Internal Energy and the First Law of Thermodynamics Revision Notes | A Level AQA Physics

1. What Is Internal Energy?

Picture a gas sealed inside a cylinder. Every single molecule is moving — bouncing off walls, spinning, vibrating. Each molecule carries kinetic energy because of that motion, and potential energy because of the forces between it and its neighbours. Add all of those energies together across every particle in the system, and you have the internal energy.

The kinetic energy contribution comes from translational motion (particles moving through space), rotation, and vibration. The potential energy contribution arises from intermolecular forces — the attractions and repulsions between particles that depend on how far apart they are.

In an ideal gas, intermolecular forces are assumed to be negligible, so the internal energy is entirely kinetic. In real substances — especially liquids and solids — the potential energy contribution is significant and cannot be ignored.

2. The First Law of Thermodynamics

Now that you know what internal energy is, the natural question is: how do you change it? There are exactly two ways to increase the internal energy of a system — transfer energy to it by heating, or do work on it.

Think of a bicycle pump. When you compress the air inside, you are doing work on the gas. The gas gets warmer — its internal energy has increased, even though no heat source was applied. Conversely, when you heat a gas in a container, energy flows in by heating and the internal energy rises again.

At A-level, AQA treats this qualitatively: you need to be able to identify whether a process increases or decreases internal energy, and state which mechanism (heating or work) is responsible. The formal equation $\Delta U = Q + W$ captures this, where $\Delta U$ is the change in internal energy, $Q$ is energy transferred by heating (positive into the system), and $W$ is work done on the system (positive when compressed).

A common scenario: a gas is compressed rapidly (work done on it, $W > 0$) and simultaneously loses a small amount of heat to the surroundings ($Q < 0$). The net change in internal energy depends on which effect dominates. If $W > |Q|$, the internal energy rises and the gas warms up.

3. Temperature, Heating, and Kinetic Energy

When you heat a substance and its temperature rises, you are increasing the average kinetic energy of its particles. For an ideal gas, the mean random kinetic energy of the particles is proportional to the absolute temperature (in kelvin) — more precisely, mean KE = (3/2)kT, where k is the Boltzmann constant and T is the absolute temperature. Raise the temperature and the particles move faster on average.

This is where the specific heat capacity equation becomes essential. The energy needed to raise the temperature of a mass $m$ by $\Delta\theta$ is given by $Q = mc\Delta\theta$, where $c$ is the specific heat capacity of the material.

Worked example: Calculating energy transferred by heating

A 2.0 kg block of aluminium is heated from 20 °C to 95 °C. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. Calculate the energy transferred to the block. [3 marks]

1. $Q = mc\Delta\theta$ — [M1] Selecting the correct equation earns the method mark.
2. $\Delta\theta = 95 - 20 = 75 \text{ K}$ — Subtract initial from final temperature. The unit K and °C give the same numerical difference for a temperature change.
3. $Q = 2.0 \times 900 \times 75$ — [M1] Correct substitution with consistent units.
4. $Q = 2.0 \times 900 \times 75 = 1800 \times 75 = 135\,000 \text{ J}$ — Working: $2.0 \times 900 = 1800$, then $1800 \times 75 = 135\,000$.
5. $Q = 135\,000 \text{ J} = 135 \text{ kJ}$ — [A1] Answer with correct unit. Both 135,000 J and 135 kJ are acceptable.

Answer: $Q = 135\,000 \text{ J}$ (or 135 kJ)

4. Changes of State: Potential Energy, Not Kinetic

Here is where the physics becomes genuinely interesting — and where many students lose marks. When a substance changes state (for example, from solid to liquid, or liquid to gas), energy is still being transferred in. Yet the temperature does not change. So where does the energy go?

The answer is that the energy goes into increasing the potential energy of the particles, not their kinetic energy. During melting, the regular lattice structure of a solid breaks down. Particles move further apart, overcoming the intermolecular attractions between them. This requires energy — and that energy is stored as increased potential energy.

Because the kinetic energy of the particles does not change during a change of state, the temperature remains constant throughout. This is why a graph of temperature against time for a substance being heated shows a flat plateau at the melting and boiling points.

On the graph above, the two flat sections correspond to melting (first plateau) and boiling (second plateau). During these plateaux, energy input is increasing the potential energy of the particle ensemble — the average separation between particles is growing as intermolecular bonds are broken or weakened.

