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

A fixed mass of ideal gas at 1.0 × 10^5\ Pa occupies 0.020\ m^3. It is compressed isothermally to 0.012\ m^3. Find the new pressure.

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What is temperature really, and how do ideal gases behave?

Internal energy and temperature, specific heat capacity and specific latent heat, the gas laws and the ideal gas equation, and the kinetic theory of gases.

A CCEA A-Level Physics answer on internal energy and temperature, specific heat capacity and specific latent heat, the gas laws and the ideal gas equation, and how the kinetic theory links temperature to molecular kinetic energy.

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

CCEA wants you to explain internal energy and temperature, use specific heat capacity and specific latent heat, apply the gas laws and the ideal gas equation, and use the kinetic theory to link temperature to molecular kinetic energy. Multi-stage heating or change-of-state calculations and an ideal gas calculation appear regularly.

The answer

Internal energy and temperature

Heat and internal energy are different: heat is energy transferred because of a temperature difference, whereas internal energy is a property the body possesses at any instant.

Heat capacity and latent heat

A standard CCEA experiment finds $c$ using an electrical heater of known power in an insulated block: $Pt = mc\Delta\theta$ , plotting temperature against time and allowing for heat losses to the surroundings.

The ideal gas and kinetic theory

The gas laws (Boyle's, Charles's and the pressure law) combine into the ideal gas equation:

pV = nRT,

where

n

is the number of moles and

R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}

the molar gas constant.

Worked example: mean speed of a gas molecule

Root-mean-square speed of nitrogen at room temperature

Nitrogen molecules ( $m = 4.7 \times 10^{-26}\ \text{kg}$ ) are at $300\ \text{K}$ . Using $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ , find the mean kinetic energy of a molecule and its root-mean-square speed.

step Find the mean kinetic energy

\overline{E_k} = \tfrac{3}{2}kT = \tfrac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\ \text{J}.

step Relate energy to the mean square speed

\tfrac{1}{2}m\overline{c^2} = 6.21 \times 10^{-21} \Rightarrow \overline{c^2} = \frac{2 \times 6.21 \times 10^{-21}}{4.7 \times 10^{-26}} = 2.64 \times 10^{5}\ \text{m}^2\,\text{s}^{-2}.

step Take the square root

c_{\text{rms}} = \sqrt{2.64 \times 10^{5}} = 5.1 \times 10^{2}\ \text{m s}^{-1}.

The molecules move at around

510\ \text{m s}^{-1}

, comparable to the speed of sound in air, which is no coincidence.

Examples in context

Example 1. A hot-water bottle. Water has an unusually high specific heat capacity ( $4200\ \text{J kg}^{-1}\,\text{K}^{-1}$ ), so a litre at $80\ ^\circ\text{C}$ stores a lot of energy and releases it slowly as it cools, keeping a bed warm for hours. The same property makes water an effective coolant in car engines and power stations.

Example 2. A car tyre on a motorway. As a tyre warms during a long drive, the gas inside is heated at roughly constant volume, so by the pressure law its pressure rises in proportion to its absolute temperature. A tyre inflated cold to $2.0 \times 10^{5}\ \text{Pa}$ at $290\ \text{K}$ rises to about $2.1 \times 10^{5}\ \text{Pa}$ when the gas reaches $310\ \text{K}$ , which is why pressures are quoted "cold".

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. $Q = mc\Delta\theta = 0.50 \times 4200 \times 60 = 126\,000\ \text{J}$ .

Q2. State what the absolute temperature of a gas measures, according to kinetic theory. [1 mark]

Cue. The mean (translational) kinetic energy of its molecules.

Q3. A fixed mass of ideal gas at $1.0 \times 10^{5}\ \text{Pa}$ occupies $0.020\ \text{m}^3$ . It is compressed isothermally to $0.012\ \text{m}^3$ . Find the new pressure. [2 marks]

Cue. Boyle's law: $p_2 = \frac{p_1 V_1}{V_2} = \frac{1.0 \times 10^{5} \times 0.020}{0.012} = 1.7 \times 10^{5}\ \text{Pa}$ .

Exam-style practice questions

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

CCEA 20196 marksAn electric kettle rated at 2.2 kW contains 0.80 kg of water at 18 degrees C. Calculate the time taken to bring the water to its boiling point, and then the additional time needed to boil away 0.10 kg of the water. Take c for water as 4200 and the specific latent heat of vaporisation as 2.3 times 10 to the 6.

Show worked answer →

Energy to heat the water to 100 degrees C:

$Q_1 = mc\Delta\theta = 0.80 \times 4200 \times (100 - 18) = 0.80 \times 4200 \times 82 = 2.75 \times 10^{5}$ J.

Time at 2200 W: $t_1 = Q_1 / P = 2.75 \times 10^{5} / 2200 = 125$ s.

Energy to boil away 0.10 kg:

$Q_2 = mL = 0.10 \times 2.3 \times 10^{6} = 2.30 \times 10^{5}$ J.

Time: $t_2 = Q_2 / P = 2.30 \times 10^{5} / 2200 = 105$ s.

Markers reward both energies (heating and the change of state), the use of power equals energy over time, and the recognition that the boiling stage uses latent heat at constant temperature. In practice the kettle is not 100 percent efficient, so the real times are a little longer.

CCEA 20215 marksA sealed cylinder contains 0.20 mol of an ideal gas at a temperature of 27 degrees C and a pressure of 1.0 times 10 to the 5 Pa. Calculate the volume of the gas. The gas is then heated at constant volume until its pressure doubles. Determine the new temperature in degrees C. Take R as 8.31.

Show worked answer →

Convert the temperature to kelvin: $T = 27 + 273 = 300$ K.

Using the ideal gas equation $pV = nRT$ :

$V = \frac{nRT}{p} = \frac{0.20 \times 8.31 \times 300}{1.0 \times 10^{5}} = \frac{498.6}{1.0 \times 10^{5}} = 5.0 \times 10^{-3}$ m cubed.

At constant volume, pressure is proportional to absolute temperature (the pressure law), so doubling the pressure doubles the absolute temperature:

$T_2 = 2 \times 300 = 600$ K $= 327$ degrees C.

Markers reward conversion to kelvin, correct substitution into $pV = nRT$ , and the constant-volume reasoning giving 327 degrees C.

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Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)