A Carnot engine has an efficiency of 50 %. If | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.Given $\eta_1 = 0.5$,so $0.5 = 1 - \frac{T_L}{T_H} \implies \frac{T_L}{

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$266.7$

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.Given $\eta_1 = 0.5$,so $0.5 = 1 - \frac{T_L}{T_H} \implies \frac{T_L}{T_H} = 0.5 \implies T_L = 0.5 T_H$.When the sink temperature is reduced by $40 \, K$,the new sink temperature is $T_L' = T_L - 40$.The new efficiency $\eta_2$ increases by $30 \%$ of the original efficiency,so $\eta_2 = 0.5 + 0.3 = 0.8$.Using the formula for the new efficiency: $0.8 = 1 - \frac{T_L - 40}{T_H}$.Substituting $T_L = 0.5 T_H$: $0.8 = 1 - \frac{0.5 T_H - 40}{T_H}$.$0.8 = 1 - 0.5 + \frac{40}{T_H}$.$0.8 = 0.5 + \frac{40}{T_H} \implies 0.3 = \frac{40}{T_H}$.$T_H = \frac{40}{0.3} = 133.33 \, K$.Wait,checking the calculation: $0.5(1.3) = 0.65$. $0.65 = 1 - \frac{T_L - 40}{T_H} \implies \frac{T_L - 40}{T_H} = 0.35$.Since $\frac{T_L}{T_H} = 0.5$,then $0.5 - \frac{40}{T_H} = 0.35 \implies \frac{40}{T_H} = 0.15 \implies T_H = \frac{40}{0.15} = 266.67 \, K$.

$A$ Carnot engine has an efficiency of $50 \%$. If the temperature of the sink is reduced by $40 \, K$,its efficiency increases by $30 \%$. The temperature of the source will be $.... \, K$.

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