Carnot Efficiency Calculator

How Carnot Efficiency Is Calculated

The Carnot cycle defines the theoretical maximum possible efficiency of any heat engine operating between two temperature reservoirs. No real engine can exceed this limit — it sets the upper bound of thermodynamic performance.

Carnot Efficiency Formula: η = 1 − (T_cold / T_hot)

Where:

• η (eta) = efficiency (0 to 1, or 0% to 100%)
• T_cold = temperature of cold reservoir in Kelvin
• T_hot = temperature of hot reservoir in Kelvin
• Converting: K = °C + 273.15

Worked Example: A steam turbine operates with steam at 550°C and exhausts at 45°C:

• T_hot = 550 + 273.15 = 823.15 K
• T_cold = 45 + 273.15 = 318.15 K
• η = 1 − (318.15 / 823.15) = 1 − 0.3865 = 0.6135 = 61.35%

In practice, real steam turbines achieve 35–45% due to friction, heat losses, and irreversibilities.

Efficiency Reference by Engine Type:

• Car gasoline engine: 20–35%
• Diesel engine: 35–45%
• Combined-cycle gas turbine: 55–62%
• Nuclear steam plant: 30–35%
• Carnot limit for typical car: ~60%

Heat pumps and refrigerators run the cycle backwards

The same two reservoirs, driven in reverse, move heat instead of producing work. Performance is quoted as a coefficient of performance, which is heat moved divided by work paid for:

COP_refrigerator = T_cold / (T_hot − T_cold) COP_heat_pump = T_hot / (T_hot − T_cold)

The two always differ by exactly 1, because the heat delivered to the hot side is the heat taken from the cold side plus the work you put in. Both are ceilings, like the efficiency above.

A heat pump warming a house to 21°C from outdoor air at 5°C has a Carnot COP of 294.15 / 16 = 18.4. Real units manage 3 to 5, which still beats a resistance heater by several times, since a resistance heater is stuck at 1.00 by definition. Drop the outdoor air to −10°C and the ceiling falls to 9.5, which is exactly why heat pumps struggle in hard winters: the temperature gap they have to bridge has doubled.

A correction worth making

It is often said that cooling the cold reservoir helps more when that reservoir is already cold. It does not. Differentiate η = 1 − T_cold/T_hot with respect to T_cold and you get −1/T_hot, so a 10 K drop on the cold side buys you 10/T_hot of efficiency no matter what T_cold happens to be. What genuinely changes the payoff is the hot side: the same 10 K of cooling is worth 1.2 points on a 823 K steam turbine and only 3.4 points on a 293 K low-grade cycle. Cooling towers earn their keep, just not for the reason usually given.