Answer
(i) 853 J/K
(ii) 670 J/K
Work Step by Step
Recall: $\Delta S=\frac{q_{rev}}{T}$
(i) $\Delta S=\frac{250\times10^{3}\,J}{(20+273.15)K}=853\,J/K$
(ii) $\Delta S=\frac{250\times10^{3}\,J}{(100+273.15)K}=670\,J/K$
Physical Chemistry: Thermodynamics, Structure, and Change
by Atkins, Peter; de Paula, Julio
Published by
W. H. Freeman
ISBN 10:
1429290196
ISBN 13:
978-1-42929-019-7
Chapter 3 - Topic 3A - Entropy - Exercises - Page 149: 3A.3(b)
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