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 201: 4-64

Answer

$T_2=90.31°C$

Work Step by Step

For the stationary insulated constant-volume closed system, the energy balance reduces to: $-W_e=\Delta U \rightarrow w_e\Delta t=mc_v\Delta T$ For an ideal gas: $PV=mRT$ Plugging in the values of $P_1=100\ kPa, V_1=48\ m³, T_1=20°C=293.15\ K, R=0.2870\ kJ/kg.K$ $m=57.052\ kg$ With $w_e=0.1\ kW, \Delta t =8h=28800s, c_v=0.718\ kJ/kg.K$ We get to $\Delta T=70.31°C$, hence $T_2=90.31°C$
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