Fundamentals of Physics Extended (10th Edition)

by Halliday, David; Resnick, Robert; Walker, Jearl

Published by Wiley

ISBN 10: 1-11823-072-8

ISBN 13: 978-1-11823-072-5

Chapter 18 - Temperature, Heat, and the First Law of Thermodynamics - Problems - Page 541: 11

Answer

$V_{30^{\circ}}=49.87cm^3$

Work Step by Step

We know that $\Delta{V}=V\beta\Delta{T}$ Putting the values, we get $\Delta{V}=50\times(3\times29\times10^{-6})(60-30)=0.13 cm^3$ $V_{30^{\circ}}=V-\Delta{V}=50-0.13=49.87cm^3$ $V_{30^{\circ}}=49.87cm^3$

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