Convert 25\ ^C to kelvin. [1 mark]

How much energy raises the temperature of 2.0\ kg of aluminium (c = 900\ J kg^-1\ K^-1) by 30\ K? [2 marks]

EnglandPhysicsSyllabus dot point

What is temperature, and how much energy does it take to heat a substance or change its state?

Thermal physics: temperature and internal energy, the kelvin scale and absolute zero, specific heat capacity and specific latent heat, and changes of state.

A focused answer to the OCR H556 thermal physics content, covering temperature as a measure of average kinetic energy, internal energy, the kelvin scale and absolute zero, specific heat capacity and specific latent heat, and the energy changes during changes of state.

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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Quick answer

Temperature is a measure of the average kinetic energy of the particles in a substance, measured on the absolute (kelvin) scale where absolute zero ( $0\ \text{K}$ , $-273\ ^{\circ}\text{C}$ ) is the temperature at which particles have minimum internal energy. Internal energy is the sum of the randomly distributed kinetic and potential energies of all the particles. To raise the temperature of a mass $m$ by $\Delta T$ takes energy $E = mc\Delta T$ , where $c$ is the specific heat capacity. To change state without a temperature change takes energy $E = mL$ , where $L$ is the specific latent heat; during a change of state the energy breaks or forms intermolecular bonds rather than changing the kinetic energy, so the temperature stays constant.

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What this dot point is asking

OCR wants you to define temperature and internal energy, use the kelvin scale and the idea of absolute zero, define and use specific heat capacity and specific latent heat, and explain the energy changes that occur during changes of state.

The answer

Temperature and internal energy

To convert between scales, $T/\text{K} = \theta/^{\circ}\text{C} + 273$ . A temperature change of one kelvin is the same size as a change of one degree Celsius.

Absolute zero and the kelvin scale

Specific heat capacity

You can measure $c$ with an electrical method, supplying a known electrical energy $E = VIt$ and measuring the temperature rise; heat losses make the measured value an overestimate unless allowed for.

Specific latent heat and changes of state

During a change of state the temperature stays constant because the supplied energy goes into changing the potential energy of the particles (breaking or forming bonds), not their kinetic energy. This is why a heating curve has flat sections at the melting and boiling points.

Mixing to find a final temperature

A $0.20\ \text{kg}$ block of metal at $90\ ^{\circ}\text{C}$ is dropped into $0.30\ \text{kg}$ of water at $15\ ^{\circ}\text{C}$ . The metal has $c = 900\ \text{J kg}^{-1}\ \text{K}^{-1}$ and water $c = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ . Find the final temperature, ignoring heat losses.

step 1: Energy balance

Energy lost by the metal equals energy gained by the water: $m_m c_m (90 - \theta) = m_w c_w (\theta - 15)$ .

step 2: Substitute

$0.20 \times 900 \times (90 - \theta) = 0.30 \times 4200 \times (\theta - 15)$ , so $180(90 - \theta) = 1260(\theta - 15)$ .

step 3: Solve

$16\,200 - 180\theta = 1260\theta - 18\,900$ , so $35\,100 = 1440\theta$ and $\theta = 24\ ^{\circ}\text{C}$ (to 2 significant figures).

Examples in context

The high specific heat capacity and latent heat of water make it a superb coolant in car engines and power stations, and let sweating cool the body efficiently as the latent heat of vaporisation is drawn from the skin. Storage heaters and concrete buildings exploit a high heat capacity to store and release energy slowly. Steam burns are more severe than hot-water burns because the steam also releases its large latent heat of vaporisation as it condenses on the skin.

Try this

Q1. Convert $25\ ^{\circ}\text{C}$ to kelvin. [1 mark]

Cue. $T = 25 + 273 = 298\ \text{K}$ .

Q2. How much energy raises the temperature of $2.0\ \text{kg}$ of aluminium ( $c = 900\ \text{J kg}^{-1}\ \text{K}^{-1}$ ) by $30\ \text{K}$ ? [2 marks]

Cue. $E = mc\Delta T = 2.0 \times 900 \times 30 = 5.4 \times 10^{4}\ \text{J}$ .

Q3. Explain why the temperature does not change while a pure substance is melting. [2 marks]

Cue. The energy supplied breaks intermolecular bonds (increasing potential energy) rather than increasing the kinetic energy of the particles, so the temperature stays constant until melting is complete.

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 20194 marksAn electric heater of power

2.0\ \text{kW}

heats

0.50\ \text{kg}

of water from

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

100\ ^{\circ}\text{C}

. The specific heat capacity of water is

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

. Calculate the minimum time this takes, and state one reason the actual time is longer.

Show worked answer →

Energy needed: $E = mc\Delta T = 0.50 \times 4200 \times (100 - 20) = 0.50 \times 4200 \times 80 = 1.68 \times 10^{5}\ \text{J}$ .

Time: $t = \frac{E}{P} = \frac{1.68 \times 10^{5}}{2000} = 84\ \text{s}$ .

The actual time is longer because some energy is lost to the surroundings (the container and the air), so not all the electrical energy raises the water temperature.

Markers reward $E = mc\Delta T$ , the time $84\ \text{s}$ , and a valid heat-loss reason.

OCR 20224 marksA

0.030\ \text{kg}

ice cube at

0\ ^{\circ}\text{C}

is added to a drink and completely melts. The specific latent heat of fusion of ice is

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

. Calculate the energy absorbed by the ice in melting, and explain why the temperature stays constant during melting.

Show worked answer →

Energy to melt: $E = mL = 0.030 \times 3.3 \times 10^{5} = 9.9 \times 10^{3}\ \text{J}$ , about $9900\ \text{J}$ .

The temperature stays constant because the energy supplied goes into breaking the bonds between molecules (increasing the potential component of internal energy) rather than increasing their kinetic energy, so there is no temperature change until all the ice has melted.

Markers reward $E = mL$ , the value about $9.9 \times 10^{3}\ \text{J}$ , and an explanation that the energy breaks bonds rather than raising the temperature.

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Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)