SATHEE: Thermodynamics - Result Question 37

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) $1 J$

Show Answer

Answer:

Correct Answer: 39. (a)

Solution:

(a) Given, efficiency of engine, $η = \frac{1}{10}$

work done on system $W = 10 J$

Coefficient of performance of refigerator

$β = \frac{Q_{2}}{W} = \frac{1 - η}{η} = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}} = 9$

Energy absorbed from reservoir

$Q_{2} = β W$

$Q_{2} = 9 \times 10 = 90 J$

A refrigerator works along the reverse direction of a heat engine.

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Thermodynamics - Result Question 37

39. A carnot engine having an efficiency of 110 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 :

Answer:

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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 :