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:
$\Delta t = \frac{20 \times 10^6 \ J }{5 \times 10^3 \ J/S} = 4000 \ s$
Essential University Physics: Volume 1 (4th Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0-134-98855-8
ISBN 13:
978-0-13498-855-9
Chapter 16 - Section 16.2 - Heat Capacity and Specific Heat - Example - Page 298: 16.1
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