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(B) Given: Heat absorbed from the hot reservoir $Q_{	ext{in}} = 300 \, J$.Heat rejected to the cold reservoir $Q_{	ext{out}} = 240 \, J$.Temperature

Vedclass · Accepted Answer

$500$

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(B) Given: Heat absorbed from the hot reservoir $Q_{	ext{in}} = 300 \, J$.Heat rejected to the cold reservoir $Q_{	ext{out}} = 240 \, J$.Temperature of the cold reservoir $T_{2} = 400 \, K$.For a reversible heat engine (Carnot engine),the efficiency is given by $\eta = 1 - \frac{Q_{	ext{out}}}{Q_{	ext{in}}}$.Substituting the values: $\eta = 1 - \frac{240}{300} = 1 - 0.8 = 0.2$.Also,the efficiency of a Carnot engine is $\eta = 1 - \frac{T_{2}}{T_{1}}$.Equating the two expressions for efficiency: $0.2 = 1 - \frac{400}{T_{1}}$.$\frac{400}{T_{1}} = 1 - 0.2 = 0.8$.$T_{1} = \frac{400}{0.8} = 500 \, K$.Thus,the minimum temperature of the hot reservoir must be $500 \, K$.

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