A Carnot engine having an efficiency of 1/10 | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$.From this,we find $\frac{T_2}{T_1} = 1 - \frac{1}{10} =

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$.From this,we find $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \frac{9}{10}$.When the same engine is used as a refrigerator,its coefficient of performance $\beta$ is given by $\beta = \frac{T_2}{T_1 - T_2}$.Dividing numerator and denominator by $T_1$,we get $\beta = \frac{T_2/T_1}{1 - T_2/T_1}$.Substituting the value $\frac{T_2}{T_1} = \frac{9}{10}$,we get $\beta = \frac{9/10}{1 - 9/10} = \frac{9/10}{1/10} = 9$.The coefficient of performance is also defined as $\beta = \frac{Q_2}{W}$,where $Q_2$ is the heat absorbed from the cold reservoir and $W$ is the work done on the system.Given $W = 10 \, J$,we have $Q_2 = \beta 	imes W = 9 	imes 10 = 90 \, J$.

$A$ Carnot engine having an efficiency of $1/10$ as a heat engine is used as a refrigerator. If the work done on the system is $10 \, J$,the amount of energy absorbed from the reservoir at a lower temperature is .......... $J$.

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