Reversed Heat Engines: Principles and Uses Revision Notes | A Level AQA Physics

1. Running an Engine Backwards

A heat engine takes thermal energy from a hot source, converts some of it into useful work, and dumps the rest into a cold sink. Now ask yourself: what happens if you reverse that process entirely? Instead of extracting work from a temperature difference, you supply work to create a temperature difference. That is the core idea behind a reversed heat engine.

Think of it like pushing water uphill. Normally water flows downhill on its own, releasing energy. To push it uphill you must put energy in. Thermal energy naturally flows from hot to cold — a reversed heat engine uses work to push it the other way, from cold to hot.

This single idea underpins two everyday devices: the refrigerator and the heat pump. They share the same operating principle but serve opposite purposes, and distinguishing between them is exactly what AQA examiners test.

2. Energy Flow in a Reversed Heat Engine

Picture a simple energy flow diagram. In a forward heat engine, thermal energy $Q_H$ flows in from the hot reservoir, work $W$ flows out, and waste heat $Q_C$ flows into the cold reservoir. In a reversed heat engine, the arrows on $W$ and $Q_C$ flip: work $W$ is supplied into the device, thermal energy $Q_C$ is extracted from the cold reservoir, and a larger quantity of thermal energy $Q_H$ is delivered to the hot reservoir.

This relationship is non-negotiable — it follows directly from conservation of energy. Notice that $Q_H$ is always greater than $W$ alone, because the device also harvests $Q_C$ from the cold side. That is why reversed heat engines can deliver more thermal energy to a space than the electrical energy they consume.

3. Refrigerators — Keeping Things Cold

A refrigerator's purpose is to maintain a cold space at a temperature below its surroundings. Work is supplied (by an electric motor, for instance) to extract thermal energy $Q_C$ from the cold interior and reject a larger quantity $Q_H$ to the warm kitchen. The cold space is the useful side here — you want to keep it cold.

The useful output of a refrigerator is $Q_C$: the thermal energy removed from the cold reservoir. You are paying for the work $W$, and you want as much cooling as possible per joule of work spent.

A domestic refrigerator might have a $\text{COP}_{\text{ref}}$ of around 2 to 4, meaning it removes 2–4 J of thermal energy from the food compartment for every 1 J of electrical energy consumed. The remaining energy, plus the work input, is rejected as heat at the back of the appliance — which is why the back of a fridge feels warm.

4. Heat Pumps — Delivering Warmth Efficiently

A heat pump uses the same physical mechanism as a refrigerator, but the goal is reversed: you want to deliver thermal energy to a warm space, not remove it from a cold one. A ground-source heat pump, for example, extracts thermal energy from the cool ground outside a building and delivers a larger quantity of thermal energy into the building's heating system.

Here the useful output is $Q_H$ — the heat delivered to the hot reservoir (the building). You are still paying for work $W$, but you benefit from both the work energy and the harvested environmental energy $Q_C$.

This is the key advantage of a heat pump over a simple electric heater. An electric heater converts 1 J of electrical energy into exactly 1 J of thermal energy — its COP is 1 by definition. A heat pump with a COP of 3 delivers 3 J of thermal energy for every 1 J of electrical energy consumed, because it harvests 2 J from the cold environment for free.

5. Comparing Refrigerators and Heat Pumps

Both devices are reversed heat engines operating on the same energy flow: work in, heat extracted from cold side, heat rejected to hot side. The difference is entirely about which side you care about. Recognising this distinction is a common exam demand.

Notice the mathematical link between the two COPs. Since $Q_H = Q_C + W$, dividing through by $W$ gives $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} + 1$. The heat pump COP is always exactly 1 more than the refrigerator COP for the same device operating under the same conditions.

6. Worked Calculation — Applying the Principles

Exam questions on reversed heat engines almost always give you two of the three quantities ($Q_H$, $Q_C$, $W$) and ask you to find the third, then calculate a COP. The step-by-step below models the method you should use every time.

Worked example: Reversed Heat Engine — Finding COP and Energy Values

A heat pump delivers 4500 J of thermal energy to a building. The work input to the pump is 1500 J. Calculate: (a) the thermal energy extracted from the cold reservoir, and (b) the coefficient of performance of the heat pump. [4 marks]

1. $Q_H = 4500 \text{ J}, \quad W = 1500 \text{ J}$ — Write down what you are given before touching any formula. This prevents substitution errors. [M1]
2. $Q_H = Q_C + W \Rightarrow Q_C = Q_H - W = 4500 - 1500 = 3000 \text{ J}$ — Rearrange $Q_H = Q_C + W$ to isolate $Q_C$. The cold reservoir supplies 3000 J — this is the environmental energy harvested. [A1]
3. $\text{COP}_{\text{HP}} = \dfrac{Q_H}{W}$ — The useful output of a heat pump is $Q_H$, so that goes on top. [M1]
4. $\text{COP}_{\text{HP}} = \dfrac{4500}{1500} = 3.0$ — 4500 divided by 1500: $4500 / 1500 = 3.0$. COP is dimensionless — no units. [A1]
5. $\text{COP}_{\text{HP}} = 3.0 > 1$ ✓ and $Q_H = Q_C + W = 3000 + 1500 = 4500 \text{ J}$ ✓ — A heat pump COP must always exceed 1. Energy conservation checks out. If your COP were less than 1 for a heat pump, recheck which quantity you placed in the numerator.

Answer: (a) $Q_C = 3000$ J; (b) $\text{COP}_{\text{HP}} = 3.0$

7. Practical Context and Uses

AQA requires you to know the basic principles and uses of heat pumps and refrigerators — detailed knowledge of the working fluid cycle or specific components is not required. What you do need is the ability to describe what each device does, explain why it is thermodynamically advantageous, and apply the energy equations.

Heat pumps are used in domestic and commercial heating. A ground-source heat pump extracts thermal energy from soil or groundwater (which stays at a relatively stable temperature year-round) and delivers it to a building's central heating system. Because the COP is typically 3–4, a household can receive 3–4 kJ of heating for every 1 kJ of electrical energy paid for.

Refrigerators and freezers maintain food at low temperatures by continuously extracting thermal energy from the cold interior and rejecting it to the kitchen. Air-conditioning systems operate on the same principle: they act as refrigerators for a room, extracting thermal energy from the indoor air and rejecting it outside.

8. Exam Technique and Common Mistakes

Reversed heat engine questions are reliable mark-earners if you approach them methodically. The most common source of lost marks is confusing which quantity is the useful output — and therefore which quantity goes in the numerator of the COP expression.

For 'explain' or 'evaluate' style questions, examiners award marks for linking the physics to the context. Saying 'the COP is 3' earns fewer marks than saying 'the COP is 3, meaning 3 J of thermal energy is delivered to the building for every 1 J of electrical energy consumed, because the device also harvests 2 J from the cold ground'.

Mastering reversed heat engines comes down to two things: keeping the energy conservation equation $Q_H = Q_C + W$ at the front of your mind, and always being clear about which side of the device is the 'useful' side before writing a COP expression — that single distinction separates the refrigerator from the heat pump at every level of exam question.