Question

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 $10J$, the amount of energy absorbed from the reservoir at lower temperature is

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

Answer 1

The correct option is B $90J$

$\text{Given:-}$ $\eta =\frac{1}{10}$

$Workdone=10J$

$\text{Solution:-}$

The efficiency of Carnot engine,

$\eta =\frac{Q_H-Q_L}{Q_L}$

Coefficient of performance of a refrigerator

$\beta =\frac{1-\eta }{\eta }$

$\beta =\frac{1-\frac{1}{10}}{1/10}=9$

Also $\beta =\frac{Q_L}{W}$ (where W is the work done) or

$Q_L=\beta \times W=9\times 10=90J$

$\text{Hence the correct option is B}$