300 , cal of heat is given to a heat engine a | vedclass

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Question

(D) For a reversible heat engine,the efficiency is given by $\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$.Given $Q_1 = 300 \, cal$ and $Q_2 = 225

Vedclass · Accepted Answer

$102$

Vedclass · Answer

(D) For a reversible heat engine,the efficiency is given by $\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$.Given $Q_1 = 300 \, cal$ and $Q_2 = 225 \, cal$.The source temperature $T_1 = 227^{\circ} C = 227 + 273 = 500 \, K$.Using the relation $\frac{Q_2}{Q_1} = \frac{T_2}{T_1}$:$\frac{225}{300} = \frac{T_2}{500}$.$T_2 = \frac{225 	imes 500}{300} = \frac{225 	imes 5}{3} = 75 	imes 5 = 375 \, K$.To convert the sink temperature to Celsius: $T_2(^{\circ} C) = 375 - 273 = 102^{\circ} C$.

$300 \, cal$ of heat is given to a heat engine and it rejects $225 \, cal$ of heat. If the source temperature is $227^{\circ} C$,then the temperature of the sink will be . . . . . . $^{\circ} C$.

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