The efficiency of a Carnot engine acting as a | vedclass

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(B) For a Carnot engine,the efficiency is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = 1/10$,we have $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \fra

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) For a Carnot engine,the efficiency is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta = 1/10$,we have $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \frac{9}{10}$.Therefore,the ratio of temperatures is $\frac{T_1}{T_2} = \frac{10}{9}$.For a refrigerator,the coefficient of performance $COP$ is defined as $\frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$.This can be rewritten as $\frac{Q_2}{W} = \frac{1}{\frac{T_1}{T_2} - 1}$.Substituting the values,we get $\frac{Q_2}{10} = \frac{1}{\frac{10}{9} - 1}$.$\frac{Q_2}{10} = \frac{1}{1/9} = 9$.Thus,$Q_2 = 10 	imes 9 = 90 \ J$.

The efficiency of a Carnot engine acting as a heat engine is $\eta = 1/10$. If it is used as a refrigerator and the work done on the system is $10 \ J$,then the heat absorbed from the low-temperature reservoir is ...... $J$.

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