5. Specific Latent Heat and Energy Calculations

The energy needed to change the state of a substance at constant temperature is calculated using the specific latent heat. There are two types: specific latent heat of fusion ($L_f$) for melting or freezing, and specific latent heat of vaporisation ($L_v$) for boiling or condensing.

Notice that $Q = mL$ contains no temperature change term — that is deliberate. The temperature is not changing during a change of state, so $\Delta\theta = 0$ and the $mc\Delta\theta$ equation gives zero. The latent heat equation handles the energy that goes into potential energy instead.

Worked example: Calculating energy for a change of state

Calculate the energy required to completely melt 0.50 kg of ice at 0 °C. The specific latent heat of fusion of water is $3.34 \times 10^5$ J kg⁻¹. [2 marks]

1. $Q = mL$ — [M1] This is a change of state at constant temperature, so use the latent heat equation, not $mc\Delta\theta$.
2. $Q = 0.50 \times 3.34 \times 10^5$ — Mass = 0.50 kg, $L_f = 3.34 \times 10^5$ J kg⁻¹.
3. $Q = 0.50 \times 334\,000 = 167\,000 \text{ J}$ — Working: $0.50 \times 334\,000 = 167\,000$. Expanding: $5 \times 334\,000 \div 10 = 1\,670\,000 \div 10 = 167\,000$.
4. $Q = 167\,000 \text{ J} = 1.67 \times 10^5 \text{ J}$ — [A1] Answer with correct unit. The temperature of the ice remains at 0 °C throughout — all energy goes into breaking the lattice structure (increasing potential energy).

Answer: $Q = 1.67 \times 10^5 \text{ J}$

A more demanding exam question will combine both equations — for example, heating a substance from room temperature up to its boiling point, then boiling it completely. You handle this in two separate stages: first $Q_1 = mc\Delta\theta$ for the temperature rise, then $Q_2 = mL$ for the change of state, and finally add them: $Q_{\text{total}} = Q_1 + Q_2$.

Worked example: Two-stage energy calculation: heating then boiling

A 0.20 kg sample of water is heated from 25 °C to 100 °C, then completely vaporised at 100 °C. Calculate the total energy required. Specific heat capacity of water: 4200 J kg⁻¹ K⁻¹. Specific latent heat of vaporisation of water: $2.26 \times 10^6$ J kg⁻¹. [4 marks]

1. $Q_1 = mc\Delta\theta$ — [M1] Temperature is changing in stage 1, so use the specific heat capacity equation.
2. $\Delta\theta = 100 - 25 = 75 \text{ K}$ — Subtract initial from final temperature.
3. $Q_1 = 0.20 \times 4200 \times 75 = 840 \times 75 = 63\,000 \text{ J}$ — Working: $0.20 \times 4200 = 840$, then $840 \times 75 = 63\,000$. [A1]
4. $Q_2 = mL_v = 0.20 \times 2.26 \times 10^6$ — [M1] Temperature is constant during boiling, so use the latent heat equation.
5. $Q_2 = 0.20 \times 2\,260\,000 = 452\,000 \text{ J}$ — Working: $0.20 \times 2\,260\,000 = 452\,000$.
6. $Q_{\text{total}} = 63\,000 + 452\,000 = 515\,000 \text{ J}$ — [A1] Total energy = $5.15 \times 10^5$ J. Notice that vaporisation requires far more energy than heating — most of the energy goes into separating the liquid molecules completely (increasing potential energy).

Answer: $Q_{\text{total}} = 515\,000 \text{ J}$ (or $5.15 \times 10^5$ J)

6. Linking Internal Energy to Changes of State

Pulling the ideas together: when you heat a substance through a change of state, the internal energy of the system is still increasing — but the increase is entirely in the potential energy component, not the kinetic energy component. The first law of thermodynamics still holds: energy is transferred in by heating, so $\Delta U > 0$.

When the substance is between state changes and its temperature is rising, the internal energy increase is entirely in the kinetic energy component. So across a full heating curve, you see the two components of internal energy taking turns: kinetic energy rises on the sloped sections, potential energy rises on the flat plateaux.

7. Exam Technique and Common Mistakes

The most reliable way to pick up marks on this topic is to use precise language. Examiners are looking for specific vocabulary, and vague answers — even if conceptually correct — often miss the mark.

For extended-response questions asking you to explain changes in internal energy during heating, a full-mark answer will typically cover: the two components of internal energy (kinetic and potential), which component changes in each phase of heating, and an explicit link to the first law — energy transferred in by heating increases internal energy.