Carnot Efficiency Calculator
Enter the hot-reservoir and cold-reservoir temperatures to find the maximum theoretical (Carnot) efficiency of a heat engine, plus the temperature ratio and COP values for a refrigerator or heat pump.
Find the Maximum Efficiency of a Heat Engine
Carnot Efficiency
Where:
- \(\eta_{max}\) — maximum possible (Carnot) efficiency of a heat engine, as a decimal fraction
- \(T_h\) — absolute temperature of the hot reservoir (K)
- \(T_c\) — absolute temperature of the cold reservoir (K)
The formula gives the upper limit on the fraction of absorbed heat that any engine operating between two thermal reservoirs can convert into work, as required by the second law of thermodynamics. Both temperatures must be on an absolute scale (kelvin); if your temperatures are in °C or °F, the calculator converts them automatically.
Carnot Efficiency – Calculation Example
Example: a heat engine works between a hot reservoir at 500 K and a cold reservoir (the surroundings) at 300 K.
\[ \eta_{max} = 1 - \frac{T_c}{T_h} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 \]
Result: the maximum efficiency is 0.40, i.e. 40 %. Even a perfectly reversible engine must reject the remaining 60 % of the absorbed heat to the cold reservoir. Running the same cycle backwards gives a Carnot coefficient of performance of \(300/(500-300) = 1.5\) as a refrigerator and \(500/(500-300) = 2.5\) as a heat pump.
How to use the calculator
- Enter the Hot-reservoir temperature (Th) — the temperature of the heat source, e.g. combustion gas, steam or a geothermal field.
- Enter the Cold-reservoir temperature (Tc) — the temperature of the cooling medium, typically ambient air or cooling water (about 300 K / 27 °C).
- Select the unit (K, °C or °F) for each temperature from the dropdown next to the input field. The calculator converts everything to kelvin internally.
- Read the result: the maximum efficiency in percent and as a decimal, the temperature ratio, and the Carnot COP values if the cycle were run as a refrigerator or a heat pump.
What is Carnot efficiency?
The Carnot efficiency is the theoretical maximum efficiency that any heat engine can achieve when operating between two thermal reservoirs at temperatures \(T_h\) and \(T_c\). It is named after Sadi Carnot, who showed in 1824 that no engine — however cleverly designed — can exceed this limit. The reason is the second law of thermodynamics: an engine must reject part of the heat it absorbs to a colder body, and the amount it must reject is fixed by the ratio \(T_c/T_h\).
Because the formula uses a ratio of temperatures, the temperatures must be absolute (kelvin or Rankine). Using Celsius or Fahrenheit directly gives wrong answers: 100 °C and 0 °C are not a 100 % efficient pair, but 373.15 K and 273.15 K give a Carnot efficiency of only about 27 %.
Real engines never reach the Carnot limit because friction, turbulence, heat loss and finite-speed heat transfer make them irreversible. Practical heat engines typically achieve 30–70 % of the Carnot value. For example, a modern coal-fired power plant with steam at ~810 K exhausting to ~300 K has a Carnot limit of about 63 %, while its actual efficiency is around 40 %. Engineers use the Carnot value as a benchmark: it tells you whether a design is worth improving further or whether it is already close to the physical ceiling.
The result also shows how much heat is inevitably wasted: the fraction \(T_c/T_h\) of every joule absorbed must be dumped to the cold reservoir. This is why power plants need cooling towers and rivers, and why the efficiency of any engine rises when you raise the source temperature or lower the sink temperature.
Assumptions, limitations and tips
- Absolute temperatures only. The formula is valid only with temperatures on an absolute scale. If you enter °C or °F, the calculator converts them to kelvin before computing the ratio.
- Two reservoirs at fixed temperatures. The Carnot model assumes the engine exchanges heat only with reservoirs that stay at constant temperature. Real sources (burning fuel, expanding steam) change temperature during the process, which lowers the achievable efficiency.
- Reversible operation. The Carnot engine is infinitely slow and frictionless. Every real process is irreversible, so treat the result as an upper bound, not a design target.
