Find the energy needed to heat 2.0\ kg of water from 15\ ^C to 100\ ^C (c = 4200\ J kg^-1\ K^-1). [2 marks]

WJEC Eduqas Eduqas 2021-style practice: Calculate the energy required to convert 0.20\ kg of ice at 0\ ^C into water at 20\ ^C. Take the specific latent heat of fusion of ice as 3.34 × 10^5\ J kg^-1 and the specific heat capacity of water as 4200\ J kg^-1\ K^-1.

Energy to melt the ice at 0\ ^C (no temperature change during melting): Q_1 = mL = 0.20 × 3.34 × 10^5 = 6.68 × 10^4\ J. Energy to warm the resulting water from 0\ ^C to 20\ ^C: Q_2 = mcθ = 0.20 × 4200 × 20 = 1.68 × 10^4\ J. Total energy: Q = Q_1 + Q_2 = 6.68 × 10^4 + 1.68 × 10^4 = 8.36 × 10^4\ J, about 84\ kJ. Markers reward the melting energy Q = mL, the warming energy Q = mcθ, and the total about 84\ kJ.

EnglandPhysicsSyllabus dot point

How do we quantify internal energy, temperature, heat capacity and the energy needed for changes of state?

Thermal physics: internal energy and the kinetic model, temperature and thermal equilibrium, specific heat capacity, and specific latent heat for changes of state.

A focused answer to the Eduqas A-Level Physics Component 1 thermal physics content, covering internal energy and the kinetic model, temperature and thermal equilibrium, specific heat capacity with Q = mc(delta theta), and specific latent heat for changes of state.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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Quick answer

The internal energy of a body is the sum of the random kinetic and potential energies of its molecules. Temperature is a measure of the average kinetic energy of the molecules; two bodies in thermal equilibrium are at the same temperature with no net energy flow. The specific heat capacity $c$ is the energy needed to raise the temperature of $1\ \text{kg}$ of a substance by $1\ \text{K}$ , so $Q = mc\Delta\theta$ . The specific latent heat $L$ is the energy needed to change the state of $1\ \text{kg}$ of a substance with no temperature change, so $Q = mL$ (fusion for melting, vaporisation for boiling). During a change of state the energy goes into breaking intermolecular bonds (potential energy), not into kinetic energy, so the temperature stays constant.

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What this dot point is asking

Eduqas wants you to define internal energy in terms of the kinetic model, explain temperature and thermal equilibrium, use specific heat capacity with $Q = mc\Delta\theta$ , and use specific latent heat with $Q = mL$ for changes of state.

The answer

Internal energy and the kinetic model

Temperature and thermal equilibrium

Specific heat capacity

Specific latent heat

Mixing hot and cold water

$0.30\ \text{kg}$ of water at $80\ ^\circ\text{C}$ is mixed with $0.50\ \text{kg}$ of water at $20\ ^\circ\text{C}$ . Find the final temperature, assuming no heat is lost. Take $c = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ .

step 1: Energy conservation

The energy lost by the hot water equals the energy gained by the cold water: $m_1 c(80 - \theta) = m_2 c(\theta - 20)$ . The $c$ cancels.

step 2: Substitute the masses

$0.30(80 - \theta) = 0.50(\theta - 20)$ , so $24 - 0.30\theta = 0.50\theta - 10$ .

step 3: Solve for the final temperature

$34 = 0.80\theta$ , so $\theta = \dfrac{34}{0.80} = 42.5\ ^\circ\text{C}$ , about $43\ ^\circ\text{C}$ .

Examples in context

Specific heat capacity governs the design of heating and cooling systems, engine coolants and storage heaters, and explains why the sea warms and cools more slowly than the land. Latent heat is exploited in refrigeration and air conditioning (a refrigerant absorbs latent heat as it evaporates), in steam heating, and in the cooling effect of sweating. The high latent heat of vaporisation of water makes scalds from steam far more dangerous than from boiling water.

Try this

Q1. Define the specific heat capacity of a substance. [1 mark]

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

Q2. Find the energy needed to heat $2.0\ \text{kg}$ of water from $15\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$ ( $c = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ ). [2 marks]

Cue. $Q = mc\Delta\theta = 2.0 \times 4200 \times 85 = 7.1 \times 10^{5}\ \text{J}$ .

Q3. State why the temperature stays constant while ice melts. [2 marks]

Cue. The energy supplied breaks intermolecular bonds (raises potential energy) rather than increasing molecular kinetic energy, so the temperature does not change.

Exam-style practice questions

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

Eduqas 20184 marksAn electric heater supplies

36\ \text{kJ}

0.50\ \text{kg}

of water, raising its temperature from

18\ ^\circ\text{C}

34\ ^\circ\text{C}

. Calculate the experimental value of the specific heat capacity of water, and suggest why it differs from the accepted value of

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

Show worked answer →

Temperature rise: $\Delta\theta = 34 - 18 = 16\ \text{K}$ (a Celsius interval equals a kelvin interval).

Rearrange $Q = mc\Delta\theta$ for $c$ : $c = \dfrac{Q}{m\Delta\theta} = \dfrac{36\,000}{0.50 \times 16} = \dfrac{36\,000}{8.0} = 4500\ \text{J kg}^{-1}\ \text{K}^{-1}$ .

The experimental value is higher because some of the supplied energy heats the container and is lost to the surroundings rather than going solely into the water, so the energy assigned to the water is overestimated. Markers reward $\Delta\theta = 16\ \text{K}$ , $c = \frac{Q}{m\Delta\theta} \approx 4500\ \text{J kg}^{-1}\ \text{K}^{-1}$ , and a valid reason (heat loss to surroundings or container).

Eduqas 20214 marksCalculate the energy required to convert

0.20\ \text{kg}

of ice at

0\ ^\circ\text{C}

into water at

20\ ^\circ\text{C}

. Take the specific latent heat of fusion of ice as

3.34 \times 10^{5}\ \text{J kg}^{-1}

and the specific heat capacity of water as

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

Show worked answer →

Energy to melt the ice at $0\ ^\circ\text{C}$ (no temperature change during melting): $Q_1 = mL = 0.20 \times 3.34 \times 10^{5} = 6.68 \times 10^{4}\ \text{J}$ .

Energy to warm the resulting water from $0\ ^\circ\text{C}$ to $20\ ^\circ\text{C}$ : $Q_2 = mc\Delta\theta = 0.20 \times 4200 \times 20 = 1.68 \times 10^{4}\ \text{J}$ .

Total energy: $Q = Q_1 + Q_2 = 6.68 \times 10^{4} + 1.68 \times 10^{4} = 8.36 \times 10^{4}\ \text{J}$ , about $84\ \text{kJ}$ .

Markers reward the melting energy $Q = mL$ , the warming energy $Q = mc\Delta\theta$ , and the total about $84\ \text{kJ}$ .

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)