A Carnot engine has an efficiency of $50\%$ when its source is at a temperature $327^\circ ext{C}$. The tempe

Question

A Carnot engine has an efficiency of $50\%$ when its source is at a temperature $327^\circ	ext{C}$. The temperature of the sink is

Accepted Answer

Efficiency $\eta = 1 - \frac{T_2}{T_1}$. Given $\eta = 0.5$ and $T_1 = 327 + 273 = 600 	ext{K}$. So, $0.5 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.5 \implies T_2 = 300 	ext{K}$. Converting to Celsius, $T_2 = 300 - 273 = 27^\circ	ext{C}$.

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