Carnot efficiency — the limit of a heat engine

What just happened?

You widened the temperature difference between the hot and cold reservoirs — the efficiency rose. The “hotter” the source and the “colder” the heat sink, the more heat we convert into work.

Carnot efficiency:

— the maximum efficiency of an ideal heat engine. $T_h$ — hot reservoir temperature, $T_c$ — cold reservoir temperature (both in kelvins). The efficiency depends only on the temperatures, not on the working substance.

What does this mean in practice?

The engine takes heat $Q_h$ from the hot reservoir, converts part of it into work $W$, and dumps the rest to the cold reservoir as $Q_c$. Hence:

Why does efficiency never reach 100%?

For $\eta = 1$ we would need $T_c = 0\,\text{K}$ (absolute zero) or $T_h \to \infty$ — both physically unreachable. That is why $\eta = 1 - T_c/T_h < 1$ always. This is a direct consequence of the second law of thermodynamics: you cannot convert all the heat into work.

Why you need this for the exam:

η = 1 − T_c/T_h plus W = η·Q_h solve most heat-engine problems. The trick: always enter temperatures in kelvins, and a real engine’s efficiency is always lower than the Carnot value.

Remember: efficiency rises with the temperature difference but never reaches 100% — η = 1 − T_c/T_h < 1. This is the limit imposed by the second law of thermodynamics.