Reversed Heat Engines, Refrigerators and Heat Pumps

A heat engine takes heat in at a high temperature, converts some of it to work, and rejects the rest at a low temperature. Reverse the cycle — run all four processes in the opposite direction — and the device becomes a refrigerator or a heat pump: it takes in work and uses that work to pump heat from cold to hot, against the natural direction of heat flow. The Clausius statement of the second law (Lesson 4) explicitly permits this provided work is supplied. The performance of such reversed cycles is measured by the coefficient of performance (COP), which (unlike efficiency, which is bounded above by 1) can be substantially greater than 1 — typical domestic heat pumps deliver 3–4 J of heat to the home for every 1 J of electrical work supplied. This lesson develops the algebra of the reversed Carnot cycle, defines the COPs of refrigerators and heat pumps, derives their Carnot upper bounds, and outlines the practical vapour-compression cycle used in essentially every fridge, freezer, air conditioner and ground-source heat pump on the market.

Spec mapping (AQA 7408 §3.11.2 — Thermodynamics, option C): This lesson covers the reversed heat engine as a refrigerator or heat pump; the coefficient of performance of a refrigerator COP_ref = Q_C/W; the coefficient of performance of a heat pump COP_HP = Q_H/W; the Carnot upper bounds COP_ref ≤ T_C/(T_H − T_C) and COP_HP ≤ T_H/(T_H − T_C); the relationship COP_HP = COP_ref + 1; and a qualitative description of the practical vapour-compression refrigeration cycle (compressor, condenser, expansion valve, evaporator). (Refer to the official AQA specification document for exact wording.)

Synoptic links: Reversed cycles complete the engine-cycle picture started in Lesson 3 and use the same four-process algebra of Lesson 2. The COP bounds derive directly from the Carnot efficiency of Lesson 4 — refrigerator and heat pump are the second-law-respecting analogues of the second-law-respecting heat engine. The vapour-compression cycle introduces phase changes (latent heat) into the analysis, connecting back to AS thermal physics §3.6.2 on specific latent heat. Real-world relevance ranges from domestic appliances (every fridge, freezer and air-conditioner uses essentially the same cycle) to building energy systems (ground-source and air-source heat pumps for low-carbon heating) to grid-scale energy storage (some proposed thermal storage systems use heat pumps to upgrade waste heat).

The Reversed Cycle — From Engine to Pump

Take any forward (clockwise) heat-engine cycle on a pV diagram. Now traverse it in the opposite direction — anticlockwise. The same four processes occur, but each runs in reverse:

The isothermal expansion at T_H (heat absorbed at T_H) becomes isothermal compression at T_H (heat rejected at T_H).
The adiabatic expansion (gas cools) becomes adiabatic compression (gas warms).
The isothermal compression at T_C (heat rejected at T_C) becomes isothermal expansion at T_C (heat absorbed at T_C).
The adiabatic compression (gas warms) becomes adiabatic expansion (gas cools).

Energy bookkeeping reverses signs throughout: net work is now done on the gas (not by it), heat flows out at T_H to the hot reservoir, and heat flows in at T_C from the cold reservoir. The cycle has become a heat pump — it uses work input to move heat from cold to hot.

The same physical hardware is interpreted differently depending on which reservoir you care about:

A refrigerator is a heat pump optimised for the cold side — its purpose is to cool the contents of the fridge by extracting heat Q_C from inside.
An air-conditioner is a heat pump optimised for the cold side of the room you occupy — it cools the room by extracting heat and dumping it outside.
A heat pump (for heating) is the same hardware optimised for the hot side — its purpose is to deliver heat Q_H to a building by gathering it from the outside (cold reservoir).
An air-source heat pump uses outside air as the cold reservoir; a ground-source heat pump uses the ground (typically 8–12 °C year-round at a few metres depth) as the cold reservoir, which gives a higher T_C and so a higher COP than air-source in winter.

