η = 1 − T_c/T_h — Carnot Efficiency Limit

Carnot efficiency — the maximum possible efficiency of any heat engine — with a worked example.

\eta = 1 - \frac{T_c}{T_h}

Maximum efficiency equals one minus the cold reservoir temperature over the hot reservoir temperature.

What it means

No heat engine can beat Carnot: the best possible efficiency depends only on the two temperatures, never on the working fluid or design. The bigger the temperature gap, the more heat you can convert — which is why power plants run as hot as materials allow.

Variables

Symbol	Name	Unit	Meaning
$η$	Efficiency	dimensionless (×100 for %)	Maximum fraction of heat convertible to work
$T_c$	Cold temperature	K	Cold reservoir temperature (absolute)
$T_h$	Hot temperature	K	Hot reservoir temperature (absolute)

Assumptions

Reversible (Carnot) cycle — the theoretical best
Temperatures in kelvin (absolute scale)
Steady hot and cold reservoirs

Worked example

Problem

A heat engine operates between 600 K and 300 K. What is its maximum theoretical efficiency?

Given:

T_h = 600 K
T_c = 300 K

Write the formula: $\eta = 1 - T_c/T_h$
Substitute: $\eta = 1 - 300/600 = 1 - 0.5$
Calculate: $\eta = 0.5 = 50\%$

50%

Even a perfect engine wastes half its heat when operating between 600 K and 300 K. Real engines do worse — this ceiling is set by the second law of thermodynamics, not by engineering.