Essential University Physics: Volume 1 (3rd Edition)

by Wolfson, Richard

Published by Pearson

ISBN 10: 0321993721

ISBN 13: 978-0-32199-372-4

Chapter 16 - Section 16.2 - Heat Capacity and Specific Heat - Example - Page 288: 16.1

Answer

4000 seconds

Work Step by Step

We first find the value of Q: $Q = mc \Delta t = (150)(4184)(50-18) \approx 20 \ MJ$ We now find the change in time: $\Delta t = \frac{20 \times 10^6 \ J }{5 \times 10^3 \ J/S} = 4000 \ s$

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