Answer
$V=0.4935m^3$
$U=2639.28kJ$
Work Step by Step
As $T\gt T_{sat}$ is a superheated vapor
From table A-13:
$\upsilon=0.061687\frac{m^3}{kg}$
$V=m\upsilon=8kg*0.061687\frac{m^3}{kg}=0.4935m^3$
$U=mu=8kg*329.91\frac{kJ}{kg}=2639.28kJ$
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 3 - Properties of Pure Substances - Problems - Page 155: 3-62
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