A Carnot engine, having an efficiency of = 1/10 as a heat engine, is used as a refrigerator. If the work done on the system is 10 J , the amount of en - Physics MCQ AIEEE 2007 2026 | ExamTest.live

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A Carnot engine, having an efficiency of η = 1/10 as a 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

For a Carnot engine: η = 1 - \frac{T _{2}}{T _{1}} = \frac{1}{10} .This gives \frac{T _{2}}{T _{1}} = 1 - \frac{1}{10} = \frac{9}{10} .When used as a refrigerator, the coefficient of performance (β) is: β = \frac{Q _{2}}{W} = \frac{T _{2}}{T _{1} - T _{2}} Divide numerator and denominator by T_{1} : β = \frac{T _{2} / T _{1}}{1 - T _{2} / T _{1}} = \frac{9/10}{1/10} = 9 Given work done W = 10 J : Q_{2} = β \times W = 9 \times 10 = 90 J

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A Carnot engine, having an efficiency of η = 1/10 as a 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:

99 J

90 J

1 J

100 J

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