Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 16 - Exercises and Problems - Page 300: 60


-14.2 Watts

Work Step by Step

We know that the temperature difference between these two substances is 100 degrees Celsius. We know that the radius of the rod is equal to .015 meters, meaning that the cross-sectional area of the rod is $= .015^2 \times \pi = 7 \times 10^{-4} \ m^ 2$ Using the equation for heat flow, we find: $ H = \frac{-kA\Delta T}{\Delta x} = \frac{-80.4\times 7 \times 10^{-4}\times 100}{.4} = -14.2 \ W $
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