The Carnot Limit
By Lucian — builder & engineer, LK Forge
There is a hard ceiling on how much of a fire’s heat any engine can turn into work, and it was found in 1824 before anyone even knew what heat was. It depends on nothing but two temperatures — how hot the source, how cold the exhaust — and no machine has ever crossed it or ever will. We took the numbers from the Carnot-cycle lab to see where it bites.
· 5 min read · exact from η = 1 − Tc/Th
Hotter is better — but 100% never comes
Fix the cold sink at a room-temperature 300 K and sweep the source hotter. The ceiling climbs fast at first, then crawls: it approaches 100% only as the source temperature runs to infinity. That is why engineers chase ever-hotter combustion and turbine-inlet temperatures — every extra hundred kelvin buys efficiency — and why a low-temperature source is a lost cause: at water’s boiling point the best conceivable engine throws away four-fifths of the heat.
Carnot ceiling η = 1 − 300/Th. The curve flattens toward 100% but never touches it — the last stretch of efficiency costs unbounded temperature.
The ceiling, and the gap below it
The Carnot value is the best case; real machines live underneath it. Here is the ceiling for four kinds of engine beside a typical figure they actually reach. An ocean-thermal plant works across a mere 22 K of sea-temperature difference, so even its ceiling is a paltry 7% — and it captures under half of that. A car has a glorious 85% ceiling but squanders most of it to friction, heat loss and haste, landing near 28%.
Carnot ceiling (amber) vs a typical achieved efficiency (grey). Real engines reach roughly a third to two-thirds of the theoretical best; the wider the temperature span, the higher the ceiling.
Where the heat goes
The limit is really a bookkeeping law. An engine draws heat Q_H from the hot reservoir, turns some into work W, and must discard the rest as waste heat Q_C into the cold one. In the ideal cycle Q_H = Th·ln r and Q_C = Tc·ln r for an expansion ratio r, so the fraction converted is W/Q_H = (Q_H − Q_C)/Q_H = 1 − Tc/Th — the two expressions agree exactly. Run the numbers for an 800 K → 300 K cycle and 878.9 units of heat in become 549.3 of work and 329.6 of waste, a clean 62.5%. You cannot shrink that waste term to zero because Tc cannot reach zero; the discarded heat is the price of running a cycle at all, and Carnot’s formula is simply its receipt.
Reproduce it yourself
One line gives the ceiling; the energy balance gives the same answer a second way.
const eta = (Th, Tc) => 1 - Tc / Th; // kelvin
eta(2000, 300); // 0.850 car combustion ceiling
eta(800, 300); // 0.625 power plant
eta(373, 300); // 0.196 steam at boiling point
// same answer from the heat balance (expansion ratio r):
const Qh = Th => Th * Math.log(r), Qc = Tc => Tc * Math.log(r);
const r = 3, Th = 800, Tc = 300;
(Qh(Th) - Qc(Tc)) / Qh(Th); // 0.625 = 1 - Tc/ThThe efficiency and heat formulas are lifted straight from the Carnot-cycle solver; the temperature sweep and engine table are the small script behind this post.
Run the cycle
Open the lab, slide the hot and cold temperatures, and watch the PV loop and the efficiency change — then widen the expansion ratio to get more work at the same efficiency.