Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 19 - Exercises and Problems - Page 351: 52


$5.877 \ hours$

Work Step by Step

a) We know that the ice is 273 degrees Kelvin. Thus, it follows that the efficiency is: $=100(1-\frac{273}{420})=35$% b) The ice weighs 1,000 kilograms, so it will take $334\times 1000 = 334,000kJ$ of energy to melt it. Using the equation for efficiency, we find: $ Q = \frac{8.5}{.35}-8.5=15.8 \ kW$ Thus, we find: $\Delta t = \frac{334,000}{15.8}=5.877 \ hours$
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