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

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

Answer 1

Answer: (A)

90 J

Answer 2

$\beta=\frac{1-\eta}{\eta}$
$=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}$
$\beta=9$
$\beta=\frac{Q_{2}}{W}$
$Q_{2}=9 \times 10=90\, J$