Ten degrees on the fire is worth 8.2 joules. Ten degrees off the river is worth 16.7.
Two ways to spend ten kelvin on the same engine. One of them returns twice as much work as the other, and the tool prints both numbers.
Set the Carnot simulator to a hot reservoir at 600 K, a cold one at 300 K and 1000 J of heat drawn per cycle. It reports 50.0% efficiency and 500.0 J of work.
Now spend ten kelvin, and choose where.
Put it on the fire — 610 K against 300 K — and the tool reports 50.8% and 508.2 J. Take it off the cooling water instead — 600 K against 290 K — and it reports 51.7% and 516.7 J.
Same ten kelvin. One placement returns 8.2 J, the other 16.7 J. The cold side is worth twice as much.
Twice is not a coincidence
Differentiate the Carnot efficiency η = 1 − TC/TH with respect to each temperature and the asymmetry falls out in one line:
∂η/∂TC = −1/TH · ∂η/∂TH = TC/TH²
Divide one by the other and everything cancels except the ratio itself:
|∂η/∂TC| ÷ |∂η/∂TH| = TH/TC
At 600 over 300 that ratio is exactly 2, which is the 16.7 against 8.2 the tool just printed. It is not a rule of thumb and it is not approximately true: the factor is the temperature ratio, whatever the temperatures are. Run a gas turbine at 1500 K against 300 K and the cold side is worth five times as much per kelvin. The hotter your fire, the less another degree of it is worth, and the more a degree off the cold end is worth.
The tool's own case guide says the cold side "matters as much as the hot". That understates it. It matters more, and by a factor you can name before you measure anything.
Which is unfortunate, because you do not own the cold side
Here is why this is not a tip for engine designers.
The hot reservoir is a thing you build. You can pick the fuel, the combustor, the blade alloy, the cooling channels. Every one of those is a decision, and the entire history of turbine engineering is the story of raising TH a few kelvin at a time against materials that would rather melt.
The cold reservoir is a river, a cooling tower, or the sky. Its temperature is the weather.
So the lever with twice the leverage is the one nobody can pull, and the lever everyone pulls is the weaker one. That inversion explains a surprising amount of practical engineering: why power stations sit on rivers and coastlines rather than near the fuel, why a condenser is a large expensive object whose only job is to keep one side cold, and why combined-cycle plants exist at all — a second engine running on the first one's waste heat is a way of buying back some of what the cold end took.
It also has a seasonal consequence that shows up in the grid. When a heatwave warms the cooling water by ten kelvin, a thermal plant does not lose a rounding error, it loses the larger of the two derivatives, and it loses it precisely when demand for air conditioning peaks. The physics runs the wrong way at the worst moment, and no amount of engineering on the hot side compensates, because the hot side is where the smaller derivative lives.
What the ceiling costs
One more thing the simulator quietly tells you, and it is worth knowing before treating any of this as an efficiency target: the entropy row reads exactly 0.0000 J/K at every setting.
That zero is the definition of the Carnot limit. A cycle with zero total entropy change is a reversible one, and reversible means infinitely slow: every step in equilibrium, no temperature difference anywhere to drive heat across, no friction. An engine at exactly this efficiency delivers exactly no power, because it takes forever to complete a stroke.
Curzon and Ahlborn worked out in 1975 what happens when you insist on power instead, and the answer is tidier than it has any right to be: at maximum power output the efficiency is 1 − √(TC/TH). For our 600 and 300 that is 29.3% rather than 50%, which is much closer to what real plants achieve. The asymmetry survives the change — the cold side still carries the bigger derivative — but the ceiling everyone quotes is a limit no working machine is trying to reach.
The number to take away is the ratio, not the efficiency. Whenever a system dumps something into an environment it does not control, look at what a unit of improvement is worth at each end before deciding which end to engineer. Here the answer is printed on the page: 16.7 against 8.2, and a formula that says why.
References (1)
- what the ceiling becomes once you insist the engine produce power at all Curzon & Ahlborn (1975). Efficiency of a Carnot engine at maximum power output. American Journal of Physics 43(1).