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 - Page 386: 17

Answer

$T=-213^oC$

Work Step by Step

Given: $l_{bi}=0.04378m$ $l_{ii}=0.04371m$ $T_i=15^oC$ $l_{bi}+\Delta l_{b}=l_{ii}+\Delta l_{i}$ $l_{bi}+\alpha_b l_{ib}\Delta T=l_{ii}+\alpha_i l_{ii}\Delta T$ $\Delta T=\frac{l_{bi}-l_{ii}}{\alpha_i l_{ii}-\alpha_b l_{ib}}=\frac{0.04378m-0.04371m}{(12\times10^{-6})(0.04371m)-(19\times10^{-6})(0.04378m)}=-227.79^oC$ $15^oC-227.79^oC=-213^oC$
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