Carnot engine · the most work heat can do

Heat flows from hot to cold — what is the MOST work an engine can wring from that flow, and what does the limit depend on?

▶ Run the simulationSee the measured result

How the lab tests it

Run a reversible Carnot cycle of an ideal gas (γ=5/3) between reservoirs T_h and T_c: isothermal expansion at T_h (drinks Q_h) → adiabatic expansion → isothermal compression at T_c (sheds Q_c) → adiabatic compression. Numerically traverse the loop, integrating the net work W=∮P dV (the enclosed P–V area) and the hot-isotherm heat Q_h by trapezoid, and recover the efficiency η=W/Q_h — never plugging into the formula. Repeat across five (T_h,T_c) pairs, sweep the expansion ratio V₂/V₁∈[1.5,12], and run an Otto cycle between the same temperature extremes as a control. ?world=carnot.

What it checks

the Carnot efficiency η = 1 − T_c/T_h — recovered from ∮P dV to machine precision (600 K/300 K ⇒ exactly 50%) and, decisively, INDEPENDENT of the working substance, the amount of gas, and the expansion ratio: η is the SAME to 1e-15 across V₂/V₁ from 1.5 to 12, so a perfect engine's quality is fixed by the two reservoir temperatures alone. And no engine beats it (Carnot's theorem = the second law): an Otto cycle between the same extremes (r_v=8, 300–1500 K) reaches only η=1−1/r_v^(γ−1)=75% against the Carnot ceiling 1−300/1500=80%. The gap is why no power plant reaches Carnot and why raising T_h, not lowering T_c, is the lever — a real coal plant at ~810 K/300 K caps near 63%. The lab's first thermodynamics world.

Carnot efficiency, COP, lost-work & maximum-power calculator

The oldest bound in engineering, and the one nobody has ever got round. Carnot asked in 1824 what the most work is that can be wrung out of heat flowing from hot to cold, and the answer turned out to depend on NOTHING about the engine — not the working substance, not the amount of gas, not the expansion ratio, not the cleverness of the linkage — only on the two reservoir temperatures. Not one number on this page is stored. The efficiency is assembled as 1 − T_c/T_h from the two temperatures you type; the heat as nRT_h·ln(V₂/V₁), with the logarithm taken here; the COPs as a temperature over a DIFFERENCE of temperatures, which is why they blow up exactly where the efficiency dies; the lost work from an entropy balance; and the maximum-power bound as a square root. Even the gas constant and the Celsius offset arrive as fields, so zeroing the lab's own readings moves not one computed digit. The defaults are this world's bench, 600 K over 300 K with a mole of monatomic gas expanding fourfold: η = 50%, Q_h = 6915.775586 J assembled analytically against the 6915.776462761074 J the simulation above trapezoid-integrated, a discretization residual of +1.27e-7 that the finding discloses and this page recovers by a road with no integration in it. Five things here the finding never computes. A real engine at 42% is running at 84% of its ceiling and throwing away 553.262047 J a cycle — and that lost work is exactly T_c times the entropy it made, which is Gouy–Stodola. A kelvin taken off the cold side is worth twice a kelvin put onto the hot one, exactly T_h/T_c, and the reason plants chase the hot side anyway is availability rather than leverage. Run the same cycle backwards and its heat-pump COP is 1/η, delivering two joules of heat per joule of electricity. Carnot's own 50% is reached only at ZERO power; at maximum power the honest bound is Curzon–Ahlborn's 1 − √(T_c/T_h) = 29.289322%. And the Otto control that falsifies the rival is not an independent fact at all: η_Otto < η_Carnot is the SAME inequality as r_v^(γ−1) < T_max/T_min, so a piston that appears to beat Carnot has simply been given a compression ratio its own peak temperature forbids. Read the temperatures in Celsius instead of kelvin and this page will tell you what that costs — it is the commonest way the formula is got wrong.