University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 17 - Temperature and Heat - Problems - Exercises - Page 575: 17.13

Answer

Death Valley: $1.9014 \mathrm{c}\mathrm{m}$ Greenland: $1.8964 \mathrm{c}\mathrm{m}$

Work Step by Step

Apply 17-6, $\Delta L=L_{0}\alpha\Delta T$ given $\alpha=2.6\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1},\quad L_{0}=1.90 \mathrm{c}\mathrm{m},$ Death Valley, $\Delta T=(48.0-20.0)^{o}C=28.0^{\mathrm{o}}\mathrm{C}$: $\Delta L=[2.6\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}](1.90 \mathrm{c}\mathrm{m})(28.0^{\mathrm{o}}\mathrm{C})=1.4\times 10^{-3} \mathrm{c}\mathrm{m}$ $L_{1}=L_{o}+\Delta L=1.9014 \mathrm{c}\mathrm{m}$ Greenland, $\Delta T=(-53.0-20.0)^{o}C=-73.0^{\mathrm{o}}\mathrm{C}$: $\Delta L=[2.6\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}](1.90 \mathrm{c}\mathrm{m})(-73.0^{\mathrm{o}}\mathrm{C})=-3.6\times 10^{-3} \mathrm{c}\mathrm{m}$ $L_{1}=L_{o}+\Delta L=1.8964 \mathrm{c}\mathrm{m}$
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