How this instrument works
Sadi Carnot published Réflexions sur la puissance motrice du feu in 1824, aged 28, and buried inside it sits one of physics' strangest results: how much heat you can convert into work depends on nothing but two absolute temperatures. Not on steam versus air, not on cylinder geometry, not on how clever your valve timing is. Carnot reached that conclusion while still believing heat was a conserved fluid called caloric — a wrong model yielding a right answer, later rebuilt on solid ground by Clausius and Kelvin during the 1850s.
Physically, η counts what fraction of heat drawn from a hot reservoir leaves as work instead of being dumped into a cold one. Raising Th helps, but dropping Tc helps more per kelvin, which is why power stations fight so hard over condenser vacuum and cooling water. A supercritical coal boiler at 873 K rejecting to a 300 K river has a ceiling near 66 percent and delivers around 45. Ocean thermal plants, working across barely 22 kelvin of seawater, are capped near 7 percent — and that figure, not engineering incompetence, explains why they stay rare.
Two assumptions carry all this weight. Reservoirs must be effectively infinite, so their temperatures never sag as heat moves; and every step must be reversible, meaning quasi-static, frictionless, infinitely slow. That last condition is brutal — a genuine Carnot cycle produces zero power, because it runs at zero speed. Push hardware fast enough to be useful and irreversibility eats your margin. Treat this output as a bound you approach, never as performance you can schedule.
- Convert both readings to kelvin first: add 273.15 to any Celsius figure. Kelvin is not optional here, since a ratio of temperatures is meaningless in any other scale.
- Enter your boiler, combustor, or solar collector figure as Hot reservoir temperature (K).
- Enter your condenser, radiator, or ambient sink figure as Cold reservoir temperature (K).
- Read Maximum efficiency (%), then compare it against what your real machine measures — that gap is what irreversibility costs you.
Worked example — 600 K boiler over a 300 K sink
A test rig raises steam at 600 K and rejects heat into a chilled loop held at 300 K. Enter 600 as Hot reservoir temperature (K) and 300 as Cold reservoir temperature (K). Substituting: η = (1 − 300 ⁄ 600) × 100 = (1 − 0.5) × 100 = 50. Maximum efficiency (%) reads 50.
Halving absolute temperature across an engine converts, at absolute best, half of incoming heat into work; whatever you build, 300 K worth must still be thrown away as waste. Should that rig measure 31 percent on a dynamometer, it is running at 62 percent of its Carnot bound — respectable for hardware obliged to turn at a useful speed.
Questions
Can I enter Celsius or Fahrenheit instead of kelvin?
No, and this is by far the most common error with this formula. Only absolute scales work, because one temperature gets divided by another. Steam at 100 °C over a 20 °C sink is not (1 − 20 ⁄ 100) = 80 percent; converted properly it becomes (1 − 293.15 ⁄ 373.15) = 21.4 percent. Add 273.15 to any Celsius reading before entering it.
Why can't efficiency reach 100 percent?
You would need Tc = 0 K, and thermodynamics' third law forbids reaching absolute zero in any finite number of steps. Every real engine also has to dump heat somewhere warmer than nothing — a river, ambient air, deep space at 2.7 K. That rejected heat pays for entropy generated along the way, and it never falls to zero.
Why do real engines fall so far short of this number?
Because Carnot's cycle is reversible, which demands infinitely slow operation and therefore zero power output. Working machines run fast, so they suffer friction, turbulence, finite-temperature heat transfer, and combustion irreversibility. A useful companion is Curzon and Ahlborn's 1975 result for efficiency at maximum power: η = 1 − √(Tc ⁄ Th), which for 600 K over 300 K gives 29.3 percent rather than 50.
Does the working fluid change the answer?
No, and that is Carnot's deep point. Steam, helium, carbon dioxide, or a magnetic solid all share one ceiling between fixed reservoirs. Were some fluid better, you could couple two engines back to back and shuttle heat from cold to hot for free, breaking thermodynamics' second law. Fluid choice affects how closely real hardware approaches this bar, never where that bar sits.
How does this relate to a refrigerator's COP?
Run a cycle backwards and those same two temperatures cap coefficient of performance instead: COP = Tc ⁄ (Th − Tc) for cooling, or Th ⁄ (Th − Tc) for a heat pump. Between 300 K and 270 K, a fridge could in principle move 9 joules of heat per joule of work supplied. Narrow temperature gaps make heat pumps thrifty for precisely the reason they leave engines feeble.
What if my cold reservoir is hotter than my hot one?
This sheet refuses that entry. Heat does not flow spontaneously up a gradient, so no work emerges from such an arrangement; algebraically you get a negative figure, and physically you have described a refrigerator needing work put in rather than an engine giving work out. Swap your two entries and try again.