Calculate the energy needed to raise the temperature of 2.0\,kg of water by 30\,^C. The specific heat capacity of water is 4200\,J/kg\,^C. [2 marks]

OCR OCR 2018-style practice: A 0.50\,kg block of aluminium is heated from 20\,^C to 70\,^C. The specific heat capacity of aluminium is 900\,J/kg\,^C. Calculate the energy transferred to the block.

A P1 Calculate question using the given equation E = m c θ. Write down the values: m = 0.50\,kg, c = 900\,J/kg\,^C, and the temperature change θ = 70 - 20 = 50\,^C (1 mark for the temperature change). Substitute: E = 0.50 × 900 × 50 (1 mark). This gives E = 22\,500\,J, which is 22.5\,kJ (2 marks for the calculation and units). Markers reward the correct temperature change, correct substitution, and the answer with a sensible unit. A common error is to use 70 instead of the change of 50.

EnglandPhysicsSyllabus dot point

What is internal energy, and how do we calculate the energy needed to heat a substance or change its state?

Internal energy as the total kinetic and potential energy of the particles, specific heat capacity and the equation linking energy, mass and temperature change, and specific latent heat for changes of state.

A focused answer to OCR Gateway GCSE Physics A topic P1 on internal energy and specific heat, covering internal energy as the total kinetic and potential energy of particles, the specific heat capacity equation, the specific latent heat equation, and the specific heat capacity practical.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Internal energy
Specific heat capacity
The specific heat capacity practical
Specific latent heat
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What this topic is asking

OCR wants you to define internal energy, use the specific heat capacity equation to calculate the energy to change a substance's temperature, and use the specific latent heat equation for changes of state. This is part of topic P1.1 of the OCR Gateway Physics A (J249) specification and links to the P1 specific-heat-capacity practical.

Internal energy

Heating a substance transfers energy to its particles and so increases its internal energy. This extra energy does one of two things:

It increases the kinetic energy of the particles, which raises the temperature, or
It increases the potential energy of the particles, which changes the state (for example by breaking the bonds that hold a solid together) without changing the temperature.

This is why temperature stays constant during melting or boiling: the supplied energy goes into potential energy to separate the particles, not into making them move faster.

Specific heat capacity

A material with a high specific heat capacity, such as water ( $c \approx 4200\,\text{J/kg}\,^{\circ}\text{C}$ ), needs a lot of energy to warm up and releases a lot when it cools, so it is good for storing and transporting thermal energy. Metals have much lower specific heat capacities, so they heat up and cool down quickly.

When rearranging, remember $\Delta\theta$ is the change in temperature (final minus initial), not the final temperature. Because the temperature scale interval is the same in $^{\circ}\text{C}$ and kelvin, a change in $^{\circ}\text{C}$ equals the change in kelvin.

The specific heat capacity practical

In the P1 practical you measure the specific heat capacity of a solid (often a metal block) or a liquid. An electrical heater is placed in the material, and a joulemeter or the equation $E = P t$ (power times time) gives the energy supplied. A thermometer measures the temperature rise, and the mass is measured on a balance. Rearranging gives $c = \dfrac{\Delta E}{m \Delta\theta}$ .

The measured value is usually a little too high because some energy is transferred to the surroundings rather than to the block; insulating the block reduces this systematic error and improves the result.

Specific latent heat

Because changes of state happen at constant temperature, the latent heat energy increases the particles' potential energy as bonds are broken (melting and boiling) or releases it as bonds form (freezing and condensing). The specific latent heat of vaporisation is usually much larger than that of fusion, because boiling separates the particles completely.

Energy to heat then melt ice

A $0.20\,\text{kg}$ block of ice at $0\,^{\circ}\text{C}$ is melted at $0\,^{\circ}\text{C}$ . The specific latent heat of fusion of ice is $3.3 \times 10^{5}\,\text{J/kg}$ . Calculate the energy needed to melt the ice.

step 1: Choose the right equation

The ice changes state at constant temperature, so use the latent heat equation $E = m L$ (not the specific heat equation, because the temperature does not change).

step 2: Write down the values

$m = 0.20\,\text{kg}$ and $L = 3.3 \times 10^{5}\,\text{J/kg}$ .

step 3: Substitute and calculate

$E = m L = 0.20 \times 3.3 \times 10^{5} = 6.6 \times 10^{4}\,\text{J}$ .

step 4: State the answer

The energy needed is $6.6 \times 10^{4}\,\text{J}$ , or $66\,\text{kJ}$ . No temperature change occurs, so all of this energy goes into breaking the bonds that hold the ice together.

Try this

Q1. Calculate the energy needed to raise the temperature of $2.0\,\text{kg}$ of water by $30\,^{\circ}\text{C}$ . The specific heat capacity of water is $4200\,\text{J/kg}\,^{\circ}\text{C}$ . [2 marks]

Cue. $\Delta E = m c \Delta\theta = 2.0 \times 4200 \times 30 = 252\,000\,\text{J}$ .

Q2. Explain why the temperature does not change while a pure substance is melting. [2 marks]

Cue. The energy supplied increases the potential energy of the particles by breaking bonds, not their kinetic energy, so the temperature stays constant.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksA

0.50\,\text{kg}

block of aluminium is heated from

20\,^{\circ}\text{C}

70\,^{\circ}\text{C}

. The specific heat capacity of aluminium is

900\,\text{J/kg}\,^{\circ}\text{C}

. Calculate the energy transferred to the block.

Show worked answer →

A P1 Calculate question using the given equation $\Delta E = m c \Delta\theta$ . Write down the values: $m = 0.50\,\text{kg}$ , $c = 900\,\text{J/kg}\,^{\circ}\text{C}$ , and the temperature change $\Delta\theta = 70 - 20 = 50\,^{\circ}\text{C}$ (1 mark for the temperature change). Substitute: $\Delta E = 0.50 \times 900 \times 50$ (1 mark). This gives $\Delta E = 22\,500\,\text{J}$ , which is $22.5\,\text{kJ}$ (2 marks for the calculation and units). Markers reward the correct temperature change, correct substitution, and the answer with a sensible unit. A common error is to use $70$ instead of the change of $50$ .

OCR 20223 marksExplain what is meant by the internal energy of a substance, and describe what happens to the internal energy when a solid is heated until it melts at its melting point.

Show worked answer →

A P1 Explain question worth three marks. Internal energy is the total kinetic and potential energy of all the particles in a substance (1 mark): the kinetic energy comes from the motion of the particles and the potential energy from their positions and the forces between them. When a solid is heated up to its melting point the particles gain kinetic energy and the temperature rises (1 mark). At the melting point the temperature stays constant while the solid melts, because the energy supplied increases the potential energy of the particles as it breaks the bonds holding them in fixed positions, not their kinetic energy (1 mark). Markers reward internal energy as total kinetic plus potential energy and the idea that during melting the energy raises potential energy at constant temperature.

Related dot points

Sources & how we know this

OCR GCSE (9-1) Physics A (Gateway Science) J249 specification — OCR (2016)