cold.
Case 4: Free expansion into a vacuum An ideal gas expanding into a vacuum does no work (there is nothing to push against), so W = 0. If the container is insulated, Q = 0 as well. Therefore ΔU = 0 — the temperature of an ideal gas does not change during free expansion. For a real gas, the temperature drops slightly because work is done against intermolecular forces.
Question: A gas is heated, receiving 500 J of energy. During this process, the gas expands and does 200 J of work on a piston. Calculate the change in internal energy of the gas.
Solution:
Using ΔU = Q − W:
ΔU = 500 − 200 = 300 J
The internal energy of the gas increases by 300 J. The remaining 200 J of the 500 J supplied was used to do work pushing the piston outwards.
Question: A bicycle pump is used to compress air. The pump does 150 J of work on the gas, and 40 J of heat escapes to the surroundings during the process. Calculate the change in internal energy of the gas.
Solution:
Work is done ON the gas, so W = −150 J (the system is compressed, not expanding). Heat is lost from the system, so Q = −40 J.
ΔU = Q − W = (−40) − (−150) = −40 + 150 = 110 J
The internal energy increases by 110 J. The gas heats up because the work done on it exceeds the heat lost to the surroundings.
Exam Tip: Be very careful with the sign convention in first law problems. If the question says work is done ON the gas, W is negative in the equation ΔU = Q − W. If heat is lost, Q is negative. Read the question carefully and assign signs before substituting.
When two objects at different temperatures are placed in thermal contact, energy flows spontaneously from the hotter object to the cooler object until they reach thermal equilibrium (the same temperature). This is a consequence of the second law of thermodynamics.
The direction of energy transfer depends on temperature, not on internal energy. A large cold object may have more total internal energy than a small hot object, but energy still flows from hot to cold.
Energy can be transferred by three mechanisms:
| Concept | Key Formula | Notes |
|---|---|---|
| Internal energy | U = total KE + total PE | Sum of randomly distributed molecular energies |
| First law | ΔU = Q − W | Conservation of energy for thermal processes |
| Change of state | Temperature constant | Only PE component of U changes |
| Heating (no state change) | Temperature rises | Mainly KE component increases |
| Ideal gas internal energy | U = total KE only | No intermolecular forces, so PE = 0 |
"Internal energy is the same as temperature." This is wrong. Temperature measures the average kinetic energy per molecule, while internal energy is the total energy of all molecules. A bath of warm water has far more internal energy than a red-hot needle, even though the needle is at a much higher temperature.
"During a change of state, no energy is being supplied." Energy is being supplied (or removed), but it does not cause a temperature change — it changes the potential energy component.
"All the heat supplied to a gas goes into raising its temperature." This is only true at constant volume. At constant pressure, some energy goes into doing work (expanding against the surroundings).
"Work is always positive." Work is positive when done BY the system; it is negative when done ON the system. The sign matters in the first law.
A fixed mass of ideal gas is taken around the following closed cycle, returning to its starting state. Calculate the heat Q and work W for each stage, and verify that ΔU_total = 0 around the cycle.
Stage A → B: Isothermal expansion at T = 300 K. Work done by gas = 400 J. Stage B → C: Adiabatic compression. Internal energy increases by 600 J. Stage C → A: Isobaric (constant pressure) cooling at the higher pressure, returning to starting state. Energy 800 J is transferred out of the gas as heat.
Stage A → B (isothermal expansion): For an isothermal process on an ideal gas, T is constant ⇒ ΔU = 0 (since U depends only on T for an ideal gas). ΔU = Q − W ⇒ 0 = Q_AB − 400 ⇒ Q_AB = +400 J. The gas absorbs 400 J of heat from a thermal reservoir at 300 K and converts it all into work.
Stage B → C (adiabatic compression): "Adiabatic" means Q = 0 (no heat exchange). ΔU = Q − W ⇒ +600 = 0 − W ⇒ W = −600 J. Work is done on the gas (W = work by gas is negative), increasing its internal energy by 600 J.
Stage C → A (isobaric cooling): Q_CA = −800 J (negative because energy leaves the gas as heat). The work done during this stage we don't know directly, but we can compute it using the closed-cycle constraint: ΔU around the closed loop = 0 (since the gas returns to its starting state, U is unchanged). ΔU_AB + ΔU_BC + ΔU_CA = 0 0 + 600 + ΔU_CA = 0 ΔU_CA = −600 J.
Apply the first law to CA: ΔU_CA = Q_CA − W_CA −600 = −800 − W_CA W_CA = −800 − (−600) = −200 J.
Negative — so work is done on the gas during the C → A stage (the gas is being compressed back to its starting volume by the external pressure as it cools).
Cycle summary table:
| Stage | Process | Q (J) | W (J) | ΔU (J) |
|---|---|---|---|---|
| A → B | Isothermal expansion | +400 | +400 | 0 |
| B → C | Adiabatic compression | 0 | −600 | +600 |
| C → A | Isobaric cooling | −800 | −200 | −600 |
| Cycle | Net | −400 | −400 | 0 ✓ |
Net cycle work: W_net = +400 − 600 − 200 = −400 J. (Negative means net work is done on the gas — this cycle is a refrigerator-like cycle in the direction shown, not a heat-engine cycle.)
Net heat: Q_net = +400 + 0 − 800 = −400 J. (Negative — net heat leaves the gas. Consistent with a refrigerator: heat is extracted from the cold reservoir during expansion and rejected to the hot reservoir during cooling.)
