Carnot's theorem

Remarks

Suppose an engine $E$ were more efficient than a reversible engine $R$. Scale the engines so that they produce the same work $W$ per cycle, and run $R$ backwards as a refrigerator driven by the work of $E$. Since $E$ is more efficient, it takes less heat from the hot reservoir than $R$ gives back to it. The combined device then moves heat from the cold reservoir to the hot one with no other effect. This contradicts the Clausius statement of the second law, and the same argument with two reversible engines gives part 2. Alternatively, the Clausius theorem gives $Q_H/T_H - Q_C/T_C \le 0$ per cycle, with equality in the reversible case, so $\eta = 1 - Q_C/Q_H \le 1 - T_C/T_H$. For an engine between $T_H = 600\,\mathrm{K}$ and $T_C = 300\,\mathrm{K}$, at most half of the absorbed heat can become work. The bound does not depend on the working substance; this is what makes it possible to define the thermodynamic temperature scale by the ratio $Q_C/Q_H$ of reversible engines.

Going deeper

For a student

The theorem is one of the first consequences of the second law, and it is equivalent to the impossibility of a perpetual motion machine of the second kind. Its proof by coupling an engine to a reversed reversible engine is the model for many arguments in classical thermodynamics. Ideal Otto, Diesel and Brayton cycles stay below the Carnot bound for their highest and lowest temperatures. The reason is that they take in heat at varying temperatures. The same reasoning bounds the coefficient of performance of refrigerators and heat pumps.