A temperature change \(\Delta T\) causes a volume change \(\Delta V\) but has no effect on the mass of an object. (a) Show that the change in density $\Delta \rho\( is given by \)\Delta \rho=-\beta \rho \Delta T$. (b) Find the fractional change in density \((\Delta \rho / \rho)\) of a brass sphere when the temperature changes from \(32^{\circ} \mathrm{C}\) to \(-10.0^{\circ} \mathrm{C}\)

Density (\(\rho\)) of an object is defined as its mass (\(m\)) divided by its volume (\(V\)): \(\rho = \frac{m}{V}\). Coefficient of volume expansion (\(\beta\)) represents how a material expands or contracts per degree change in temperature.

The relationship between volume change \(\Delta V\), initial volume \(V\), coefficient of volume expansion \(\beta\) and temperature change \(\Delta T\) is given by the formula: \(\Delta V = V \beta \Delta T\)

The final volume \(V'\) after the temperature change can be calculated using the initial volume \(V\) and the volume change \(\Delta V\): \(V' = V + \Delta V = V + (V \beta \Delta T)\)

The change in density \(\Delta \rho = \rho' - \rho\) can be calculated using the initial and final densities, which in turn depend on the initial and final volumes. The mass doesn't change, so we can write: \(\rho = \frac{m}{V}\) and \(\rho' = \frac{m}{V'}\) Subtracting the two expressions to calculate \(\Delta \rho\): \(\Delta \rho = \rho' - \rho = \frac{m}{V'} - \frac{m}{V} \) Now substitute \(V'\) using the expression from step 3: \(\Delta \rho = \frac{m}{V + (V \beta \Delta T)} - \frac{m}{V} \) Factor out \(m\) from both terms: \(\Delta \rho = \frac{m}{V} \left(\frac{1}{1 + (\beta \Delta T)} - 1\right) \) Now, remember that \(\rho = \frac{m}{V}\): \(\Delta \rho = \rho \left(\frac{1}{1 + (\beta \Delta T)} - 1\right) \) Finally, expand the bracket to get the desired expression: \(\Delta \rho = \rho \left(\frac{1 - (1 + \beta \Delta T)}{1 + \beta \Delta T}\right) = -\beta \rho \Delta T \) Now we have shown the relationship between change in density, coefficient of volume expansion, and change in temperature.

Now for part (b), we have to find the fractional change in density \((\Delta \rho / \rho)\). Using the result we derived above \(\Delta \rho = -\beta \rho \Delta T\), we can simply divide by the density \(\rho\): \(\frac{\Delta \rho}{\rho} = -\beta \Delta T\) Since the temperature change is from \(32^{\circ} \mathrm{C}\) to \(-10.0^{\circ} \mathrm{C}\), the temperature change is \(\Delta T = -42.0^{\circ} \mathrm{C}\). The coefficient of volume expansion for brass is approximately \(\beta = 60.0 \times 10^{-6} \mathrm{C}^{-1}\). Substitute these values in the above equation: \(\frac{\Delta \rho}{\rho} = - (60.0 \times 10^{-6} \mathrm{C}^{-1})\times (-42.0^{\circ} \mathrm{C}) = 2.52 \times 10^{-3}\) The fractional change in density is \(2.52 \times 10^{-3}\).