Conservation check: Q_net − W_net = −400 − (−400) = 0 ✓. Internal energy returns to starting value, as required for a closed cycle.
Exam Tip: For closed-cycle thermodynamic problems, always tabulate Q, W, and ΔU for each stage and verify ΔU_total = 0 around the cycle. This is a routine sanity check that catches sign errors. The fact that net work and net heat are equal (in magnitude and sign) is a consequence of the first law applied to a complete cycle: ∮ΔU = 0 ⇒ ∮Q = ∮W.
Specimen question modelled on the AQA paper format (6 marks).
A sealed container of water at 50 °C is brought into thermal contact with an identical sealed container of water at 10 °C. Both containers are thermally insulated from the surroundings. The system is left until thermal equilibrium is reached.
Distinguish carefully between heat, temperature and internal energy in this scenario, and explain what is conserved and what changes during the process. [6 marks]
On 6-mark "distinguish and explain" items of this type, examiners typically allocate the marks across two assessment objectives. AO1 (knowledge and understanding) is generally worth around 2 marks for definitions — for the precise meaning of internal energy as the sum of randomly distributed kinetic and potential energies, for temperature as a measure of mean molecular kinetic energy, and for heat as energy in transit between systems at different temperatures. AO2 (application) typically carries the remaining 4 marks for applying those definitions to the specific scenario: identifying the direction of heat transfer, recognising that total internal energy of the closed system is conserved while the distribution shifts, and concluding with an equilibrium temperature analysis.
When two containers of water at different temperatures are placed in contact, heat will flow from the hotter water at 50 °C to the cooler water at 10 °C. Heat is the energy that moves because of a temperature difference. The hot water loses internal energy and its temperature falls. The cold water gains internal energy and its temperature rises. Eventually they reach the same temperature in the middle, which is 30 °C since the masses are equal. Internal energy is the total energy of all the molecules in each container. Temperature is different from internal energy because temperature tells you how hot something is. The total energy of the whole system stays the same because no energy escapes — the insulation stops heat being lost to the surroundings. So energy is conserved overall but it is shared out differently between the two containers.
Examiner commentary: This response earns roughly 3-4 of 6 marks. It correctly identifies the direction of heat flow (M1), distinguishes heat as energy in transit (M2 — partial), and applies energy conservation to deduce an equilibrium temperature of 30 °C (M3). It loses marks for treating internal energy as simply "total energy" rather than the sum of randomly distributed kinetic and potential energies, and for not linking temperature explicitly to mean molecular kinetic energy. A Grade C candidate captures the macroscopic picture but not the molecular precision an A-Level mark scheme rewards.
Three distinct quantities must be separated. Internal energy is the sum of the randomly distributed kinetic and potential energies of all the molecules in a sample. Temperature is a measure of the mean translational kinetic energy per molecule; the kelvin scale assigns T = 0 to the state of minimum molecular kinetic energy. Heat (denoted Q) is not a property of a body — it is energy that is transferred between two bodies because they are at different temperatures, and it ceases to exist as "heat" the moment it has entered or left a system, becoming part of the recipient's internal energy.
In this scenario the warmer water (50 °C, 323 K) initially has a higher mean molecular kinetic energy than the cooler water (10 °C, 283 K). At the interface, faster-moving molecules transfer kinetic energy to slower ones through collisions, so net energy flows from the hotter water to the cooler water. We label this transferred energy Q.
Apply the first law ΔU = Q − W to each container. Because the containers are rigid and sealed, no work is done by either system on the surroundings: W = 0 for each. Therefore ΔU = Q for each container. The hot water loses internal energy ΔU_hot = −Q (negative because energy leaves), and the cold water gains internal energy ΔU_cold = +Q. Since the masses and substance are identical and Q_hot = −Q_cold, the magnitudes balance: total internal energy of the combined system is conserved (no heat escapes because the system is insulated). The equilibrium temperature is therefore exactly the mean, 30 °C, where the mean molecular kinetic energy across both bodies is uniform and no further net transfer occurs.
What is conserved: the total internal energy of the closed two-container system. What changes: the distribution of that internal energy — it shifts from a non-equilibrium configuration (hot/cold) to a uniform equilibrium configuration. What is not conserved separately: the internal energy of each individual container.
Examiner commentary: Full 6/6. Three precise definitions delivered at the molecular level (M1, M2), correct first-law decomposition with W=0 reasoning (M3, M4), clear distinction between system-level conservation and sub-system change (M5), and a clean equilibrium-temperature deduction (M6). The A*-band move is the explicit appeal to the first law and to mean molecular kinetic energy rather than the GCSE-level "temperature tells you how hot".
For a deeper treatment of internal energy and the statistical roots of temperature, Schroeder's An Introduction to Thermal Physics (Addison-Wesley) develops U from the multiplicity of microstates and is the standard undergraduate gateway from A-Level thermal physics into statistical mechanics. Atkins' Physical Chemistry presents the same first law from a chemist's perspective with explicit treatment of enthalpy H = U + pV that the A-Level syllabus leaves implicit. Kittel and Kroemer's Thermal Physics (2nd ed.) is the harder but more elegant treatment, building U from the partition function. Oxford and Cambridge physics interviewers commonly ask candidates to explain why internal energy of an ideal monatomic gas is purely kinetic — a question that links A-Level §3.6.2.1 directly to undergraduate equipartition.