Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 17 - Exercises and Problems: 59

Answer

Please see the work below.

Work Step by Step

We know that $L=L_{\circ}(1+\alpha \Delta T)$ This can be rearranged as: $\Delta T=\frac{(\frac{L}{L_{\circ}})-1}{\alpha}$ We plug in the known values to obtain: $\Delta T=\frac{(\frac{1.000}{0.997})-1}{12\times 10^{-6}}$ $\Delta T=251^{\circ}K$
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