Coefficient of Performance: Refrigerators and Heat Pumps Revision Notes | A Level AQA Physics

1. Why COP Matters: Getting More Than You Pay For

Here is a striking fact: a well-designed heat pump can deliver three or four times more heat energy to a room than the electrical energy it consumes. That sounds like it breaks the laws of thermodynamics — but it does not. The device is not creating energy; it is moving energy from a cold place to a warm one, and the work you supply pays for that transfer.

The coefficient of performance (COP) is the number that captures this idea precisely. It tells you how much useful output you get per joule of work put in. A higher COP means a more efficient device — and understanding how to calculate and interpret it is a core A-level engineering physics skill.

2. The Energy Bookkeeping: $Q_C$, $Q_H$, and $W$

Picture any refrigerator or heat pump as a pump sitting between two temperature reservoirs. On one side is the cold reservoir at temperature $T_C$; on the other is the hot reservoir at temperature $T_H$. The device extracts heat $Q_C$ from the cold side, receives work input $W$ from an external source (the compressor motor), and rejects heat $Q_H$ to the hot side.

Energy conservation — the first law of thermodynamics — demands that nothing is lost. Every joule that leaves the cold reservoir and every joule of work done on the system must end up in the hot reservoir. This gives the fundamental bookkeeping equation:

This single equation is the foundation for both COP formulae. Notice that $W = Q_H - Q_C$: the work you supply is exactly the difference between the heat dumped at the top and the heat collected at the bottom. Keep this in mind — it is the key to understanding why the two COP expressions look the way they do.

3. COP of a Refrigerator

A refrigerator's job is to keep the inside cold. Its useful output is therefore the heat removed from the cold space, $Q_C$. The cost is the work $W$ supplied to the compressor. The COP of a refrigerator is defined as the ratio of these two quantities:

The middle expression, $\frac{Q_C}{Q_H - Q_C}$, comes directly from substituting $W = Q_H - Q_C$ into the basic ratio. The rightmost expression, $\frac{T_C}{T_H - T_C}$, applies only to an ideal Carnot device — one operating reversibly between two fixed temperatures. Real refrigerators always have a lower COP than this ideal limit.

Notice what the temperature formula tells you about design. If $T_C$ is close to $T_H$ — meaning the temperature difference is small — the denominator $T_H - T_C$ is small and COP is large. A refrigerator working hard to maintain a very low temperature (large $T_H - T_C$) has a much lower COP. This is why industrial freezers are far less efficient than a standard kitchen fridge.

4. COP of a Heat Pump

A heat pump's purpose is the opposite of a refrigerator's: it heats a space. The useful output is therefore the heat delivered to the hot reservoir, $Q_H$. The cost is still the work $W$. Swapping the numerator from $Q_C$ to $Q_H$ gives the heat pump COP:

There is an elegant relationship between the two COP values. Because $Q_H = Q_C + W$, dividing both sides by $W$ gives $COP_{\text{hp}} = COP_{\text{ref}} + 1$. This means the heat pump COP is always greater than the refrigerator COP by exactly 1 when both devices operate between the same two temperatures. A heat pump COP is therefore always greater than 1 — you always get more heat out than the work you put in.

5. Worked Calculations

Exam questions on COP come in two flavours: those that give you heat transfer values, and those that give you temperatures. The worked examples below cover both. Work through each step yourself before revealing the next one.

Worked example: Calculating COP from heat transfer values

A refrigerator extracts 900 J of heat from its interior and rejects 1200 J to the room. Calculate (a) the work input $W$, (b) $COP_{\text{ref}}$, and (c) $COP_{\text{hp}}$ if the same device were used as a heat pump.

