Carnot Efficiency Calculator
Calculate max heat-engine efficiency with the Carnot formula η = 1 − Tc/Th. Supports Kelvin and Celsius, with work and rejected-heat outputs.
Carnot Efficiency Calculator
Background.
Carnot efficiency is the maximum possible efficiency any heat engine can achieve while operating between two temperature reservoirs — a hot source it draws heat from and a cold sink it rejects heat to — and this calculator computes it directly from the formula η = 1 − T_c/T_h, where both temperatures must be on an absolute (Kelvin or Rankine) scale. No real engine, however cleverly engineered, can beat this number for a given pair of reservoir temperatures; it is a hard ceiling imposed by the second law of thermodynamics, not an engineering target that better materials or manufacturing could ever exceed.
The result derives from Nicolas Léonard Sadi Carnot's 1824 analysis of an idealized, fully reversible heat-engine cycle. Carnot showed that the efficiency of such a reversible cycle depends only on the two reservoir temperatures, and on nothing else about the engine's design, working fluid, or size — a genuinely surprising result for 1824, decades before the first and second laws of thermodynamics were formally stated. Every irreversibility a real engine introduces — friction, turbulence, heat leaking across a finite temperature difference, imperfect insulation — only ever reduces efficiency below this Carnot ceiling, never toward it from the other direction.
This calculator is built to head off the single most common mistake in applying the formula: using Celsius (or Fahrenheit) temperatures directly instead of converting to an absolute scale first. Because η = 1 − T_c/T_h is a ratio of the two absolute temperatures, plugging in Celsius values directly gives a wrong, often wildly wrong, answer — and can even produce an apparent efficiency over 100% or a negative number, both physically meaningless. To prevent that mistake, this calculator lets you choose Kelvin or Celsius from a dropdown and performs the Kelvin conversion internally and automatically, while guarding against every input combination that would otherwise produce a non-physical result: a reservoir at or below absolute zero, or a cold reservoir hotter than the hot reservoir.
Beyond the bare efficiency figure, the calculator also reports work output and rejected heat for a specified heat input per cycle, since a raw percentage is often less immediately useful than seeing it applied to an actual energy quantity — a power-plant engineer sizing turbine output, or a student checking a textbook energy-balance problem, usually wants both the efficiency and what it implies for a specific heat-input scenario. The two together, work output plus rejected heat, must sum back to the original heat input by conservation of energy, which this calculator's outputs satisfy exactly.
Real-world heat engines never approach the idealized reservoir spreads used in textbook examples like 600 K and 300 K. Steam-turbine power plants, gas turbines, and automobile engines all operate with much smaller practical temperature differences between their hot and cold sides than the theoretical maximum their working fluid could tolerate, which is why actual thermal-efficiency figures for real power plants (typically 30–45%) sit well below what a naive Carnot calculation using peak combustion temperature and ambient temperature would suggest — friction, heat leakage, and the practical engineering limits on how close an engine can get to a truly reversible cycle all chip away at the theoretical ceiling this calculator computes.
What is carnot efficiency calculator?
A heat engine is any device that converts heat flowing from a hot reservoir to a cold reservoir into mechanical work — a steam turbine, an internal-combustion engine, a jet engine. The second law of thermodynamics establishes that no heat engine can convert 100% of the heat it draws from a hot reservoir into useful work; some fraction must always be rejected to a cold reservoir. The Carnot efficiency, η = 1 − T_c/T_h, is the specific upper bound on how much of that heat can be converted, for an idealized, perfectly reversible engine operating between reservoirs at absolute temperatures T_h (hot) and T_c (cold).
Because the formula is a ratio of absolute temperatures, it inherently requires both temperatures to be measured on a scale where zero means the complete absence of thermal energy — the Kelvin scale (or, equivalently, the Rankine scale used in some US engineering contexts). The Kelvin scale's zero point, absolute zero, corresponds to −273.15°C, and the SI kelvin is defined so that the Celsius scale's numeric size and the Kelvin scale's numeric size are identical (a 1°C step equals a 1 K step) — the two scales differ only by that fixed 273.15 offset, which the Carnot formula's *ratio* (not difference) of temperatures does not automatically cancel out the way a temperature-difference calculation would.
