Q. A Carnot engine, having an efficiency of $\eta = 1/10$ 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) 99 J
B) 90 J
C) 1 J
D) 100 J
Explanation
(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, $\beta = \frac{Heat\ absorbed(Q_{2})}{Work\ done\ (W)}$
$\Rightarrow \ \ 9 = \frac{Q_{2}}{10}$
$or\ \ \ Q_{2} = 90\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.