College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 13 - Problems - Page 496: 11


The glass rod must be heated to a temperature of $116.7^{\circ}C$

Work Step by Step

We can find an expression for the increase in length of the lead rod when it is heated: $\Delta L = \alpha~\Delta T~L$ $\Delta L = (29\times 10^{-6}~K^{-1})(30.0~K)~L$ $\Delta L = 870~L~\times 10^{-6}$ We can find the required change in temperature of the glass rod so that the increases in length of the two rods are equal: $\Delta L = \alpha~\Delta T~L$ $870~L~\times 10^{-6} = (9\times 10^{-6}~K^{-1})~\Delta T~L$ $\Delta T = \frac{870~L~\times 10^{-6}}{9~L\times 10^{-6}}$ $\Delta T = 96.7~K$ $\Delta T = 96.7^{\circ}C$ Since the initial temperature was $20.0^{\circ}C$, the glass rod must be heated to a temperature of $116.7^{\circ}C$.
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