Physics: Principles with Applications (7th Edition)

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

Chapter 13 - Temperature and Kinetic Theory - Problems: 12

Answer

The brass rod must be heated to ${8.1\times10^{2}}^{\circ}C$ to be $1.5%$ longer than it was at $25^{\circ}C$.

Work Step by Step

Let $l=$ the length of the brass rod at $25^{\circ}C$. $$\Delta l=al_{0}\Delta T$$$$1.015l-l=(19\times10_{-6}\frac{1}{^{\circ}C})(l)(T_{f}-25^{\circ}C)$$$$.015=(19\times10_{-6}\frac{1}{^{\circ}C})(T_{f}-25^{\circ}C)$$ $$T_{f}=814.47^{\circ}C$$
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