CARNOT
Q_h = W + Q_c to the last joule, η = 1 − T_c/T_h — the ceiling nobody beats.
Use the simulation above to change the variables and play through the guided stages. The explanation below describes the default starting values; the simulation updates its explanation as you experiment.
Setup
Two reservoirs — one hot, one cold — and a cylinder of gas willing to run in circles between them. Sadi Carnot asked, in 1824, the most consequential engineering question ever posed: of the heat drawn from the fire, how much can POSSIBLY become work?
Four legs
Four reversible legs, each in closed form: expand isothermally in contact with the fire, drinking heat; coast adiabatically down to the cold temperature; compress isothermally, exhaling heat to the cold sink; coast home. The two coasts do equal and opposite work — they cancel exactly — and the cycle closes because the adiabatic volume ratios conspire: V₃/V₄ = V₂/V₁, an identity this page checks.
The verdict
The net work is the enclosed area, exactly — and the efficiency collapses to two numbers: η = 1 − T_c/T_h. The expansion ratio, the gas amount, every engineering choice cancels out. At 600 K over 300 K that is exactly one half, and no engine between these reservoirs — built or unbuilt — will ever beat it.
One cycle
One full cycle, quasi-static, every frame an exact state: amber while drinking from the fire, violet on the silent coasts, cyan while paying the cold reservoir its unavoidable share. The gas returns to its exact starting state and remembers nothing — only the reservoirs remember, and the flywheel keeps the difference.
Audit
The ledger of ledgers, closed to zero residual: Q_h = W + Q_c joule for joule; the coasts cancel; the cycle closes bitwise; η by the work route equals η by the temperature route. And the deepest line the site owns: Q_h/T_h = Q_c/T_c — the entropy drawn from the fire is returned, in full, to the cold. That equality is reversibility; its refusal to ever tilt the other way is the second law.