First slide

Heat engine

Question

A Carnot engine, having an efficiency of $η = \frac{1}{10}$ 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

Difficult

A

90 J
B

1 J
C

100 J
D

99 J

Solution

For Carnot engine
$Efficiency, η = 1 - \frac{T_{1}}{T_{2}}; \frac{1}{10} = 1 - \frac{T_{1}}{T_{2}}$
$\frac{T_{1}}{T_{2}} = 1 - \frac{1}{10} = \frac{9}{10} . . . . (i)$
For refrigerator
$∴ \frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}$
$or \frac{Q_{1} + W}{Q_{1}} = \frac{T_{2}}{T_{1}}$
$\frac{Q_{1} + 10}{Q_{1}} = \frac{10}{9}$
$1 + \frac{10}{Q_{1}} = \frac{10}{9}; \frac{10}{Q_{1}} = \frac{10}{9} - 1 = \frac{1}{9}$
$Q_{1} = 90 J$
$So, 90 J heat is absorbed at lower temperature.$

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