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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = \frac{1}{5}$,we have $\frac{1}{5} = 1 - \frac{T_2}{T_1}$,

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$150 \ K, 120 \ K$

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = \frac{1}{5}$,we have $\frac{1}{5} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{4}{5}$ or $T_2 = 0.8 T_1$.When $T_2$ is lowered by $45 \ K$,the new temperature is $T_2' = T_2 - 45$.The new efficiency is $\eta' = \frac{1}{2}$,so $\frac{1}{2} = 1 - \frac{T_2 - 45}{T_1}$.This simplifies to $\frac{T_2 - 45}{T_1} = \frac{1}{2}$,or $T_2 - 45 = 0.5 T_1$.Substitute $T_2 = 0.8 T_1$ into the equation: $0.8 T_1 - 45 = 0.5 T_1$.$0.3 T_1 = 45$,which gives $T_1 = \frac{45}{0.3} = 150 \ K$.Now,find $T_2$: $T_2 = 0.8 	imes 150 = 120 \ K$.Thus,the temperatures are $150 \ K$ and $120 \ K$.

An engine operating between temperatures $T_1$ and $T_2$ has efficiency $\eta = \frac{1}{5}$. When $T_2$ is lowered by $45 \ K$,its efficiency becomes $\eta' = \frac{1}{2}$. Temperatures $T_1$ and $T_2$ are respectively:

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