Question

By how much should the temperature of an aluminum sample
change if its volume is to change by 1.5 % of its volume.
(Coefficient of linear expansion of aluminum is 2.4e-5 ⁄ C)


The answer is: 208.333 C but I do not know how to get there without having the volume or radius.

Answer 1

In this task, you must assume that we have an aluminum cube volume 1 m³.

The dependence of the volume on the length of the edge of the cube is calculated by the formula:

$$V = S^3$$

$$S = \sqrt[3]{V}$$

$$S = 1 m$$

Cube volume after heating:

$$V_1 = V + V*0.015$$

$$V_1 = 1 + 0.015 = 1.015m^3$$

$$S_1 = \sqrt[3]{1.015}m$$

$$S_1 = 1.005 m$$

Finally:

$$\alpha = {\Delta S \over S\Delta T}$$

$$\Delta S = S_1 - S$$

$$\Delta S = 1.005 - 1 = 0.005 = 5*10^{-3}$$

$$\Delta T = {\Delta S \over \alpha S }$$

​

$$\Delta T = {5*10^{-3} \over 2.4*10^{-6}} = {0.005 \over 0.000024} = 208.333 K$$

​

Answer: 208.333 K