The Carnot cycle is the most efficient possible thermodynamic cycle operating between two temperatures T_H and T_C. It consists of four reversible steps: (1) isothermal expansion at T_H (absorbs Q_H), (2) adiabatic expansion (temperature drops to T_C), (3) isothermal compression at T_C (rejects Q_C), (4) adiabatic compression (temperature returns to T_H). Because every step is reversible, the Carnot cycle generates no net entropy. It is an idealization — real cycles are irreversible and less efficient.
Sketch the Carnot cycle on both a PV diagram and a TS (temperature-entropy) diagram. On the TS diagram, the cycle is a perfect rectangle, making it immediately clear that the enclosed area represents net work and the efficiency depends only on the two temperatures.
You know from the second law of thermodynamics that heat flows spontaneously from hot to cold, and that any irreversible process generates entropy. You also know from isothermal and adiabatic processes what happens to a gas during each type of step individually. The Carnot cycle chains four of these steps together in a particular order to construct the most efficient possible heat engine — and the argument for why it is most efficient is itself a proof by contradiction using the second law.
The cycle runs as follows. The gas starts at temperature T_H. Step 1 (isothermal expansion): the gas expands while in thermal contact with the hot reservoir at T_H, absorbing heat Q_H and doing work. Temperature stays at T_H because the process is quasi-static and isothermal. Step 2 (adiabatic expansion): thermal contact is broken and the gas continues expanding with no heat flow, doing more work as its temperature drops from T_H to T_C. Step 3 (isothermal compression): the gas is placed in contact with the cold reservoir at T_C and compressed, rejecting heat Q_C while temperature stays at T_C. Step 4 (adiabatic compression): thermal contact is broken again and the gas is compressed back to the starting state as temperature rises from T_C to T_H. Net result: work W = Q_H − Q_C was extracted, and the gas has returned to its initial state (it is a cycle).
The efficiency η = W/Q_H = 1 − Q_C/Q_H. For the Carnot cycle specifically, because every step is reversible, no entropy is generated anywhere. Entropy enters with the hot reservoir (ΔS_H = −Q_H/T_H) and leaves with the cold reservoir (ΔS_C = +Q_C/T_C). Zero net entropy means Q_H/T_H = Q_C/T_C, so Q_C/Q_H = T_C/T_H, giving Carnot efficiency η_C = 1 − T_C/T_H. This depends only on the two reservoir temperatures, not on the working substance.
Why is this the maximum? Suppose a better engine existed with efficiency η > η_C. Run the Carnot cycle backwards as a refrigerator (pumping heat from cold to hot using work) and power it with the hypothetical better engine. The net effect would be a machine that transfers heat from a cold reservoir to a hot one with no other effect — a violation of the second law's Clausius statement. Therefore no engine operating between T_H and T_C can exceed Carnot efficiency. The T-S diagram makes this especially clean: on T-S axes, the Carnot cycle is a perfect rectangle, net work equals the enclosed area, and the efficiency is immediately visible as (T_H − T_C)/T_H = 1 − T_C/T_H. Every real engine's cycle, when plotted on T-S axes, will have less enclosed area relative to the heat input, confirming the Carnot bound.