Answer
(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:
$\Delta \rho = -\beta~\rho~\Delta T$
$\frac{\Delta \rho}{\rho} = -\beta~\Delta T$
$\frac{\Delta \rho}{\rho} = -(56\times 10^{-6}~K^{-1})~(-42~K)$
$\frac{\Delta \rho}{\rho} = 2.35\times 10^{-3}$
