A Carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J , the amount of energy - Physics MCQ NEET 2017 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Efficiency η = \frac{1}{10} ⟹ η = \frac{W}{Q _{1}} ⟹ Q_{1} = 10 W = 10 \times 10 = 100 J .Heat rejected to sink in engine mode Q_{2} = Q_{1} - W = 100 - 10 = 90 J .Coefficient of performance of refrigerator β = \frac{Q _{2}}{W} = \frac{1 - η}{η} = \frac{9/10}{1/10} = 9 . 9 = \frac{Q _{ab sor b e d}}{10} ⟹ Q_{ab sor b e d} = 90 J .

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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:

1 J

90 J

99 J

100 J

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