Thermodynamics - Result Question 37

39. 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: 39. (a)

Solution:

  1. (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.