Chapter 13, Problem 14P

An aluminum sphere is 8.75 cm in diameter. What will be its % change in volume if it is heated from 30”C to 160°C?

To determine

The % change in volume of the aluminum sphere when it is heated

Answer to Problem 14P

Solution:

The % change in volume of the aluminum sphere when it is heated is 0.98%.

Explanation of Solution

Given:

Initial diameter of the sphere, $d_i=8.75cm$

Therefore, initial radius, $r_i=4.375cm$

Initial temperature, $T_i=30°C$

Final temperature, $T_f{}_{}=160°C$

Formula used:

$\Delta V=\beta V_i\Delta T,$ Where, V_i is the initial volume and β is the coefficient of volume expansion

Calculation:

The relation between change in volume of a substance (ΔV)and change in temperature (ΔT)is given by,

$\Delta V=\beta V_i\Delta T$--- (i),

Where, V_i is the initial volume and β is the coefficient of volume expansion

Here, $\Delta T=T_f–T_i$

For aluminum, $\beta =75\times {10}^{-6}/°C$

For the given sphere, $V_i=$$\frac{4}{3}\pi r_i{}^3$

= $\frac{4}{3}\pi {(4.375cm)}^3$

$=350.6c{m}^3$

Substituting the known values in equation (i),

$\begin{array}{l}\Delta V={\left(75\times {10}^{-6}/°C)\times (350.6c{m}^3\right)}^{ }\times \left(160°C-30°C\right)\hfill \\ =3.42c{m}^3\hfill \end{array}$

Therefore, % change in volume = $\frac{\Delta V}{V_i}x100\%$

= $\frac{3.42c{m}^3}{350.6c{m}^3}\times 100\%=0.98\%$