Carnot's result is remarkable precisely because it is independent of engine design: two engines built from completely different technologies, but operating between the same two reservoir temperatures, share the identical theoretical efficiency ceiling. Real engines fall short of that ceiling because real processes involve friction, turbulence, and heat transfer across a finite (non-infinitesimal) temperature difference — all of which are thermodynamically irreversible and therefore, by the second law, strictly reduce achievable efficiency below the reversible Carnot limit.
How to use this calculator.
- Choose your temperature unit — Kelvin or Celsius.
- Enter the hot-reservoir temperature.
- Enter the cold-reservoir temperature (it must not exceed the hot-reservoir temperature).
- Enter the heat input per cycle, if you want work-output and rejected-heat figures alongside the efficiency percentage.
- Read the Carnot efficiency — remember this is a theoretical maximum, not what any real engine actually achieves.
The formula.
For Carnot's idealized reversible cycle, the heat rejected to the cold reservoir (Q_c) and the heat drawn from the hot reservoir (Q_h) are in the same ratio as the two reservoirs' absolute temperatures: Q_c/Q_h = T_c/T_h. Since the work output of any heat engine, by conservation of energy, is simply the heat drawn in minus the heat rejected (W = Q_h − Q_c), the efficiency — work output divided by heat input — works out to η = W/Q_h = (Q_h − Q_c)/Q_h = 1 − Q_c/Q_h = 1 − T_c/T_h, the formula this calculator implements directly.
When a temperature unit of Celsius is selected, the calculator converts both hot and cold temperatures to Kelvin using the exact SI relationship T(K) = T(°C) + 273.15 before computing anything else — so a hot reservoir entered as 327°C becomes 600.15 K internally, and the ratio T_c/T_h is always computed from two genuinely absolute temperatures, never from raw Celsius values. Given the heat input per cycle, work output follows directly as W = η × Q_h, and rejected heat as Q_rejected = Q_h − W (equivalently, Q_h × T_c/T_h) — the two must sum back to the original heat input exactly, which is simply energy conservation restated.
A worked example.
A textbook heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K, drawing 1000 J of heat from the hot side each cycle. η = 1 − 300/600 = 1 − 0.5 = 0.5, so the Carnot efficiency is exactly 50% — meaning that even a perfectly reversible, idealized engine operating between these two temperatures could convert at most half of the heat it draws into useful work, no matter how well built. With 1000 J drawn from the hot reservoir, the maximum possible work output is W = 0.5 × 1000 = 500 J, and the remaining 500 J must be rejected to the cold reservoir (Q_rejected = 1000 − 500 = 500 J) — the two figures sum back to the original 1000 J input exactly, as conservation of energy requires. Any real engine operating between these same two reservoirs would extract less than 500 J of work per 1000 J of heat input, because real cycles are never perfectly reversible.
Frequently asked questions.
Why must the temperatures be in Kelvin (or another absolute scale)?
Can a real engine ever reach Carnot efficiency?
What happens if the cold reservoir temperature equals the hot reservoir temperature?
What happens as the cold reservoir approaches absolute zero?
Does raising the hot-reservoir temperature help efficiency more than lowering the cold-reservoir temperature?
References& sources.
- [1]OpenStax College Physics, via University of California, Davis LibreTexts. "Carnot's Perfect Heat Engine: The Second Law of Thermodynamics Restated." Eff_c = 1 − T_c/T_h, temperatures must be in kelvins or another absolute scale.
- [2]Bureau International des Poids et Mesures (BIPM). "The kelvin." SI unit of thermodynamic temperature, historically and currently defined relative to absolute zero.
- [3]Carnot, N.L.S. (1824). Réflexions sur la puissance motrice du feu, et sur les machines propres à développer cette puissance. Bachelier, Paris. Original formulation of the reversible-cycle efficiency limit.
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