Question

The temperature of a Pyrex disk is changed from$\mathbf{10}\mathbf{.0}\mathbf{°}\mathbf{\text{Cto60.0°C}}$. Its initial radius is$\mathbf{8}\mathbf{.00}\mathbf{}\mathbf{\text{cm}}$; its initial thickness is$\mathbf{0}\mathbf{.500}\mathbf{}\mathbf{\text{cm}}$. Take these data as being exact.What is the change in the volume of the disk?

Short answer

The change in volume of disk is $4.83\times {10}^{-8}{\text{m}}^{\text{3}}$.

Step-by-step solution

Identification of given data

i) Initial radius is$R=8.00\text{cmor0.08m}$

ii) Initial thickness is$t=0.500\text{cmor0.005m}$

iii) Initial temperature is$T_1=10.0°\text{C}$

iv) Final temperature is$T_2=60.0°\text{C}$ .

Understanding the concept of thermal and volume expansion

When an object is heated or cooled, its length changes by an amount proportional to the original length and the temperature change.This process is called the linear expansion of the given substance. But when the expansion takes place along all the three dimensions of the substance, then the expansion is referred to as volume expansion. The coefficient of volume expansion is related to the linear coefficient.

Formulae:

Area of the disk,$A=\pi R^2$ …(i)

where$R$ is the radius of the disk.

Volume of a body,$V=A\times t$…(ii)

where$A$ is the area of the disk and$t$ is the thickness of the disk.

Coefficient of volume expansion due to thermal radiation$\beta =\frac{\Delta V}{V\times \Delta T}$,…(iii)

where V is initial volume,$\Delta T$is change in temperature and$\Delta V$is change in volume.

Determining the change in volume of disk

Using equation (i) and the value of inner radius, the area of the disk is found to be

$\begin{array}{c}A=\pi \times {0.08}^2{\text{m}}^2\\ =0.02010{\text{m}}^{\text{2}}\end{array}$

Now, using this value of area and formula of equation (ii), we can get the volume of the body as

$\begin{array}{c}V=0.02010{\text{m}}^2\times 0.005\text{m}\\ =1.0053\times {10}^{-4}{\text{m}}^3\end{array}$

Now coefficient of volume expansion can be found using the value of the coefficient of linear expansion as

$\begin{array}{c}\beta =3\times \alpha \\ =3\times 3.2\times {10}^{-6}/\text{°C}\\ =9.6\times {10}^{-6}/°\text{C}\end{array}$

Now, using equation (iii) and the given values, we get the change in volume as

$\begin{array}{c}9.6\times {10}^{-6}{/}^0\text{C}=\frac{\Delta V}{1.0053\times {10}^{-4}{\text{m}}^3\times ({60}^0\text{C}-{10}^0\text{C})}\\ \Delta V=(9.6\times {10}^{-6}\times 1.0053\times {10}^{-4}\times 50){\text{m}}^3\\ =482.544\times {10}^{-10}{\text{m}}^{\text{3}}\\ =4.83\times {10}^{-8}{\text{m}}^{\text{3}}\end{array}$

Hence, the volume change is.$4.83\times {10}^{-8}{\text{m}}^{\text{3}}$