How to use it
- Choose the temperature unit: °C, K or °F.
- Enter the hot reservoir temperature (the boiler, the combustion gas, the room a heat pump warms).
- Enter the cold reservoir temperature (the cooling water, the outside air, the inside of a fridge).
- Optionally enter how much heat the engine takes in, in joules (or any energy unit — the work comes out in the same unit).
How it works
Both temperatures are converted to kelvin: K = °C + 273.15, K = (°F − 32) × 5/9 + 273.15. Then:
- Carnot efficiency η = 1 − Tc ÷ Th
- Refrigerator COP = Tc ÷ (Th − Tc): heat removed from the cold side per unit of work
- Heat pump COP = Th ÷ (Th − Tc): heat delivered to the warm side per unit of work
- Maximum work W = η × Qh, and the rest, Qh − W, goes to the cold reservoir
Worked example
A steam cycle takes heat at 500 °C and rejects it at 25 °C. In kelvin that is 773.15 K and 298.15 K, a difference of 475 K.
- η = 475 ÷ 773.15 = 61.44 %
- Of 1000 J of heat, at most 614.37 J become work and 385.63 J leave with the cooling water.
- A refrigerator between the same temperatures could have COP 298.15 ÷ 475 = 0.628, a heat pump 773.15 ÷ 475 = 1.628.
Tips and common mistakes
- Small temperature differences give big COPs. A heat pump moving heat from 0 °C outside to 20 °C inside has an ideal COP of 293.15 ÷ 20 = 14.66; real units reach a fraction of that.
- Efficiency is not COP. Efficiency is below 100 % for engines; COP of fridges and heat pumps is often above 1, and that is not a contradiction.
- Use the reservoir temperatures, not the fuel’s heating value. The limit depends only on the two temperatures, not on the fuel or the working fluid.