What is specific heat capacity?

The specific heat capacity c is the energy needed to raise the temperature of 1\,kg of a substance by 1\,K, measured in J kg^-1\,K^-1. Water has an unusually high value of 4200\,J kg^-1\,K^-1, which is why it is used as a coolant and why coastal climates are mild.

How much energy is needed to heat 0.50\,kg of water from 20^C to 80^C? Take c = 4200\,J kg^-1K^-1. [2 marks]

WJEC WJEC 2020-style practice: An electric heater of power 2.0\,kW is used to convert 0.30\,kg of ice at 0^C into water at 20^C. The specific latent heat of fusion of ice is 3.34 × 10^5\,J kg^-1 and the specific heat capacity of water is 4200\,J kg^-1\,K^-1. Calculate the total energy required and the minimum time, assuming no losses.

Energy has two stages: melting the ice at constant temperature, then heating the water. Melting: E_1 = mL = 0.30 × 3.34 × 10^5 = 1.00 × 10^5\,J. Heating the water from 0^C to 20^C: E_2 = mcθ = 0.30 × 4200 × 20 = 2.52 × 10^4\,J. Total energy: E = E_1 + E_2 = 1.00 × 10^5 + 2.52 × 10^4 = 1.25 × 10^5\,J. Time: t = EP = 1.25 × 10^52.0 × 10^3 = 63\,s. Markers reward separating melting from heating, using mL and mcθ, and the time from t = E/P.

WalesPhysicsSyllabus dot point

What is internal energy, and how do specific heat capacity and latent heat describe heating and changes of state?

Internal energy and temperature, specific heat capacity, specific latent heat, and thermal equilibrium.

A focused answer to WJEC A-Level Physics Unit 3 thermal physics, covering internal energy and its link to temperature, specific heat capacity, specific latent heat for changes of state, and thermal equilibrium.

Generated by Claude Opus 4.811 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

WJEC wants you to define internal energy and distinguish it from temperature, use specific heat capacity and specific latent heat, and explain thermal equilibrium. The specific heat and latent heat calculations are reliable exam marks, and the conceptual distinction between internal energy and temperature is a favourite explanation question.

The answer

Internal energy and temperature

A large cool object can hold more internal energy than a small hot one, because internal energy depends on the number of particles as well as the energy of each.

Specific heat capacity

The specific heat capacity $c$ is the energy needed to raise the temperature of $1\,\text{kg}$ of a substance by $1\,\text{K}$ , measured in $\text{J kg}^{-1}\,\text{K}^{-1}$ . Water has an unusually high value of $4200\,\text{J kg}^{-1}\,\text{K}^{-1}$ , which is why it is used as a coolant and why coastal climates are mild.

Specific latent heat

Heating and melting in one calculation

Find the energy needed to turn $0.20\,\text{kg}$ of ice at $-10^{\circ}\text{C}$ into ice at $0^{\circ}\text{C}$ and then melt it. The specific heat capacity of ice is $2100\,\text{J kg}^{-1}\,\text{K}^{-1}$ and its latent heat of fusion is $3.34 \times 10^{5}\,\text{J kg}^{-1}$ .

step Warm the ice to its melting point

$E_1 = mc\Delta\theta = 0.20 \times 2100 \times 10 = 4200\,\text{J}$ .

step Melt the ice at constant temperature

$E_2 = mL = 0.20 \times 3.34 \times 10^{5} = 6.68 \times 10^{4}\,\text{J}$ .

step Add the two stages

$E = 4200 + 66\,800 = 7.1 \times 10^{4}\,\text{J}$ . Note the melting stage takes far more energy than the warming stage, which is typical of latent heat.

Thermal equilibrium

Two bodies in thermal contact are in thermal equilibrium when they reach the same temperature, so there is no net flow of thermal energy between them.

Examples in context

Example 1. Sweating to cool down. When sweat evaporates from your skin it absorbs its latent heat of vaporisation from your body, removing a large amount of energy at constant temperature. Because water's latent heat of vaporisation is high, even a thin film of sweat carries away a lot of heat, which is why evaporation is such an effective cooling mechanism on a hot day.

Example 2. Storage heaters. Night-storage heaters contain dense ceramic bricks with a high specific heat capacity. They are heated overnight on cheap electricity, storing a large amount of internal energy via $E = mc\Delta\theta$ , then release it slowly through the day. The large mass and high specific heat let them bank enough energy to warm a room for hours.

Try this

Q1. How much energy is needed to heat $0.50\,\text{kg}$ of water from $20^\circ\text{C}$ to $80^\circ\text{C}$ ? Take $c = 4200\,\text{J kg}^{-1}\text{K}^{-1}$ . [2 marks]

Cue. $E = mc\Delta\theta = 0.50\times4200\times60 = 1.26\times10^5\,\text{J}$ .

Q2. Explain why temperature stays constant while a solid melts. [2 marks]

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

Exam-style practice questions

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

WJEC 20206 marksAn electric heater of power

2.0\,\text{kW}

is used to convert

0.30\,\text{kg}

of ice at

0^{\circ}\text{C}

into water at

20^{\circ}\text{C}

. The specific latent heat of fusion of ice is

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

and the specific heat capacity of water is

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

. Calculate the total energy required and the minimum time, assuming no losses.

Show worked answer →

Energy has two stages: melting the ice at constant temperature, then heating the water.

Melting: $E_1 = mL = 0.30 \times 3.34 \times 10^{5} = 1.00 \times 10^{5}\,\text{J}$ .

Heating the water from $0^{\circ}\text{C}$ to $20^{\circ}\text{C}$ : $E_2 = mc\Delta\theta = 0.30 \times 4200 \times 20 = 2.52 \times 10^{4}\,\text{J}$ .

Total energy: $E = E_1 + E_2 = 1.00 \times 10^{5} + 2.52 \times 10^{4} = 1.25 \times 10^{5}\,\text{J}$ .

Time: $t = \dfrac{E}{P} = \dfrac{1.25 \times 10^{5}}{2.0 \times 10^{3}} = 63\,\text{s}$ .

Markers reward separating melting from heating, using $mL$ and $mc\Delta\theta$ , and the time from $t = E/P$ .

WJEC 20183 marksExplain, in terms of particles, the difference between internal energy and temperature, and why a swimming pool can store more internal energy than a cup of boiling water.

Show worked answer →

Internal energy is the total of the random kinetic and potential energies of all the particles in a body. Temperature is a measure only of the average kinetic energy per particle, not the total.

A cup of boiling water has a higher temperature (its particles have more average kinetic energy), but a swimming pool contains vastly more particles. The sum of all those particle energies, the internal energy, can therefore be far greater for the cool pool than for the hot cup, even though each pool particle carries less energy.

Markers reward defining internal energy as the total particle energy and temperature as the average per particle, and applying this to the large number of particles in the pool.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)