Thermodynamic Cycles And Efficiency

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A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. If it absorbs 800 J of heat from the hot reservoir in each cycle, what is the work output and the heat rejected to the cold reservoir per cycle?

RankGuru · Accepted Answer

The Carnot efficiency is the maximum efficiency achievable by any heat engine operating between two thermal reservoirs: $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 = 50\%$. This follows from the Second Law of Thermodynamics, which establishes that no engine can be more efficient than the Carnot engine operating between the same reservoirs. The work output per cycle is: $W = \eta \times q_H = 0.5 \times 800 = 400$ J. By energy conservation (First Law applied to a complete cycle where $\Delta U_{cycle} = 0$): $q_C = q_H - W = 800 - 400 = 400$ J rejected to the cold reservoir. Option B (600 J work) would require an efficiency of 75%, which exceeds the Carnot limit. Option C (267 J work) corresponds to an efficiency of 33%, below the Carnot efficiency. Option D (200 J work) gives only 25% efficiency. The Carnot engine represents the theoretical upper bound on thermodynamic efficiency — a direct consequence of the Second Law. Plausibility check: with $T_C = T_H/2$, efficiency must be exactly 50%, and $q_C = q_H$ follows — magnitude and direction both consistent with energy conservation.

Thermodynamic Cycles And Efficiency

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