Learn7408Paper 3B · Option: Engineering physicsThermodynamics and enginesHeat engines and heat pumps

Thermodynamics and engines · Subtopic 2 of 2

Heat Engines and Heat Pumps

The first law $Q=\Delta U+W$ tracks energy in a gas. Isothermal changes keep $pV$ constant; adiabatic changes ( $Q=0$ ) keep $pV^\gamma$ constant. Work done is the area under a p–V curve, so the area enclosed by a cycle is the net work per cycle. No engine can turn all heat into work: maximum efficiency is $\frac{T_H-T_C}{T_H}$ . Heat pumps and refrigerators run engines backwards.

AQA 7408 Paper 3B≈ 50 minRevision 1 · updated 2026-10-27

Section 1 of 14

Learning objectives

Apply $Q=\Delta U+W$ to gas processes.
Use $pV=$ constant, $pV^\gamma=$ constant and $W=p\Delta V$ .
Find work from p–V diagrams and indicator diagrams.
Calculate efficiencies and coefficients of performance.

Pause and think

A car engine burns fuel at over $2000$ °C, yet only about a third of the energy reaches the wheels. Where does the rest go?

Section 2 of 14

Core idea

Start hereWhere the heat goes

A car engine burns fuel at thousands of degrees, but most of that energy leaves through the exhaust and the radiator. This isn't bad engineering: the second law of thermodynamics says any engine must throw some heat away.

The big idea

$Q=\Delta U+W$ . Work is the area under a p–V curve; a cycle's enclosed area is net work. Efficiency $\le\frac{T_H-T_C}{T_H}$ .

1. Gas processes

Adiabatic compression heats a gas

$Q=0$
Adiabatic: no heat exchanged.
$\Delta U=-W$
First law.
$W<0\Rightarrow\Delta U>0$
Work is done on the gas when compressed.

T

rises: diesel engines ignite fuel this way.

Worked exampleHalving the volume

Air ( $\gamma=1.4$ ) at $1.0\times10^5$ Pa compressed adiabatically to half its volume.

Goal: Final pressure.
Strategy: $pV^\gamma=$ constant.

$p_2=p_1\left(\frac{V_1}{V_2}\right)^\gamma$
Rearrange.
$=1.0\times10^5\times2^{1.4}=2.6\times10^5\text{ Pa}$
Calculate.

Answer

2.6\times10^5

Pa (more than the isothermal

2.0\times10^5

✓ CheckTemperature also rises, so pressure rises more ✓.

2. Engines and heat pumps

Worked exampleRefrigerator

Removes $2000$ J from inside using $500$ J of work.

Goal: COP and heat released.
Strategy: Energy conservation.

$COP_{ref}=\frac{2000}{500}=4.0$
Useful heat removed per joule of work.
$Q_H=2000+500=2500\text{ J}$
All energy ends up in the kitchen.

AnswerCOP

4.0

;

2500

J released.

✓ CheckA fridge with its door open warms the kitchen ✓.

Energy flows

Drawing diagram…

Engines and heat pumps are the same machine run in opposite directions.

Common misconceptionKelvin

\eta_{\max}

uses kelvin:

800

K to

300

K gives

0.625

, not

\frac{527-27}{527}

Key takeawayIndicator diagrams Real engines lose work to friction, valves and incomplete combustion, so their loops are smaller and rounded.

Pause and think

Why is a heat pump's COP higher on a mild day?

Summary

$Q=\Delta U+W$ .
$pV$ / $pV^\gamma$ constant.
Area = work.
$\eta_{\max}=\frac{T_H-T_C}{T_H}$ .

Adiabatic: No heat transfer.
Isothermal: Constant temperature.
Coefficient of performance: Useful heat moved per unit of work.

HistoryCarnot. Sadi Carnot, a young French engineer, showed in 1824 that an engine's maximum efficiency depends only on its hot and cold temperatures, before the first law was even known.

Fun factHeat pumps beat heaters. A heat pump with COP

3

delivers

3

kJ of heat for every

1

kJ of electricity, because it moves heat from outside rather than making it.

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