Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 20 - Entropy and the Second Law of Thermodynamics - Problems - Page 604: 6b

Answer

$\Delta S=30.24\frac{J}{K}$

Work Step by Step

We can find the entropy change as; $\Delta S=\frac{Q}{T}$......eq(1) Also, $Q=mL_v$ where $L_v$ is water heat of vaporization and its value is $2256000\frac{J}{kg}$. Thus, $Q=(0.012)(2256000)=11280J$ We plug in the known values in eq(1) to obtain: $\Delta S=\frac{11280}{373}=30.24\frac{J}{K}$
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