A heat engine operates with the cold reservoi | vedclass

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(B) For a reversible (Carnot) heat engine,the efficiency is given by $\eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{T_L}{T_H}$.Given:Temperature of the cold

Vedclass · Accepted Answer

$540$

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(B) For a reversible (Carnot) heat engine,the efficiency is given by $\eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{T_L}{T_H}$.Given:Temperature of the cold reservoir,$T_L = 324 \; K$.Heat taken from the hot reservoir,$Q_H = 300 \; J$.Heat delivered to the cold reservoir,$Q_L = 180 \; J$.Using the relation for a reversible engine: $\frac{Q_L}{Q_H} = \frac{T_L}{T_H}$.Substituting the values:$\frac{180}{300} = \frac{324}{T_H}$.$T_H = \frac{324 	imes 300}{180}$.$T_H = \frac{324 	imes 5}{3} = 108 	imes 5 = 540 \; K$.Thus,the minimum temperature of the hot reservoir is $540 \; K$.

$A$ heat engine operates with the cold reservoir at a temperature of $324 \; K$. If the heat engine takes $300 \; J$ of heat from the hot reservoir and delivers $180 \; J$ of heat to the cold reservoir per cycle,what is the minimum temperature of the hot reservoir in $K$?

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