Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 19 - Exercises and Problems - Page 351: 66


Please see the work below.

Work Step by Step

As We know that $\Delta S=\int_{25}^{10}n C\frac{dT}{T}$ We plug in the known values to obtain: $\Delta S=40\times \frac{31}{(343)^3}\int_{25}^{10}T^2dT$ $=\frac{3\times 10^{-5}}{3}((10)^3-(25)^3)=-.148\frac{J}{K}$ The negative sign shows that copper is cooled.
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