Question

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]​

Answer 1

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Answer 2

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