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: 15b


$\Delta S=+943\frac{J}{K}$

Work Step by Step

$m_{i1}+m_{i2}=m_{f1}+m_{f_2}$ Since $m_{f1}=m_{f2}$ $\implies m_{i1}+m_{i2}=2m_{f1}$ This can be rearranged as: $m_{f1}=\frac{m_{i1}+m_{i2}}{2}$ $m_{f1}=\frac{1.773+0.227}{2}=1Kg$ Now, $m=m_{i1}-m_{f1}=1.773-1=0.773Kg$ $0.773Kg$ of ice must be melted that is why heat must be added so heat is positive. We also know that: $Q=mL_F$ We plug in the known values to obtain: $Q=(0.773)(333000)=257409J$ Now, $\Delta S=\int _{Q_i} ^{Q_f}\frac{dQ}{T}=\frac{Q}{T}$ We plug in the known values to obtain: $\Delta S=\frac{257409}{273}=+943\frac{J}{K}$
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