College Physics (4th Edition)

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

Chapter 13 - Problems - Page 497: 26


(a) $\Delta \rho = -\beta~\rho~\Delta T$ (b) $\frac{\Delta \rho}{\rho} = 2.35\times 10^{-3}$

Work Step by Step

(a) We can find the change in volume when the temperature changes: $\Delta V = \beta~\Delta T~V$ The new volume is: $~V+ \beta~\Delta T~V$ The new density is: $~\frac{m}{V+ \beta~\Delta T~V}$ We can find the change in density: $\Delta \rho = \frac{m}{V+ \beta~\Delta T~V}-\frac{m}{V}$ $\Delta \rho = \frac{m~V}{(V)(V+ \beta~\Delta T~V)}-\frac{(m)(V+ \beta~\Delta T~V)}{(V)(V+ \beta~\Delta T~V)}$ $\Delta \rho = -\frac{(m)(\beta~\Delta T~V)}{(V)(V+ \beta~\Delta T~V)}$ $\Delta \rho = -\frac{m~\beta~\Delta T}{V~(1+ \beta~\Delta T)}$ $\Delta \rho = -\frac{\beta~\rho~\Delta T}{1+ \beta~\Delta T}$ $\Delta \rho \approx -\frac{\beta~\rho~\Delta T}{1}$ $\Delta \rho \approx -\beta~\rho~\Delta T$ (b) We can find the fractional change in density: $\Delta \rho = -\beta~\rho~\Delta T$ $\frac{\Delta \rho}{\rho} = -\beta~\Delta T$ $\frac{\Delta \rho}{\rho} = -(56\times 10^{-6}~K^{-1})~(-42~K)$ $\frac{\Delta \rho}{\rho} = 2.35\times 10^{-3}$
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