Heat Engines, Efficiency & the Carnot Limit

You’ll be able to

Compute the efficiency of a heat engine from heat input and work output
Compute the maximum theoretical efficiency from reservoir temperatures
Explain why no engine can reach 100% efficiency

An engine takes heat from hot, dumps some to cold, keeps the difference

A heat engine absorbs heat Q_H from a hot reservoir, converts part of it to work W, and rejects the rest as Q_C to a cold reservoir. Energy conservation requires W = Q_H − Q_C. The rejected heat is not a design flaw to be engineered away — the second law says it is unavoidable, because an engine needs a temperature difference to run and that means somewhere colder to dump heat into.

Efficiency

e = W / Q_H = 1 − Q_C/Q_H · maximum (Carnot): e_max = 1 − T_C/T_H, with temperatures in KELVIN

Kelvin is not optional. Using Celsius in the Carnot formula produces a nonsense answer, sometimes above 1.

The Carnot ceiling

The Carnot efficiency 1 − T_C/T_H is the maximum any engine operating between those two temperatures can achieve, regardless of design, working substance or engineering skill. It is reached only by a reversible engine, which requires infinitely slow operation and therefore produces no useful power. Real engines fall well short. Two consequences worth stating: efficiency rises with a hotter hot reservoir or a colder cold one, and 100% would require T_C = 0 K, which is unattainable.

Refrigerators run the cycle backward

A refrigerator or heat pump uses work to move heat from cold to hot — the direction heat does not go by itself. On a PV diagram it is a counterclockwise cycle. Its performance is measured not by efficiency but by coefficient of performance, the heat moved per unit of work, which can exceed 1 without violating anything: you are relocating heat rather than creating energy.

Worked example

An engine absorbs 2400 J per cycle from a reservoir at 500 K and rejects heat to a reservoir at 300 K, producing 600 J of work. Find its actual efficiency, the Carnot maximum, and the heat rejected.

1.Actual efficiency: e = W/Q_H = 600/2400 = 0.25, or 25%.
2.Carnot maximum: e_max = 1 − T_C/T_H = 1 − 300/500 = 1 − 0.60 = 0.40, or 40%.
3.Heat rejected: Q_C = Q_H − W = 2400 − 600 = 1800 J.
4.Check: 1 − Q_C/Q_H = 1 − 1800/2400 = 0.25. Consistent.

Answer: Actual efficiency 25%, Carnot maximum 40%, and 1800 J rejected per cycle. The engine achieves well under the theoretical ceiling, which is normal — the ceiling requires reversible operation.

Watch out

Convert to Kelvin before using the Carnot formula. With 227 °C and 27 °C, using Celsius gives 1 − 27/227 = 0.88, while the correct 500 K and 300 K give 0.40. The error is large and always in the flattering direction.

Checkpoint

An engine operates between 600 K and 300 K. Its maximum possible efficiency is:

Checkpoint

An engine absorbs 1000 J and rejects 700 J per cycle. Its efficiency is:

Checkpoint

A refrigerator can have a coefficient of performance greater than 1 because it:

On the exam

If a question gives you both the actual and the Carnot efficiency, it usually wants you to comment on the gap. Say that the shortfall reflects irreversibilities — friction, finite-rate heat transfer, turbulence — rather than a violation of anything.

Answer the 3 checkpoints as you read.