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Q. The increase in entropy in $JK ^{-1}$ of a substance when it absorbs $1\, kJ$ of heat energy at $3\,K$ is

TS EAMCET 2019

Solution:

Given,

Heat$(q)=1 kJ =1000 \,J$

Temperature $=3 \,K $

Entropy $(\Delta S) =\frac{\text { Absorb heat }(q)}{\text { Temperature }(T)} $

$=\frac{1000 \,J }{3 K }$ Entropy $(S)=333.3 J / K$