Question

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

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

Answer 1

The correct option is A $90J$

efficiency of carnot cycle is given as
$\eta =\frac{W}{Q_H}$
where W is work done and $Q_H$ is amount of heat added to system.
$Q_H=\frac{W}{\eta }=100J$
$Q_C=Q_H-W=100-10=90J$