College Physics (4th Edition)

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

Chapter 13 - Problems - Page 496: 10

Answer

(a) The space that should be left between the slabs is $3.60~mm$ (b) The gap between the slabs is $10.8~mm$

Work Step by Step

(a) We can find the increase in length of each concrete slab when it is heated: $\Delta L = \alpha~\Delta T~L$ $\Delta L = (12\times 10^{-6}~K^{-1})(20.0~K)(15~m)$ $\Delta L = 3600\times 10^{-6}~m$ $\Delta L = 3.60\times 10^{-3}~m$ $\Delta L = 3.60~mm$ Since we can assume that each slab expands by half of this distance in each direction, the space that should be left between the slabs is $3.60~mm$ (b) We can find the decrease in length of each concrete slab when it is cooled from $40.0^{\circ}C$ to $-20.0^{\circ}C$: $\Delta L = \alpha~\Delta T~L$ $\Delta L = (12\times 10^{-6}~K^{-1})(-60.0~K)(15~m)$ $\Delta L = -10,800\times 10^{-6}~m$ $\Delta L = -10.8\times 10^{-3}~m$ $\Delta L = -10.8~mm$ Since we can assume that each slab decreases by half of this distance in each direction, the gap between the slabs is $10.8~mm$
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.