Why Can No Engine Be 100% Efficient?

No heat engine can be 100% efficient because the second law of thermodynamics requires some energy to be transferred as heat to a colder reservoir. Therefore, not all the energy supplied by heating can be converted into useful work.

The thermodynamic mechanism

A heat engine operates cyclically between a hot reservoir at temperature $T_h$ and a cold reservoir at temperature $T_c$. It absorbs energy $Q_h$, performs useful work $W$, and transfers energy $Q_c$ to the cold reservoir.

Over one complete cycle, the engine returns to its original state, so $\Delta U=0$. Conservation of energy therefore gives

$$Q_h=W+Q_c.$$

Its efficiency is

$$\eta=\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}.$$

For $\eta=1$, the engine would need $Q_c=0$. However, this would convert all heat extracted from one reservoir into work, violating the second law. Heat rejection is necessary because the total entropy of an isolated system cannot decrease.

For an ideal reversible engine, the maximum possible efficiency is the Carnot efficiency:

$$\eta_{\text{Carnot}}=1-\frac{T_c}{T_h},$$

where temperatures must be measured in kelvin. Since $T_c>0\text{ K}$ for any real cold reservoir, $\eta_{\text{Carnot}}<1$.

A common misconception is that engines fail to reach 100% efficiency only because of poor design or friction. Removing practical losses could approach the Carnot limit, but it could never remove the thermodynamic requirement to reject heat.

IB exam technique

This is B.4 Thermodynamics (HL only). In an Explain response, link the second law to necessary heat rejection and then use $\eta=1-Q_c/Q_h$. If temperatures are provided, calculate the maximum theoretical efficiency using kelvin, not degrees Celsius.