Question

The specific heat capacity of a metal at low temprature (T) is given as Cₚ(kjk⁻¹kg⁻¹) = 32(T/400)³. A 100g vessel of this metal is to be cooled from 20k to 4k by a special refrigerator operating at room temprature (27°C) . The amount of work required to cool the vessel is [AIEEE 2011] ​

Answer 1

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Answer 2

Explanation:- Heat required to chsbhe the temprature of vessel by a small amount dT ,

-dQ = mCₚdT

Total heat required

-Q = m ₄∫²⁰32(T/400) ³dt

= 100× 10⁻³×32 [T⁴/4]⁴₂₀ /(400)³

⟹ Q = 0.001996KJ

work done required to maintain the temprature of sink to T₂

W = Q₁ -Q₂ /Q₂ =(T₁/T₂ -1) Q₂

⟹ W= (T₁-T₂/T₂)Q₂

For , T₂ = 20k

W₁ = 300-20/20× 0.001996 = 0.028 KJ

For T₂ = 4k

W₂ = 300-4/4 ×0.001996 = 0.148KJ

As temprature is changing from 20 k to 4K work done required will be more than W₁ but less than W₂

Hence, work required to cool the vessel is between 0.148Kj and 0.028KJ