Calculate the energy to raise 2.0 kg of water by 5.0 K (c = 4200 J kg^-1 K^-1). [2 marks]

EnglandPhysicsSyllabus dot point

How much energy does it take to warm something up or to melt it?

Internal energy, the distinction between temperature change and change of state, specific heat capacity and specific latent heat, and continuous flow and method of mixtures experiments.

A focused answer to AQA A-Level Physics 3.6.2.1, covering internal energy, specific heat capacity, specific latent heat of fusion and vaporisation, and the energy required to change temperature or state.

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 dot point is asking
Internal energy
Temperature change: specific heat capacity
Change of state: specific latent heat
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What this dot point is asking

AQA specification point 3.6.2.1 wants you to describe internal energy as the sum of randomly distributed kinetic and potential energies of molecules, distinguish a temperature change from a change of state, and use specific heat capacity $c$ and specific latent heat $l$ in calculations.

Internal energy

The word "randomly" is important: it distinguishes internal energy from the ordered kinetic energy of a body moving as a whole. A flying brick has ordered kinetic energy but its internal energy is set by the random jiggling of its molecules.

Temperature change: specific heat capacity

Water has an unusually high specific heat capacity ( $4200 \text{ J kg}^{-1} \text{ K}^{-1}$ ), which is why it is used as a coolant and why coastal climates are mild: a lot of energy is needed to change its temperature. Note that a temperature change in degrees Celsius is numerically equal to the change in kelvin, so $\Delta\theta$ can be in either.

Change of state: specific latent heat

During a change of state the energy goes into increasing molecular potential energy (separating molecules and breaking bonds), not into kinetic energy, so the temperature stays constant. This is why a mixture of ice and water stays at $0 \text{ }^{\circ}\text{C}$ until all the ice has melted.

Heating then warming melted ice

How much energy turns $0.50 \text{ kg}$ of ice at $0 \text{ }^{\circ}\text{C}$ into water at $20 \text{ }^{\circ}\text{C}$ ? Take $l_{\text{fusion}} = 3.34 \times 10^5 \text{ J kg}^{-1}$ and $c_{\text{water}} = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$ .

step 1: Identify the two stages

The ice must first melt at constant temperature, then the resulting water must be warmed from $0 \text{ }^{\circ}\text{C}$ to $20 \text{ }^{\circ}\text{C}$ .

step 2: Energy to melt the ice

$Q_1 = ml = (0.50)(3.34 \times 10^5) = 1.67 \times 10^5 \text{ J}$ .

step 3: Energy to warm the water

$Q_2 = mc\Delta\theta = (0.50)(4200)(20) = 4.2 \times 10^4 \text{ J}$ .

step 4: Add the stages

$Q = Q_1 + Q_2 = 1.67 \times 10^5 + 4.2 \times 10^4 = 2.1 \times 10^5 \text{ J}$ .

Try this

Q1. Define specific heat capacity. [2 marks]

Cue. The energy needed to raise the temperature of $1 \text{ kg}$ of a substance by $1 \text{ K}$ .

Q2. Why does temperature stay constant during boiling? [2 marks]

Cue. Energy goes into increasing molecular potential energy by separating molecules, not into kinetic energy.

Q3. Calculate the energy to raise $2.0 \text{ kg}$ of water by $5.0 \text{ K}$ ( $c = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$ ). [2 marks]

Cue. $Q = mc\Delta\theta = 2.0 \times 4200 \times 5.0 = 4.2 \times 10^4 \text{ J}$ .

Exam-style practice questions

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

AQA 20184 marksAn electric heater of power

50 \text{ W}

is used to warm

0.30 \text{ kg}

of water in an insulated container. Assuming no energy is lost, calculate the time taken to raise the temperature of the water from

18 \text{ }^{\circ}\text{C}

42 \text{ }^{\circ}\text{C}

. The specific heat capacity of water is

4200 \text{ J kg}^{-1} \text{ K}^{-1}

Show worked answer →

Find the energy needed using $Q = mc\Delta\theta$ , with $\Delta\theta = 42 - 18 = 24 \text{ K}$ (a change of $24 \text{ }^{\circ}\text{C}$ equals a change of $24 \text{ K}$ ).

$Q = (0.30)(4200)(24) = 3.02 \times 10^4 \text{ J}$ .

The heater supplies energy at $P = 50 \text{ W}$ , so the time is $t = \dfrac{Q}{P} = \dfrac{3.02 \times 10^4}{50} = 6.0 \times 10^2 \text{ s}$ (about $10$ minutes).

Markers reward $Q = mc\Delta\theta$ , the temperature difference, and dividing the energy by the power to find the time.

AQA 20224 marksExplain, in terms of the molecules, why the temperature of a pure substance stays constant while it melts, even though energy is being supplied.

Show worked answer →

Energy supplied during melting is used to break the bonds (overcome the forces) holding the molecules in the rigid solid lattice, increasing the molecular potential energy.

The mean kinetic energy of the molecules does not increase, and since temperature is a measure of the mean molecular kinetic energy, the temperature stays constant.

Once all the bonds are broken and the substance is fully liquid, further energy raises the kinetic energy again and the temperature rises.

Markers reward energy going into potential energy (bond breaking), no change in kinetic energy, and linking constant temperature to constant mean kinetic energy.

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Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)