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A heat engine operates with the cold reservoir at temperature 324 K.
The minimum temperature of the hot reservoir, if the heat engine takes 300 J heat from the hot reservoir and delivers 180 J heat to the cold reservoir per cycle, is ___________K.

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answer is 540.

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Detailed Solution

T_c = 324 k
T_H = ?
Q_H = 300 J
Q_L = 180 J

$\begin{array}{l} 1 - \frac{Q_{L}}{Q_{H}} = 1 - \frac{T_{L}}{T_{H}} \\ \frac{Q_{L}}{Q_{H}} = \frac{T_{L}}{T_{H}} \\ T_{H} = \frac{Q_{H}}{Q_{L}} \times T_{L} = \frac{300}{180} \times 324 = 540 K \end{array}$

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