Question

A Carnot engine, having an efficiency of $\eta=1/10$ 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

• A. $99\,J$
• B. $90\,J$
• C. $1\,J$
• D. $100\,J$

Answer 1

Answer: (B)

Relation between coefficient of performance and efficiency of refrigerator is
$\beta =\frac{1 - \eta}{\eta}$
$\therefore \, \beta =\frac{1 - \frac{1}{10}}{\frac{1}{10}}=9$
Coefficient of performance, $\beta =\frac{\text{Heat absorbed} \left( Q_{2} \right)}{\text{Work done} \left( W \right)}$
$\Rightarrow 9=\frac{Q_{2}}{10}$
$Q_{2}=90 \, J$