The Thermodynamics of Refrigeration — Coefficient of Performance Explained
A refrigerator does something that looks, at first glance, like it should be impossible: it moves heat from a cold space to a warmer one, uphill against the natural direction heat always flows on its own. The second law of thermodynamics does not forbid this — it only says it cannot happen for free. Every refrigerator, air conditioner and heat pump is a machine built around exactly this idea, and the number that tells you how well it does the job is called the coefficient of performance, or COP. This article derives the theoretical limit on that number, works it out for a real fridge and a real heat pump, and explains the one thing about COP that trips up almost every student meeting it for the first time.
The formula: coefficient of performance
A refrigerator is a heat engine run in reverse. Instead of taking in heat at a high temperature and delivering work, it takes in work and uses it to pump heat from a cold reservoir to a hot one. For the ideal, fully reversible case — a reversed Carnot cycle — the coefficient of performance depends only on the two absolute temperatures involved:
Both quantities are ratios of "useful heat" to "work you had to pay for" — a refrigerator's useful output is the heat removed from the cold space, while a heat pump's useful output is the heat delivered to the warm space. Because the same compressor cycle does both jobs at once (Qh = Qc + W by the first law), the two COPs are always exactly 1 apart.
What each term means
| Term | Meaning | Unit |
|---|---|---|
| Qc | Heat absorbed from the cold reservoir per cycle | J (or W as a rate) |
| Qh | Heat rejected to the hot reservoir per cycle | J (or W as a rate) |
| W | Work input, usually the compressor's electrical work | J (or W as a rate) |
| Tc, Th | Absolute temperatures of the cold and hot reservoirs — always kelvin | K |
| COP | Ratio of useful heat transfer to work input — not an efficiency, and routinely greater than 1 | dimensionless |
Worked example 1 — a domestic refrigerator
A refrigerator keeps its interior at 4 °C while the kitchen around it is at 25 °C. What is the maximum possible COP?
Convert to kelvin: Tc = 4 + 273 = 277 K, Th = 25 + 273 = 298 K
COPrefrigerator = Tc ÷ (Th − Tc) = 277 ÷ (298 − 277) = 277 ÷ 21
COPrefrigerator = 13.19
Read this as: for every 1 joule of compressor work, the ideal cycle removes 13.19 joules of heat from the interior. That is the theoretical ceiling — no real fridge reaches it, because real compressors, heat exchangers and expansion valves all introduce irreversibilities that a Carnot cycle assumes away.
Worked example 2 — a heat pump heating a house in winter
The same hardware, run to heat rather than cool, keeps a house interior at 20 °C while the outside air is at −5 °C.
Tc = −5 + 273 = 268 K, Th = 20 + 273 = 293 K
COPheat pump = Th ÷ (Th − Tc) = 293 ÷ (293 − 268) = 293 ÷ 25
COPheat pump = 11.72
Cross-check with the refrigerator formula on the same two temperatures:
COPrefrigerator = 268 ÷ 25 = 10.72, and 10.72 + 1 = 11.72 ✓ — matches exactly, as the
theory demands.
This is the reason heat pumps are pitched as more efficient than direct electric resistance heating: a resistance heater converts 1 joule of electricity into at most 1 joule of heat (COP = 1), while a heat pump moves several joules of heat for the same electrical input by relocating heat that already exists in the outside air rather than manufacturing new heat.
Where this is actually used
A real vapour-compression refrigeration cycle carries out four steps: a compressor raises the pressure and temperature of a refrigerant vapour; the hot vapour rejects heat and condenses to liquid in the condenser (this is Qh); the liquid expands rapidly through a throttling valve, cooling sharply as it does so; and the cold, low-pressure liquid absorbs heat and evaporates in the evaporator (this is Qc), before returning to the compressor to repeat the cycle. Every domestic fridge, split air conditioner, industrial cold-storage plant and heat pump water heater runs some version of this same four-step loop; only the refrigerant, the scale and the engineering detail change.
Real machines never reach the Carnot COP calculated above. Actual domestic refrigerator and air-conditioner COPs typically sit somewhere in the range of 2 to 4, not because the physics is wrong but because compressors are not perfectly reversible, heat exchangers need a finite temperature difference to transfer heat at a useful rate, and the throttling step itself is inherently irreversible — it destroys the availability of the refrigerant to do work without extracting any. The Carnot value is not a design target that clever engineering will eventually reach; it is a hard upper bound set by the second law, and every real irreversibility can only push the actual COP below it.
Common mistakes that cost marks
- Using Celsius instead of kelvin. Because the formula is a ratio involving a difference of temperatures in the denominator but an absolute temperature in the numerator, plugging in 4 and 25 instead of 277 and 298 gives a completely wrong answer, not just a slightly wrong one.
- Treating COP as an efficiency and expecting it to be less than 1. A heat engine's thermal efficiency is bounded above by 1 (you can never get out more work than the heat you put in). A refrigerator's COP is a different kind of ratio — heat moved divided by work paid for — and there is no such bound. A COP of 13 does not violate energy conservation; it simply means most of the heat delivered came from the cold reservoir, not from the work input, and Qh = Qc + W still balances exactly.
- Mixing up COPrefrigerator and COPheat pump. They describe the same physical cycle but ask what you are paying for versus what you actually want — cooling in one case, heating in the other — and they are never equal.
- Assuming the Carnot COP is achievable. It is the reversible-cycle ceiling. State it as the theoretical maximum unless the question explicitly gives you a real-cycle or actual COP to work with.
Exam relevance
| Exam | Typical use |
|---|---|
| IIT-JAM / CUET-PG Physical Chemistry | Reversed heat engines, second-law bounds, entropy of the universe for a refrigeration cycle |
| GATE (Chemistry / allied engineering) | Carnot COP numericals, heat pump vs refrigerator COP relationships |
| CSIR-NET Physical Chemistry | Second-law statements (Clausius form), reversible cycle limits, entropy arguments |
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