graph LR
    HR["Hot reservoir<br/>T_H (kitchen, outdoors)"] -->|"Q_H out"| C["Cycle<br/>(anticlockwise)"]
    C -->|"Q_C in"| CR["Cold reservoir<br/>T_C (fridge interior, ground)"]
    W["Work input<br/>W (electrical)"] -->|"into cycle"| C

    style HR fill:#e74c3c,color:#fff
    style CR fill:#3498db,color:#fff
    style W fill:#27ae60,color:#fff

Exam Tip: The Q-direction sign convention is critical here. In a refrigerator/heat pump, Q_C flows into the gas (from the cold reservoir), Q_H flows out of the gas (to the hot reservoir), and W is done on the gas. The first law on the closed cycle gives Q_C + W = Q_H, or equivalently Q_H = Q_C + W. This says: the heat dumped at the hot side equals the heat taken from the cold side plus the work input.

Coefficient of Performance

Because the purpose of a refrigerator or heat pump is different from a heat engine, the figure of merit is also different. Engines convert heat into work (efficiency η = W/Q_in, bounded between 0 and 1). Refrigerators and heat pumps use work to move heat (the figure of merit is what useful heat you get for each joule of work, which can be — and usually is — greater than 1).

Refrigerator COP

COP_ref = Q_C / W

The "purpose" of a refrigerator is to remove heat from the cold reservoir, so the useful output is Q_C and the cost is W. COP_ref is dimensionless. A typical domestic fridge has COP_ref ≈ 2–4.

Heat Pump COP

COP_HP = Q_H / W

The "purpose" of a heat pump (for heating) is to deliver heat to the hot reservoir, so the useful output is Q_H and the cost is W. COP_HP > 1 always — because Q_H = Q_C + W ≥ W. A typical domestic heat pump for space heating has COP_HP ≈ 3–4 under favourable conditions (mild outdoor temperatures, modest indoor target).

The Relationship COP_HP = COP_ref + 1

By the first law on the closed cycle: Q_H = Q_C + W.

Divide by W: Q_H/W = Q_C/W + 1, i.e.

COP_HP = COP_ref + 1

This is exact and follows directly from energy conservation. If you have an air-conditioner with COP_ref = 3.0, the same hardware acts as a heat pump with COP_HP = 4.0 (provided you reverse the direction of heat flow). This is why reversible-cycle "heat pumps" double up as air-conditioners with the flip of a valve — the same compressor, the same refrigerant, the same heat exchangers.

Carnot Bounds on COP

The Carnot cycle is reversible and represents the maximum performance for any cycle operating between two given reservoirs. For a reversed Carnot cycle between T_H and T_C (in kelvin):

ΔS_hot = +Q_H/T_H (heat flows into the hot reservoir reversibly at T_H).
ΔS_cold = −Q_C/T_C (heat flows out of the cold reservoir reversibly at T_C; gas's S decreases).
Total ΔS around the cycle = 0 (reversible) ⇒ Q_H/T_H = Q_C/T_C ⇒ Q_H/Q_C = T_H/T_C.

From the first law W = Q_H − Q_C:

COP_ref, Carnot = T_C / (T_H − T_C)

COP_HP, Carnot = T_H / (T_H − T_C)

These are upper bounds for any real device. Two key qualitative features:

COP increases as ΔT = T_H − T_C decreases. Pumping heat across a small temperature difference is easy; pumping across a large temperature difference requires more work per joule moved.
COP_HP > COP_ref always (as it must from Q_H > Q_C).

graph TD
    A["Carnot reversed cycle"] --> B["ΔS_cycle = 0"]
    B --> C["Q_H/T_H = Q_C/T_C"]
    C --> D["Q_H/Q_C = T_H/T_C"]
    D --> E["COP_ref = T_C/(T_H − T_C)"]
    D --> F["COP_HP = T_H/(T_H − T_C)"]

    style E fill:#27ae60,color:#fff
    style F fill:#27ae60,color:#fff

Worked Example — Carnot Bounds at Two Temperature Spreads

Question: Compare a heat pump operating (a) between T_C = 5 °C and T_H = 20 °C (mild winter, modest indoor target), with (b) between T_C = −10 °C and T_H = 50 °C (cold winter, hot water tank).

