SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 103 of 200

Marks: +1, -0

1 kJ heat is transferred to the surroundings from a reservoir at a temperature of 527°C. The change in entropy of the reservoir is:

Solution:

Heat transferred to the surrounding (dQ) = -1 kJ = -1000 J
Temperature of the reservoir (T) = 527°C = 527 + 273 = 800 K
Entropy (dS) = dQ/T

{ds}={-1000}/{800}=-1.25 {J} \/ {K}

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