Thermodynamics Ques 34
34. 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 :
[2017, 2015]
(a) $90 J$
(b) $99 J$
(c) $100 J$
(d) $1 J$
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Answer:
Correct Answer: 34.(a)
Solution:
- (a) Given, efficiency of engine, $\eta=\frac{1}{10}$
work done on system $W=10 J$
Coefficient of performance of refigerator
$\beta=\frac{Q_2}{W}=\frac{1-\eta}{\eta}=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}=9$
Energy absorbed from reservoir
$Q_2=\beta W$
$ Q_2=9 \times 10=90 J $
A refrigerator works along the reverse direction of a heat engine.
BETA
