Answer
c) $47kW$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=mgV$
$P_{p}=400kg*9.81\frac{m}{s^2}*12\frac{m}{s}=47.09kW$
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 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 107: 2-129
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