The Carnot Engine & Refrigerators
The best engine nature allows, and the same machine run backwards.
Lesson at a glanceShow the key points
- Carnot cycle (ideal gas): isothermal expansion at T₁ (Q₁ in), adiabatic expansion to T₂, isothermal compression at T₂ (Q₂ out), adiabatic compression.
- On a T–S diagram it is a rectangle: Q₁/T₁ = Q₂/T₂, so η = 1 − T₂/T₁ (26.7).
- Carnot's theorem: all reversible engines between T₁ and T₂ have this efficiency; none can exceed it.
- Refrigerator: takes Q₂ from cold, work W done on it, delivers Q₁ = Q₂ + W to hot; minimum W = Q₂(T₁/T₂ − 1).
- Clausius: no cyclic device can only transfer heat from a body to a hotter one.
In 1824 Sadi Carnot imagined the perfect heat engine. It can’t be built, but it sets the best efficiency any engine between two temperatures can reach.
The Carnot cycle
An ideal gas fills a cylinder with a diathermic (heat-conducting) base, adiabatic walls and an adiabatic piston. There are two large bodies at fixed temperatures, hot (T₁) and cold (T₂). Starting compressed at T₁ (state a):
- a → b: on the hot body, expand isothermally at T₁; the gas does work and takes in heat Q₁.
- b → c: on an insulating stand, expand adiabatically; the gas does work and cools to T₂.
- c → d: on the cold body, compress isothermally at T₂; work is done on the gas and it rejects heat Q₂.
- d → a: on the insulating stand, compress adiabatically back to T₁.
With a frictionless piston moved very slowly, the whole cycle is reversible.
Its efficiency
On a temperature–entropy diagram the cycle is a rectangle. Along a → b the entropy rises from S₁ to S₂ while Q₁ flows in at T₁; the adiabats keep S fixed; along c → d it falls back while Q₂ leaves at T₂:
What is the highest efficiency of an engine working between 500 K and 300 K? If it takes in 2000 J per cycle, how much work can it do at most?
Carnot’s theorem
It follows from the second law, and caps every heat engine at η = 1 − T₂/T₁.
Refrigerators (heat pumps)
Run the engine backwards: take heat Q₂ from a cold body, have work W done on the working substance, and deliver Q₁ = Q₂ + W to a hot body. Heat moves from cold to hot. Done reversibly between single temperatures, this is a Carnot refrigerator, which needs the least possible work:
A freezer keeps food at 270 K in a 300 K kitchen. At least how much work removes 1000 J of heat from inside?
- 1Carnot cycle: isothermal expansion (T₁, Q₁ in), adiabatic expansion, isothermal compression (T₂, Q₂ out), adiabatic compression.
- 2Q₂/Q₁ = T₂/T₁; η = 1 − T₂/T₁.
- 3Carnot's theorem: no engine between two temperatures beats a reversible one.
- 4Refrigerator: W = Q₂(T₁/T₂ − 1) at minimum. Clausius: heat can't flow cold → hot by itself.
Check yourself
Keep exploring
Hand-picked videos, simulations and readings that explain this idea from a different angle.