- Efficiency cannot reach 100 %. A finite cold reservoir above absolute zero always forces some heat rejection. 100 % would require \(T_c = 0\) K, which is unattainable (third law of thermodynamics).
- COP values. The refrigerator and heat-pump coefficients of performance shown in the details refer to the same two reservoirs run in reverse. Note that a heat pump's COP is always 1 greater than the refrigerator's COP for the same temperatures.
- To compare how well a real machine converts input work into useful output (rather than comparing it with a thermodynamic limit), see the Mechanical Efficiency calculator.
Typical Carnot efficiency values
| Application | Th (approx.) | Tc (approx.) | Carnot efficiency |
|---|---|---|---|
| Car engine (gasoline) | 800 K | 300 K | 62.5 % |
| Steam power plant | 810 K | 300 K | 63 % |
| Combined-cycle gas plant | 1600 K | 300 K | 81 % |
| Geothermal (low enthalpy) | 360 K | 300 K | 17 % |
| Ocean thermal energy (OTEC) | 300 K | 278 K | 7 % |
These are theoretical ceilings; actual efficiencies are always lower. For other thermodynamic and physics quantities, you may also find our Pressure and Density calculators useful.
Frequently Asked Questions
How do I calculate Carnot efficiency by hand?
Convert both reservoir temperatures to kelvin (add 273.15 to °C, or use \\(K = (°F + 459.67) \times 5/9\) for Fahrenheit), then apply \\(\eta_{max} = 1 - T_c/T_h\). For example, with a hot reservoir at 600 K and a cold reservoir at 300 K: \\(\eta_{max} = 1 - 300/600 = 0.5\), i.e. 50 %. Multiply the decimal by 100 to get a percentage.
Why must temperatures be in kelvin (or Rankine), not °C or °F?
The formula uses a ratio of temperatures, and ratios are only meaningful on absolute scales where zero means true zero thermal energy. If you plug in 100 °C and 0 °C you would wrongly get 100 % efficiency; converted correctly to 373.15 K and 273.15 K, the true Carnot efficiency is only about 27 %. Rankine also works because it is absolute — just use °R for both temperatures consistently.
What is a good or typical Carnot efficiency value?
It depends entirely on the temperature gap. Narrow-gap applications like ocean thermal energy conversion (~7 %) or low-enthalpy geothermal (~17 %) have low ceilings, while a combined-cycle gas plant (~81 %) or a car engine (~62 %) allows much more. A useful rule: real machines typically reach only 30–70 % of their Carnot limit, so a high Carnot value alone does not guarantee a high actual efficiency.
What happens if the cold temperature is higher than the hot temperature?
The calculation is invalid: an engine cannot run in that case, and the formula would return a negative efficiency. The calculator rejects such inputs with an error message. However, the reverse situation is physically meaningful — a refrigerator or heat pump deliberately moves heat from cold to hot by consuming work, which is why the COP values are shown alongside the efficiency.
Can Carnot efficiency ever be 100 %?
No, not in practice. The formula reaches 1 only when \\(T_c = 0\) K or \\(T_h\) becomes infinite. The first is forbidden by the third law of thermodynamics (absolute zero is unattainable), and the second is physically impossible. Every real engine between finite temperatures must reject the fraction \\(T_c/T_h\) of its absorbed heat to the cold reservoir.
Why is the heat pump COP always 1 higher than the refrigerator COP?
Because the two machines use the same cycle but count different outputs as useful. A refrigerator's COP is \\(T_c/(T_h - T_c)\), counting only the heat removed from the cold space; a heat pump's COP is \\(T_h/(T_h - T_c)\), counting all heat delivered to the warm space — which includes the input work converted to heat. The difference between the two numerators is exactly \\(T_h - T_c\), making COP heat pump = COP refrigerator + 1.
What are common mistakes when using the Carnot formula?
The biggest error is entering Celsius or Fahrenheit values as if they were absolute temperatures. Others include swapping \\(T_h\) and \\(T_c\) (which gives an efficiency above 100 %, a clear red flag), forgetting that real-world source temperatures drop during combustion or steam expansion, and treating the Carnot value as a design target rather than an unattainable upper bound. Always sanity-check: the result must lie between 0 % and 100 %.