Question

An aluminum sphere is 8.55 cm in diameter. What will be it's % change in volume if it is heated from 30 C to 155 C?

Answer 1

We know that

• The sphere diameter is 8.55 cm.

,

• The temperature change is from 30 C to 155 C.

First, we have to find the radius of the sphere. The radius is the half diameter.

$\begin{gathered} r=\frac{d}{2}=\frac{8.55\operatorname{cm}}{2} \\ r=4.275\operatorname{cm} \end{gathered}$

Now we have to find the volume of the sphere using the following formula.

$V=\frac{4}{3}\pi(r)^3$

Where r = 4.275 cm.

$\begin{gathered} V=\frac{4}{3}\pi(4.275\operatorname{cm})^3 \\ V\approx327.26(cm)^3 \end{gathered}$

Then, we use the following formula

$V_2-V_1=BV_0(T_2-T_1)$

Where the initial volume is 327.26 cubic cm, B is a constant about thermal expansion for aluminum, and we have to find the final volume to then calculate the percentage change.

$\begin{gathered} \Delta V=75\times10^{-6}\cdot327.26(155-30) \\ \Delta V=75\times10^{-6}\cdot327.26(125) \\ \Delta V\approx3.07(cm)^3 \end{gathered}$

This means that the volume change is 3.07 cubic centimeters.

At last, we have to divide the volume change by the initial volume, and then we have to multiply it by 100% to express it as a percentage.

$\frac{\Delta V}{V_1}\times100=\frac{3.07}{327.26}\times100\approx0.938$

Therefore, the percentage change is 0.938%.