Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 4 - Energy Analysis of Closed Systems - Problems - Page 206: 4-123

Answer

$\Delta t=456s=7.60\ min$

Work Step by Step

With $W_e=\Delta H\rightarrow\dot{W}_e\Delta t=m\Delta h_v$: Given $\Delta t=25\ min, m=0.5\ kg,\ \Delta h_v=2256.4\ kJ/kg$ we get: $\dot{W}_e=0.752\ kW$ For $\dot{W}_e\Delta t=mc\Delta T$ Given $m=1\ kg,\ \Delta T=(100-18)K, c_v=4.18\ kJ/kg.K$: $\Delta t=456s=7.60\ min$
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