Question

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

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

Answer 1

Answer: (C)

$90J$

Efficiency of heat engine $\eta =\frac{Q_H-Q_L}{Q_H}=0.1$ (Given)
Coefficient of performance of refrigerator $\beta =\frac{Q_L}{Q_H-Q_L}$ .....(2)
Or $\beta =\frac{Q_L}{Q_H\eta }$ ......(3)
Given : $W=Q_H-Q_L=10J$
Putting in (2) we get $\beta =\frac{Q_L}{10}$ ...(4)
From (4) and (5), we get
$\therefore$ $\frac{Q_L}{10}=\frac{Q_L}{Q_H\times 0.1}$
$⟹Q_H=100J$
Thus heat absorbed at lower temperature $Q_L=Q_H-W$
$\therefore$ $Q_L=100-10=90J$