Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 17 - Exercises and Problems - Page 325: 40

Answer

73 seconds

Work Step by Step

We find the total heat that is consumed: $ Q = mL + mc\Delta T \\ Q=(.12)(334,000)+(.12)(4184)(37) \\ Q= 59 \times 10^3 J$ Thus, we find: $ \Delta t = \frac{ 59 \times 10^3 J}{800W}=73s$
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