A cyclic engine absorbs 1200 J per cycle from a reservoir at 540 K and rejects heat only to a reservoir at 290 K. Its designer claims work output 615.5556 J per cycle. (a) Calculate the maximum possible efficiency and work output. (b) Use the claimed work to calculate the total reservoir entropy change per cycle. (c) Decide whether the claim is possible and explain the role of reversibility. Use reservoir entropy change ΔS = Q/T, with signed heat entering each reservoir.

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Maximum efficiency 46.3%; maximum work 555.6 J; claimed total reservoir ΔS = -0.2069 J K⁻¹. The claim is impossible. For two reservoirs the reversible Carnot limit is η_max = 1 − T_c/T_h = 46.3%. Temperatures must be absolute, because the ratio follows from thermal entropy. Maximum work is η_max Q_h = 555.6 J. The corresponding minimum rejected heat is Q_hT_c/T_h = 644.4 J. The claimed work implies Q_c = Q_h − W_claim = 584.4 J. The hot reservoir loses entropy Q_h/T_h, while the cold reservoir gains Q_c/T_c. Thus ΔS_total = −Q_h/T_h + Q_c/T_c = -0.2069 J K⁻¹, which is negative. Since the engine returns to its initial state, it cannot hide a net entropy increase internally. The claim violates the second law even though its energy accounting can satisfy the first law. A reversible engine reaches the Carnot limit with zero total entropy change; real irreversible operation produces positive total entropy and lower efficiency.

Audit a proposed heat engine · set 2

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