Audited ·Last updated 27 Jul 2026·3 citations·Tier 2·0 uses

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

The temperature of the hot reservoir the engine draws heat from, in the selected unit.
The temperature of the cold reservoir the engine rejects heat to, in the selected unit.
Temperature unit
The heat drawn from the hot reservoir in one cycle, used to compute work output and rejected heat.
Carnot efficiency
50.00
The maximum theoretical efficiency of a heat engine operating between these two reservoirs.
Carnot efficiency (decimal)
0.5
Work output
500
Heat rejected to cold reservoir
500
Hot reservoir (Kelvin)
600
Cold reservoir (Kelvin)
300

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.

  1. Choose your temperature unit — Kelvin or Celsius.
  2. Enter the hot-reservoir temperature.
  3. Enter the cold-reservoir temperature (it must not exceed the hot-reservoir temperature).
  4. Enter the heat input per cycle, if you want work-output and rejected-heat figures alongside the efficiency percentage.
  5. Read the Carnot efficiency — remember this is a theoretical maximum, not what any real engine actually achieves.

The formula.

η = 1 − Tc ⁄ Th

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.

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.

heat Input1,000
cold Temp300
temp UnitK
hot Temp600

Frequently asked questions.

Why must the temperatures be in Kelvin (or another absolute scale)?
The Carnot formula η = 1 − T_c/T_h is a ratio of the two reservoir temperatures, and that ratio is only physically meaningful when zero on the scale actually represents zero thermal energy — which is exactly what the Kelvin scale's zero point (absolute zero, −273.15°C) means. Plugging Celsius values directly into the ratio gives a wrong answer, because 0°C does not mean zero thermal energy; it is an arbitrary reference point (the freezing point of water) that happens to sit 273.15 K above true absolute zero. For example, treating 27°C and 327°C as if they were the absolute temperatures in the ratio would give 1 − 27/327 ≈ 91.7%, a badly wrong result — the correct Kelvin-based calculation (300.15 K and 600.15 K) gives about 50.0%, essentially the same efficiency as the classic 300 K/600 K textbook case. This is exactly why this calculator forces you to pick Kelvin or Celsius explicitly and performs the Kelvin conversion internally before computing anything, rather than trusting whatever raw number you type in.
Can a real engine ever reach Carnot efficiency?
No — Carnot efficiency is a theoretical upper bound for a perfectly reversible cycle, and every real process involves some irreversibility (friction, turbulence, heat transfer across a finite temperature gap, imperfect insulation, finite-time operation) that the second law of thermodynamics guarantees will reduce actual efficiency below the Carnot ceiling. Real engines can, and do, get closer to their Carnot limit through better engineering — reduced friction, better insulation, more efficient heat exchangers — but a truly reversible cycle running in zero time and producing zero entropy is a physical idealization, not an achievable engineering target.
What happens if the cold reservoir temperature equals the hot reservoir temperature?
Efficiency is exactly zero. With T_c = T_h, the ratio T_c/T_h equals 1, so η = 1 − 1 = 0 — which makes physical sense: if there is no temperature difference between the two reservoirs, there is no thermal gradient available to extract work from, regardless of how the engine is designed. This calculator treats this as a perfectly valid (not an error) input, returning zero work output and reporting that all of the heat input is rejected unchanged.
What happens as the cold reservoir approaches absolute zero?
Efficiency approaches, but never actually reaches, 100%. As T_c approaches 0 K, the ratio T_c/T_h approaches 0, so η = 1 − T_c/T_h approaches 1 — but the third law of thermodynamics establishes that absolute zero itself is unreachable by any finite process, so a real cold reservoir can only ever approach, never equal, 0 K, and Carnot efficiency can only ever approach, never equal, 100%. This calculator's guard against a cold-reservoir input of exactly 0 K (or below) reflects that same physical impossibility.
Does raising the hot-reservoir temperature help efficiency more than lowering the cold-reservoir temperature?
It depends on the starting point, but in most practical engineering situations, raising T_h tends to be more effective, mainly because there is usually much more room to push T_h upward (limited chiefly by material strength and combustion chemistry) than there is to push T_c downward (usually limited to somewhere near the ambient environment's temperature, since actively refrigerating the cold reservoir below ambient typically consumes more energy than it recovers). Mathematically, both changes move the T_c/T_h ratio in the efficiency-improving direction, but which one delivers a larger efficiency gain per degree changed depends on the specific numeric values of T_h and T_c in a given scenario — this calculator makes it easy to test both directions by simply changing one input at a time and comparing the resulting efficiency.

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