College Physics (4th Edition)

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

Chapter 13 - Problems - Page 497: 24

Answer

The materials should be heated to a temperature of $423^{\circ}C$

Work Step by Step

We can find the required change in diameter of the hole relative to the change in diameter of the bolt: $\Delta L = 1.0000~cm-0.9980~cm = 0.002~cm$ We can find an expression for increase in diameter of the hole $\Delta L_h$ when it is heated: $\Delta L_h = \alpha_h~\Delta T~L_h$ We can find an expression for increase in diameter of the bolt $\Delta L_b$ when it is heated: $\Delta L_b = \alpha_b~\Delta T~L_b$ Note that when the washer fits over the bolt, $\Delta L_h-\Delta L_b = 0.002~cm$, We can find the required change in temperature: $\Delta L_h-\Delta L_b = 0.002~cm$ $\alpha_h~\Delta T~L_h - \alpha_b~\Delta T~L_b = 0.002~cm$ $\Delta T = \frac{0.002~cm}{\alpha_h~L_h - \alpha_b~L_b}$ $\Delta T = \frac{0.002~cm}{(17\times 10^{-6}~K^{-1})(0.9980~cm) - (12\times 10^{-6}~K^{-1})~(1.0000~cm)}$ $\Delta T = 403~K$ $\Delta T = 403^{\circ}C$ The required increase in temperature is $403^{\circ}C$, so the materials should be heated to a temperature of $423^{\circ}C$
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