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 301: 76


The proof is below.

Work Step by Step

We start with the given equation: $ H = -kA \frac{dT}{dr}$ We know that the area of a cylinder is $2\pi rL$, so we find: $ \frac{H}{2\pi R} dr = -kLdT $ Applying the integral on both sides, it follows: $ \frac{HlnR_2 - HlnR_1}{2\pi}= -kL(T_1-T_2)$ $H = \frac{2\pi kL(T_1-T_2)}{ln(R_2/R_1)}$
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