Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 464: 1


$1930~cm^3$ of water has the same mass as $100~cm^3$ of gold.

Work Step by Step

We can find the mass $M$ of $100~cm^3$ of gold as: $M = \rho~V$ $M = (19.3\times 10^3~kg/m^3)(100\times 10^{-6}~m^3)$ $M = 1.93~kg$ We then find the volume of this mass of water; $V = \frac{M}{\rho}$ $V = \frac{1.93~kg}{1.00\times 10^3~kg/m^3}$ $V = 1.93\times 10^{-3}~m^3$ $V = (1.93\times 10^{-3}~m^3)(\frac{1~cm^3}{10^{-6}~m^3})$ $V = 1930~cm^3$ $1930~cm^3$ of water has the same mass as $100~cm^3$ of gold.
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