College Physics (4th Edition)

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

Chapter 13 - Problems - Page 498: 28


$\frac{\Delta V}{V_0} = 3~\alpha~\Delta T$ $\beta = 3~\alpha$

Work Step by Step

Let the original area $V_0 = s_0^3$ We can find an expression for the new volume after the material expands: $V = (s_0+\Delta s)^3$ $V = s_0^3+3s_0^2~\Delta s+3s_0 (\Delta s)^2+(\Delta s)^3$ $V \approx s_0^3+3s_0^2~\Delta s$ We can find an expression for the ratio $\frac{\Delta V}{V_0}$: $\Delta V = V-V_0$ $\Delta V = (s_0^3+3s_0^2~\Delta s)-s_0^3$ $\Delta V = 3s_0^2~\Delta s$ $\Delta V = 3s_0^2~(\alpha~\Delta T~s_0)$ $\Delta V = 3s_0^3~\alpha~\Delta T$ $\Delta V = 3~V_0~\alpha~\Delta T$ $\frac{\Delta V}{V_0} = 3~\alpha~\Delta T$ Since $\frac{\Delta V}{V_0} = \beta~\Delta T$, we can see that $\beta = 3~\alpha$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.