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A Carnot engine having an efficiency of $\eta = \frac{1}{10}$ as 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 lower temperature is
A
100 J
B
99 J
C
90 J
D
1 J
Detailed Solution
Coefficient of performance $\beta = \frac{Q_2}{W}$, and $\beta = \frac{1-\eta}{\eta}$
$\beta = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = 9$
$Q_2 = \beta W = 9\times 10 = 90$ J
$\beta = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = 9$
$Q_2 = \beta W = 9\times 10 = 90$ J