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: 14

Answer

$6.588$ mL of water can be added to the glass jar.

Work Step by Step

We first calculate the change in volume of the glass due to thermal expansion. $$\Delta V=bV_{0}\Delta T $$ $$V_{f}-450mL=(27\times10^{-6})(450mL)(20.0^{\circ}C-100^{\circ}C)$$ $$V_{f}=449.028 mL$$ Now, we calculate the change in volume of the water due to thermal expansion. $$\Delta V=bV_{0}\Delta T $$ $$V_{f}-450mL=(210\times10^{-6})(450mL)(20.0^{\circ}C-100^{\circ}C)$$ $$V_{f}=442.44 mL$$ Therefore, the amount of water that could be added is $449.028 mL-442.44 mL=6.588mL$.
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