Question

Find the change in volume of an aluminum sphere with an initial radius of 10 cmwhen the sphere is heated from$\mathbf{0}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}\mathbf{C}\mathbf{}\mathbf{to}\mathbf{}\mathbf{100}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}\mathbf{C}$.

Short answer

The change in volume of an aluminum sphere is$29{\mathrm{cm}}^3$

Step-by-step solution

The given data

1. Initial radius of sphere ${\mathrm{r}}_1=10\mathrm{cm}$at${\mathrm{T}}_1=0^{\mathrm{o}}\mathrm{C}$
2. Final temperature at which sphere expands${\mathrm{T}}_2={100}^{\mathrm{o}}\mathrm{C}$
3. Linear expansion coefficient for aluminum is$\mathrm{\alpha }=23\times {10}^{-6}/{}^{\mathrm{o}}\mathrm{C}$

Understanding the concept of linear expansion

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. Thermal expansion can occur due to an increase in temperature. For the given problem, we have to use the formula for volume expansion. Change in volume expansion depends on the original volume, the temperature change, and the volume expansion coefficient.

Formula:

The linear expansion of a body,$∆\mathrm{L}=\mathrm{L\alpha }∆\mathrm{T}$ …(i)

Whereis the coefficient of linear expansion of body.

The volume change in expansion of a body,$∆\mathrm{V}=\mathrm{V\beta }∆\mathrm{T}$ …(ii)

Whereis the coefficient of volume expansion of the rod

The volume of the sphere, $\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3$ …(iii)

Calculation of the volume change

So, the coefficient of volume expansion of the rod is given as:

$\begin{array}{rcl}\mathrm{\beta }& =& 3\mathrm{\alpha }\\ & =& 3\times 23\times {10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\\ & =& 69\times {10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\end{array}$

Therefore, we can calculate the initial volume of the sphere by using the formula of equation (iii) as given:

$\begin{array}{rcl}\mathrm{V}& =& \frac{4}{3}\times 3.14\times {\left(10\mathrm{cm}\right)}^3\\ & =& 4186.66{\mathrm{cm}}^3\end{array}$

We can now find the change in volume using equation (ii) as:

$\begin{array}{rcl}∆\mathrm{V}& =& 4186.66{\mathrm{cm}}^3\times 69\times {10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\times {\left(100-0\right)}^{\mathrm{o}}\mathrm{C}\\ & =& 28.88{\mathrm{cm}}^3\\ & \approx & 29{\mathrm{cm}}^3\end{array}$

Hence, the change in volume is$29{\mathrm{cm}}^3$