Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - General Problems - Page 441: 65

Answer

a) ${43.64}^\circ \rm C$ b) $0.043\;\rm J/K$

Work Step by Step

a) Since the system is isolated, so the heat lost by water will be gained by aluminum. Thus, $$Q^-_{water}=Q^+_{Al}$$ $$m_{water}c_{water}(T_{i,water}-T_f)=m_{Al}c_{Al}(T_f-T_{i,Al})$$ Solving for $T_f$; $$m_{water}c_{water}T_{i,water}-m_{water}c_{water}T_f=m_{Al}c_{Al} T_f-m_{Al}c_{Al}T_{i,Al} $$ $$m_{water}c_{water}T_{i,water}+ m_{Al}c_{Al}T_{i,Al}=m_{Al}c_{Al} T_f+m_{water}c_{water}T_f $$ $$T_f=\dfrac{m_{water}c_{water}T_{i,water}+ m_{Al}c_{Al}T_{i,Al}}{m_{Al}c_{Al} +m_{water}c_{water} }$$ Plugging the known; $$T_f=\dfrac{ (0.15\cdot 4186\cdot 45)+ (0.11\cdot 900\cdot 35)}{[0.11\cdot 900] +[0.15\cdot 4186] }$$ $$T_f=\color{red}{\bf 43.64}^\circ \rm C$$ b) Now we can calculate the entropy of the system but remember that the temperature of the mixture is not constant all the time, so we need to take the average temperature of each one of them. Thus, $$T_{avg,Al}=\dfrac{T_{i,Al}+T_f}{2} =\dfrac{35+43.64}{2}=39.32\;\rm C$$ $$T_{avg,Al}=\rm \color{blue}{\bf 312.32}\;K$$ $$T_{avg,water}=\dfrac{T_{i,water}+T_f}{2} =\dfrac{45+43.64}{2}=44.32\;\rm C$$ $$T_{avg,water}=\rm \color{blue}{\bf 317.32}\;K$$ The entropy change of the system is given by $$\Delta S_{sys}=\Delta S_{water}+\Delta S_{AL}$$ $$\Delta S_{sys}=\dfrac{Q^+_{Al}}{T_{avg,Al}}+\dfrac{ \overbrace{ Q_{water}^- }^{{\color{red}{=-Q^+_{Al} }}}}{T_{avg,water}}$$ $$\Delta S_{sys}=m_{Al}c_{Al}(T_f-T_{i,Al})\left[\dfrac{1 }{T_{avg,Al}}-\dfrac{ 1}{T_{avg,water}}\right]$$ Plugging the known; $$\Delta S_{sys}=0.11\cdot 900\cdot 8.6\cdot\left[\dfrac{1 }{312.32}-\dfrac{ 1}{ 317.32}\right]$$ $$\Delta S_{sys}=\color{red}{\bf0.043}\;\rm J/K$$
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