Carnot Efficiency Calculator

The Carnot cycle defines the theoretical upper limit of efficiency for any heat engine operating between two temperature reservoirs. Named after French engineer Sadi Carnot, who described it in 1824, the cycle consists of two isothermal steps and two adiabatic steps. No real engine can convert heat to work more efficiently than a Carnot engine running between the same two temperatures, because the second law of thermodynamics forbids it. The formula is simple: efficiency eta equals 1 minus T_cold divided by T_hot, where both temperatures are measured in kelvin. An efficiency of 40 percent means that 40 percent of the heat drawn from the hot reservoir is converted to useful work, and the remaining 60 percent is rejected to the cold reservoir. This calculator has two modes. The first calculates efficiency from the two reservoir temperatures. The second solves for the required cold or hot reservoir temperature that would produce a given efficiency. You also get the heat rejected to the cold reservoir for a given heat input. Understanding Carnot efficiency helps engineers set realistic benchmarks for power stations, refrigeration cycles and heat pumps. Enter temperatures in kelvin; the tool includes a Celsius-to-kelvin converter for convenience. Results are the ideal theoretical maximum and will exceed any real engine's performance.

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K
K
%
J
40.00%
Carnot efficiency
Work output W400.00 J
Heat rejected Q_cold600.00 J
COP (refrigerator)1.50

Temperatures must be in kelvin (K = C + 273.15). This is the theoretical maximum; real engines achieve lower efficiencies due to friction and irreversibilities.

How it works

The Carnot efficiency formula is eta = 1 - T_cold / T_hot. To find the required cold reservoir temperature for a given efficiency: T_cold = T_hot x (1 minus eta). To find the required hot reservoir: T_hot = T_cold / (1 minus eta). With heat input Q_in, useful work W = eta x Q_in and heat rejected Q_cold = Q_in minus W. The Carnot coefficient of performance for a refrigerator running in reverse is COP = T_cold / (T_hot minus T_cold).

Worked example

A heat engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Carnot efficiency: eta = 1 minus 300/500 = 1 minus 0.6 = 40.00%. With a heat input of 1,000 J, useful work W = 0.40 x 1000 = 400.00 J and heat rejected to the cold reservoir = 1000 minus 400 = 600.00 J. These match the defaults pre-filled above.

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