A carnot engine , having an Efficiency of η = 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 temprature is [AIEEE 2007]
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Explanation:- For carnot engine using as
refrigerator
W = Q₂(T₁/T₂- 1) _______(1)(For heat engine)
We know that efficiency of carnot cycle
η = 1-T₂/T₁
- ∵ given η = 1/10
- ∴T₂/T₁ = 1-1/10
- T₂ /T₁ = 9/10
- T₁/T₂ = 10/9
- so , put the value of T₂/T₁ in first then we will get energy absorbed from the reservoir .
given work done = 10J
10= Q₂ × (T₁/T₂ -1)
10= Q₂(10/9-1) =
Q₂ = 90 J Answer
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