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(D) Let the initial temperature of the source be $T_1$ and the sink be $T_2$. The initial efficiency is $\eta = 1 - \frac{T_2}{T_1}$.When the source t

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$36$

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(D) Let the initial temperature of the source be $T_1$ and the sink be $T_2$. The initial efficiency is $\eta = 1 - \frac{T_2}{T_1}$.When the source temperature is increased by $25 \%$,the new source temperature $T_1' = 1.25 T_1$.The new efficiency is $\eta' = 1 - \frac{T_2}{1.25 T_1}$.According to the problem,$\eta' = \eta + 0.80 \eta = 1.8 \eta$.Substituting the expressions: $1 - \frac{T_2}{1.25 T_1} = 1.8 \left( 1 - \frac{T_2}{T_1} ight)$.Assuming $T_1 = 100 	ext{ K}$ for simplicity,$\eta = 1 - \frac{T_2}{100}$.Then $1 - \frac{T_2}{125} = 1.8 \left( 1 - \frac{T_2}{100} ight)$.$1 - \frac{T_2}{125} = 1.8 - 0.018 T_2$.$0.018 T_2 - 0.008 T_2 = 1.8 - 1 \Rightarrow 0.01 T_2 = 0.8 \Rightarrow T_2 = 80 	ext{ K}$.The new efficiency is $\eta' = 1 - \frac{80}{125} = \frac{125 - 80}{125} = \frac{45}{125} = 0.36 = 36 \%$.

When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$,its efficiency increases by $80 \%$. The new efficiency of the engine is (in $\%$)

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