Maximum Efficiency of a Heat Engine and Carnot's Cycle

Converting mechanical work into heat (as happens, for example, in a refrigerator, Section 4.9.2) is an inherently irreversible process. It follows that a heat engine converts the MAXIMUM possible amount of its absorbed heat into work only if every irreversible process within its cycle can somehow be avoided — in that ideal limiting case, the engine's efficiency would be as large as the laws of physics permit. Sadi Carnot, in 1824, proposed exactly such a hypothetical, maximally-efficient ideal engine, now called the Carnot engine.

A Carnot engine's cycle is built from exactly two kinds of reversible steps, chosen very deliberately:

1. Heat exchange steps (corresponding to legs A-to-B and C-to-D on a p-V diagram like Fig. 4.21/4.27) — for a heat-exchange step to be reversible, it must occur isothermally: the working substance must be at the SAME temperature as whichever reservoir it is exchanging heat with at that moment — at temperature $T_H$ (the source's temperature) while absorbing heat from the hot reservoir, and at temperature $T_C$ (the sink's temperature) while rejecting heat to the cold reservoir. This exactly matches the requirement, established in Section 4.7.3.2, that an isothermal process keeps the system in continuous thermal equilibrium with its (fixed-temperature) surroundings throughout.
2. Work-only steps (legs B-to-C and D-to-A) — for a work-transfer step (with no heat exchange) to be reversible, it must be adiabatic ($Q = 0$ throughout, Section 4.7.3.5) — this is exactly what is needed to move the working substance's temperature between $T_H$ and $T_C$ without letting any heat leak across a finite temperature difference (which would itself be an irreversible process). Putting these together: the Carnot cycle consists of exactly two isothermal steps (at $T_H$ and $T_C$) and two adiabatic steps connecting them — this is the specific four-leg structure shown in Fig. 4.27.

Deriving the Carnot efficiency. Using the isothermal work formula, Eq. (4.7)/(4.9), for the two isothermal legs, and the adiabatic work formula, Eq. (4.20), for the two adiabatic legs, and working through the algebra (using $p_iV_i^{\gamma} = p_fV_f^{\gamma}$ to relate the endpoints of each adiabatic leg to the isothermal legs on either side of it), the net efficiency of a full Carnot cycle comes out to the remarkably clean result:

$\eta = \frac{W}{Q_H} = \frac{Q_H + Q_C}{Q_H} = 1 - \frac{T_C}{T_H} \qquad \text{--- (4.29)}$ …