Question

A rectangular plate of glass initially has the dimensions $\mathbf{0}\mathbf{.200}\mathbf{}\mathbf{\text{m}}$by.$\mathbf{0}\mathbf{.300}\mathbf{}\mathbf{\text{m}}$ The coefficient of linear expansion for the glass is $\mathbf{9}\mathbf{.00}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\text{/K}}$.What is the change in the plate’s area if its temperature is increased by$\mathbf{20}\mathbf{.0}\mathbf{}\mathbf{\text{K}}$?

Short answer

The change in the plate’s area when its temperature increases by $20.0\text{K}$will be$2.16\times {10}^{-5}{\text{m}}^{\text{2}}\text{.}$

Step-by-step solution

Stating the given data 

i) The initial dimensions are.$0.2\text{m}\text{\hspace{0.17em}\hspace{0.17em}by\hspace{0.17em}\hspace{0.17em}}0.3\text{m}$

ii) Coefficient of linear expansion,.$\alpha =9.0\times {10}^{-6}/\text{K}$

iii) Increase in volume,$\Delta T=20.0\text{K}$ .

Understanding the concept of thermal expansion

The propensity of matter to alter its form, area, volume, and density in reaction to a change in temperature is known as thermal expansion. As we know the coefficient of thermal expansion, we can calculate the increase in the dimensions of the plate. Using those dimensions, we can find the area, and by comparing it to the original area, we can find changes in the area.

Formula:

Linear expansion due to thermal expansion$\Delta L=L\alpha \Delta T$,…(i)

Change in the area due to expansion,.$\Delta A=A_{final}-A_{\text{initial}}$…(ii)

Calculation of change in area of the plate 

The initial area of the plate before the expansion is given by

$\begin{array}{c}{A}_{\text{initial}}=0.2\text{\hspace{0.17em}m}\times 0.3\text{\hspace{0.17em}m}\\ =0.06{\text{m}}^{\text{2}}\end{array}$

Using equation (i), the expansion for$L=0.2\text{\hspace{0.17em}m}$is found to be

$\begin{array}{c}\Delta L=0.2\text{\hspace{0.17em}m}\times 9\times {10}^{-6}/\text{K}\times 20\text{\hspace{0.17em}K}\\ =3.6\times {10}^{-5}\text{m}\end{array}$

So, the changed dimension for the side$L=0.2\text{\hspace{0.17em}m}$ is given by

$\begin{array}{c}{L_1}^{\text{'}}=0.2\text{\hspace{0.17em}m}+3.6\times {10}^{-5}\text{m}\\ =0.200036\text{m}\end{array}$

Using equation (i), the expansion for$L=0.3\text{\hspace{0.17em}m}$is found to be

$\begin{array}{c}\Delta L=0.3\text{\hspace{0.17em}m}\times 9\times {10}^{-6}/\text{K\hspace{0.17em}}\times 20\text{\hspace{0.17em}m}\\ =5.4\times {10}^{-5}\text{m}\end{array}$

So, the changed dimension for the side $L=0.3\text{\hspace{0.17em}m}$is given by

$\begin{array}{c}{L_2}^{\text{'}}=0.3\text{\hspace{0.17em}m}+5.4\times {10}^{-5}\text{\hspace{0.17em}m}\\ =0.300054\text{m}\end{array}$

The final area for the expanded lengths is

$\begin{array}{c}{A}_{\text{final}}=0.200036\text{\hspace{0.17em}m}\times 0.300054\text{\hspace{0.17em}m}\\ =0.0600216{\text{\hspace{0.17em}m}}^{\text{2}}\end{array}$

The change in area, foundusing equation (ii),is

$\begin{array}{c}\Delta A=0.0600216{\text{\hspace{0.17em}m}}^{\text{2}}-0.06{\text{\hspace{0.17em}m}}^{\text{2}}\\ =2.16\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

Hence, the increase in the area is.$2.16\times {10}^{-5}{\text{m}}^{\text{2}}$