1. $Q_C = 900$ J, $Q_H = 1200$ J — Always list what you know before selecting a formula. [B1]
2. $W = Q_H - Q_C = 1200 - 900 = 300$ J — First-law energy balance: the work supplied equals the difference in heat transfers. [M1]
3. $COP_{\text{ref}} = \frac{Q_C}{W} = \frac{900}{300} = 3.0$ — Useful output for a refrigerator is $Q_C$. COP is dimensionless — no units. [M1, A1]
4. $COP_{\text{hp}} = \frac{Q_H}{W} = \frac{1200}{300} = 4.0$ — Useful output for a heat pump is $Q_H$. Notice $COP_{\text{hp}} = COP_{\text{ref}} + 1 = 3.0 + 1 = 4.0$. [A1]
5. $COP_{\text{hp}} - COP_{\text{ref}} = 4.0 - 3.0 = 1.0$ ✓ — The difference must always equal 1. If it does not, recheck your arithmetic.

Answer: (a) $W = 300$ J, (b) $COP_{\text{ref}} = 3.0$, (c) $COP_{\text{hp}} = 4.0$

Worked example: Calculating ideal COP from temperatures

A heat pump operates between an outdoor temperature of $-5\,^\circ$C and an indoor temperature of $20\,^\circ$C. Calculate the ideal $COP_{\text{hp}}$ and the ideal $COP_{\text{ref}}$ for this device.

1. $T_C = -5 + 273 = 268$ K $T_H = 20 + 273 = 293$ K — The temperature formulae require absolute (kelvin) temperatures. Forgetting this conversion is the most common error in COP questions. [M1]
2. $T_H - T_C = 293 - 268 = 25$ K — This denominator appears in both COP expressions. Calculate it once and reuse it.
3. $COP_{\text{hp}} = \frac{T_H}{T_H - T_C} = \frac{293}{25} = 11.72$ — Numerator is $T_H$ for a heat pump. [M1, A1]
4. $COP_{\text{ref}} = \frac{T_C}{T_H - T_C} = \frac{268}{25} = 10.72$ — Numerator is $T_C$ for a refrigerator. Alternatively, $COP_{\text{ref}} = COP_{\text{hp}} - 1 = 11.72 - 1 = 10.72$. [A1]
5. $11.72 - 10.72 = 1.00$ ✓ — Difference equals 1 as required. Both values are greater than 1, which is physically correct.

Answer: Ideal $COP_{\text{hp}} = 11.72$; ideal $COP_{\text{ref}} = 10.72$ (both to 4 significant figures)

6. Comparing Refrigerators and Heat Pumps

The same physical device — a vapour-compression cycle with a compressor, condenser, expansion valve, and evaporator — can function as either a refrigerator or a heat pump depending on which heat transfer you designate as the useful output. The table below summarises the key differences in how you define and calculate COP for each mode.

The minimum COP for a heat pump deserves attention. Because $Q_H = Q_C + W$ and $Q_C \geq 0$, we always have $Q_H \geq W$, so $COP_{\text{hp}} = Q_H / W \geq 1$. A heat pump can never deliver less heat than the work you put in — at worst it equals the work input, which is what a simple electric heater achieves. Any COP above 1 represents a genuine thermodynamic advantage over direct electrical heating.

7. How Temperature Affects COP

The temperature-based formulae reveal a powerful design insight. For both devices, the denominator is $T_H - T_C$. As this temperature difference increases, the denominator grows and COP falls. Conversely, a smaller temperature difference means a higher COP and a more efficient device.

This trend has real engineering consequences. A ground-source heat pump, which extracts heat from soil at a relatively stable temperature, maintains a smaller $T_H - T_C$ than an air-source heat pump on a cold winter day. The ground-source device therefore achieves a higher COP and lower running costs — a direct application of the temperature formula.

8. Exam Technique and Common Mistakes

COP questions appear regularly in AQA engineering physics papers and tend to reward students who are precise about which formula to use and which temperature scale to apply. The following guidance consolidates the most mark-winning habits and the traps that cost students marks most often.

A final point on exam language: when a question uses the command word explain, you must link cause to effect. For example, 'the COP decreases because the temperature difference $T_H - T_C$ increases, making the denominator larger and the ratio smaller' scores full marks. Stating only 'the COP decreases' without the reasoning scores nothing for explanation marks.