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 411: 64

Answer

a. $44C^{\circ}$ b. The bullet doesn't melt

Work Step by Step

a. There is more KE in the beginning; half of the KE lost goes toward heating the bullet. $$\frac{1}{2}(\frac{1}{2}m(v_i^2-v_f^2))=mc_{lead}\Delta T$$ $$\Delta T=\frac{(v_i^2-v_f^2)}{4 c_{lead}}=\frac{(220m/s)^2-(160m/s)^2}{4 (130 J/(kg\cdot C^{\circ}))}=44C^{\circ}$$ b. The final temperature of the bullet is far below the melting temperature of lead. The bullet doesn’t melt at all.
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