A heat engine operates between a hot reservoir at temperature T_H and a cold reservoir at temperature T_C. The engine performs one complete cycle. Explain why the thermal efficiency of this engine must always be less than 1, even if all mechanical components are frictionless.

The second law of thermodynamics states that the total entropy of an isolated system cannot decrease. A heat engine is part of an isolated system together with the two reservoirs, so the entropy change of the engine plus reservoirs must be ≥ 0. Therefore some heat must always be transferred to the cold reservoir, i.e. Q_C > 0.
Because Q_C > 0, not all of the heat absorbed from the hot reservoir can be converted into work: W = Q_H – Q_C, so W < Q_H.
The thermal efficiency is defined as η = W/Q_H = 1 – Q_C/Q_H. Since Q_C/Q_H is a positive number, η must be less than 1.
The Carnot efficiency η_C = 1 – T_C/T_H gives the theoretical upper limit for any engine operating between T_H and T_C. For any finite T_C > 0, η_C < 1, which confirms that a 100 % efficient heat engine is impossible.

Examiner_tips: 1) Mention the second law and entropy increase. 2) Show the algebraic relation η = 1 – Q_C/Q_H. 3) State that Q_C > 0. 4) Reference Carnot efficiency as the upper bound.

Common_mistakes: 1) Forgetting that Q_C must be >0. 2) Confusing η = Q_H/W instead of W/Q_H. 3) Ignoring the Carnot limit and claiming η can reach 1 for some engines.