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: 9

Answer

$\Delta$$S=0.75\frac{J}{K}$

Work Step by Step

Given: $m = 0.01 Kg$ $T_o = -10^{\circ}C = 263K$ $T = 15^{\circ}C = 288K$ $c_{ice}=2220\frac{J}{Kg*K}$ $c_{water}=4187\frac{J}{Kg*K}$ $L_f=3.33*10^{5}\frac{J}{Kg}$ The total energy absorbed by the ice can be given by: $Q = mc_{ice}\Delta$$T + mL_f+mc_{water}\Delta$$T$ Plugging in: $Q_{ice} = (0.01 Kg)(2220\frac{J}{Kg*K})(273K-263K) + (0.01 Kg)(3.33*10^{5}\frac{J}{Kg})+(0.01Kg)(4187\frac{J}{Kg*K})(288K-273K) = 4180.05 J$ $Q_{lake} = -Q_{ice} = -4180.05$ Since the temperature of the lake will remain essentially unchanged, we can easily find the entropy change of the lake: $\Delta$$S_{lake}=\frac{Q_{lake}}{T}=\frac{-4180.05J}{288K}=-14.514\frac{J}{K}$ Since ice at $-10^{\circ}C$ has to be be heated, then melted, then heated more before it reaches the temperature of the lake: $\Delta$$S_{ice}=\int_i^f\frac{mc_{ice}dT}{T}+\frac{mL_f}{T}+\int_i^f\frac{mc_{water}dT}{T}$ Since mass and specific heat are not variables, we can pull them out of the integral: $\Delta$$S_{ice}=mc_{ice}\int_{T_{o}}^T\frac{dT}{T}+\frac{mL_f}{T}+mc_{water}\int_{T_o}^T\frac{dT}{T}$ Please note that the temperatures are relative to the stages. So for example, at the ice stage, $T=273$ and $T_o=263$ Integrating: $\Delta$$S_{ice}=mc_{ice}ln\frac{T}{T_o}+\frac{mL_f}{T}+mc_{water}ln\frac{T}{T_o}$ Plugging in: $\Delta$$S_{ice}=(0.01 Kg)(2220\frac{J}{Kg*K})ln(\frac{273K}{263K}) + \frac{(0.01 Kg)(3.33*10^{5}\frac{J}{Kg})}{273K}+(0.01Kg)(4187\frac{J}{Kg*K})ln(\frac{288K}{273K})=15.266\frac{J}{K}$ Finally, add the two entropy values to find the total entropy: $\Delta$$S=\Delta$$S_{ice}+\Delta$$S_{lake}=15.266-14.514\approx0.75\frac{J}{K}$ I only use $\approx$ because I round to two significant figures.
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