Solution:

Convert to kelvin: (a) T_C = 278 K, T_H = 293 K, ΔT = 15 K. (b) T_C = 263 K, T_H = 323 K, ΔT = 60 K.

(a) COP_HP, Carnot = 293/15 = 19.5. Very high.

(b) COP_HP, Carnot = 323/60 = 5.4. Much lower — and a real heat pump would achieve perhaps 2.5–3.5 here.

This is why heat pumps work best for space heating with low-temperature radiators or underfloor heating (small ΔT ⇒ high COP) and worst for high-temperature applications like domestic hot water at 60 °C (large ΔT ⇒ low COP). Modern heat-pump installations therefore typically pair a heat pump with underfloor heating and add an immersion heater or auxiliary boost for hot water.

Worked Example — Domestic Heat Pump

Question: A domestic air-source heat pump is to heat a house to maintain an indoor temperature of 20 °C when the outdoor temperature is 5 °C. The house loses heat to the outside at a steady rate of 4.5 kW. Assume the heat pump achieves 50% of its Carnot COP.

(a) Calculate the Carnot COP for heating. (b) Calculate the real COP_HP at 50% of Carnot. (c) Calculate the electrical power input required. (d) Compare the running cost with a direct-electric heater (which converts electricity to heat 1:1).

Solution:

(a) T_C = 278 K, T_H = 293 K. COP_HP, Carnot = 293/(293−278) = 293/15 = 19.5.

(b) Real COP_HP = 0.5 × 19.5 = 9.77.

(Note: 50% of Carnot is typical for very well-designed heat pumps at favourable conditions. Real domestic heat pumps in UK winter typically deliver real COPs of 2.5–4 under more challenging conditions — colder outdoors, indoor radiators rather than underfloor — and this textbook idealisation gives the headline upper-end figure.)

(d) For the same 4.5 kW of heating, a direct-electric heater consumes 4.5 kW of electricity — about 10 times the heat pump's input. At typical UK electricity prices the heat pump's running cost is roughly one-tenth that of a resistive electric heater, which is the headline economic argument for heat-pump installation.

In practice, heat pumps lose much of this advantage in deep winter (cold outdoor T pushes ΔT up, COP down) and during defrost cycles (extracting heat from cold air causes condensation/frost on the outdoor coil, requiring periodic defrost cycles that consume energy). Seasonal performance factors (SPF) for UK air-source heat pumps typically come out around 2.5–3.5 for whole-year averaged performance — still substantially better than direct electric heating, but not as flattering as the textbook idealisation.

Exam Tip: When a question gives a "fraction of Carnot" multiplier, multiply the Carnot COP by that fraction — it does not mean a different value of W or Q. The fraction represents real-world irreversibilities reducing performance below the ideal bound.

The Vapour-Compression Cycle

Real refrigerators, freezers, air conditioners and heat pumps use a clever trick to make the four-stage reversed cycle practical: the working fluid (called a refrigerant) is chosen so that its phase changes happen at convenient temperatures and pressures. The cycle becomes:

Reversed Heat Engines, Refrigerators and Heat Pumps

Reversed Heat Engines, Refrigerators and Heat Pumps

The Reversed Cycle — From Engine to Pump

Coefficient of Performance

Refrigerator COP

Heat Pump COP

The Relationship COP_HP = COP_ref + 1

Carnot Bounds on COP

Worked Example — Carnot Bounds at Two Temperature Spreads

Worked Example — Domestic Heat Pump

The Vapour-Compression Cycle

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