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 302: 81



Work Step by Step

We know that $H=\frac{dQ}{dt}=mc\frac{dT}{dt}=-kA\frac{dT}{dx}$ $\implies mc\frac{dT}{dt}=-\frac{Ak}{L}(T+15C^{\circ})$ $(mcR)\frac{dT}{T+15C^{\circ}}=-dt$ $(mcR)\int_{T=20C^{\circ}}^{0C^{\circ}} \frac{dT}{T+15C^{\circ}}=-\int _0 ^tdt$ We plug in the known values to obtain: $6.5\times 10^6\times 6.67\times 10^{-3}ln(T+15)|^0_{20}=-t$ This simplifies to: $t=36735s$ $t=\frac{36735}{3600}=10hr$
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