The following figure shows a Carnot engine th | vedclass

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(D) For the Carnot engine, the efficiency $\eta$ is given by $\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}$.Substituting the values, $\frac{W}{Q_1} = 1

Vedclass · Accepted Answer

$1.75$

Vedclass · Answer

(D) For the Carnot engine, the efficiency $\eta$ is given by $\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}$.Substituting the values, $\frac{W}{Q_1} = 1 - \frac{200}{400} = 1 - 0.5 = 0.5$, so $W = 0.5 Q_1$.For the Carnot refrigerator, the coefficient of performance $COP$ is given by $COP = \frac{Q_4}{W} = \frac{T_4}{T_3 - T_4}$.Substituting the values, $\frac{Q_4}{W} = \frac{250}{350 - 250} = \frac{250}{100} = 2.5$, so $W = \frac{Q_4}{2.5} = 0.4 Q_4$.Since the engine drives the refrigerator, the work $W$ produced by the engine is equal to the work $W$ consumed by the refrigerator.Therefore, $0.5 Q_1 = 0.4 Q_4$, which implies $\frac{Q_4}{Q_1} = \frac{0.5}{0.4} = 1.25$.For a Carnot refrigerator, $\frac{Q_3}{T_3} = \frac{Q_4}{T_4}$, so $Q_3 = Q_4 \left( \frac{T_3}{T_4} ight) = Q_4 \left( \frac{350}{250} ight) = 1.4 Q_4$.Now, we find the ratio $\frac{Q_3}{Q_1} = \frac{Q_3}{Q_4} 	imes \frac{Q_4}{Q_1} = 1.4 	imes 1.25 = 1.75$.

The following figure shows a Carnot engine that works between temperatures $T_1=400 \text{ K}$ and $T_2=200 \text{ K}$ and drives a Carnot refrigerator that works between temperatures $T_3=350 \text{ K}$ and $T_4=250 \text{ K}$. The quantity $\frac{Q_3}{Q_1}$ will be

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