Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 14 - Heat - General Problems - Page 410: 56

Answer

$0.68C^{\circ}$

Work Step by Step

Half of the initial gravitational potential energy goes toward heating the boulder. $$\frac{1}{2}mgh=mc_{marble}\Delta T$$ $$\Delta T=\frac{gh}{2 c_{marble}}=\frac{(9.80m/s^2)(120m)}{2 (860 J/(kg\cdot C^{\circ}))}=0.68C^{\circ}$$
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