Answer
$Q=86.88kW$
Work Step by Step
At $40^{\circ}C$, from table A-4:
$h_{fg}=2406\frac{kJ}{kg}$
$Q=mh_{fg}=130\frac{kg}{h}*2406\frac{kJ}{kg}=312780\frac{kJ}{h}*\frac{1h}{3600s}=86.88kW$
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 154: 3-44
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