Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 16 - Section 16.4 - Exercises and Problems - Page 299: 27

Answer

Please see the work below.

Work Step by Step

(a) We can find the heat capacity as $heat\space capacity =\frac{2520}{15}$ $heat\space capacity =168\frac{J}{K}$ (b) We know that $Q=mc\Delta T$ This can be rearranged as: $c=\frac{Q}{m\Delta T}$ We plug in the known values to obtain: $c=\frac{2520}{(0.350)(15)}$ $c=480\frac{J}{Kg\space C}$
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