The area Aof a rectangular plate is$\mathit{a}\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.4}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$. Its coefficient of linear expansion is$\mathit{\alpha }\mathbf{}\mathbf{=}\mathbf{}\mathbf{32}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}{\mathbf{\text{/}}}^{\mathbf{\text{0}}}\mathbf{\text{C}}$. After a temperature rise$\mathbf{}\mathit{\Delta }\mathit{T}\mathbf{}\mathbf{=}\mathbf{89}\mathbf{\text{°C}}$, side ais longer byand side bis longer by $\mathit{\Delta }\mathit{b}\mathbf{}$(Fig. 18-61). Neglecting the small quantity $\mathbf{(}\mathbf{\Delta }\mathbf{a}\mathbf{}\mathbf{\Delta }\mathbf{b}\mathbf{)}\mathbf{/}\mathit{a}\mathit{b}$, find$\mathit{\Delta }\mathit{A}$.

1. Area A of the rectangular plate, given by$ab=1.4{\text{m}}^{\text{2}}$
2. Temperature rise,$\Delta T=89\text{°C}$
3. Coefficient of linear expansion,$\alpha =32\times {10}^{-6}{\text{/}}^{\text{0}}\text{C}$ .

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particlesWe use the concept of linear thermal expansion to solve this problem. We use the equation of thermal expansion to get the ratio of change in length and original length. Using that ratio and neglecting the small term, we can find the change in the area.

Formula:

Linear expansion of the body due to thermal radiation,$\Delta L=L\times \alpha \times \Delta T$ …(i)

where L is the length and$\Delta T$ is change in temperature.

Using equation (i) and the given values, we get the ratio of change in length to the original length of the body is given as:

$\begin{array}{c}\frac{\Delta L}{L}=32\times {10}^{-6}{\text{/}}^{\text{0}}\text{C}\times 89\text{°C}\\ =0.003\end{array}$ …(ii)

Now, when the temperature rises by$\Delta T=89°C$, both sides of the rectangular plate will change by$\Delta \text{aand}\Delta \text{b}$as given.

So, the new area of rectangular plate using equation (i) is given by

$A_{new}=(a+\Delta a)\times (b+\Delta b)$

Taking the ratio of the new area and the original area, we get

$\begin{array}{c}\frac{A_{new}}{A}=\frac{(a+\Delta a)\times (b+\Delta b)}{ab}\text{(}\because {\text{originallength,A=abm}}^2\text{)}\\ =1+\frac{\Delta a}{a}+\frac{\Delta b}{b}\text{(}\because \frac{\Delta a\Delta b}{ab}\text{isveryverysmall,socanbeneglected)}\\ \frac{A_{new}}{1.4}=1+0.003+0.003=1.006\text{(}\because {\text{originallength,ab=1.4m}}^2\text{andfromequation(ii))}\\ A_{new}=1.006\times 1.4{\text{\hspace{0.17em}m}}^{\text{2}}\\ =1.408{\text{\hspace{0.17em}m}}^{\text{2}}\end{array}$

Now, the change in area is given by

$\begin{array}{c}\Delta A=A_{new}-A\\ =1.408{\text{\hspace{0.17em}m}}^{\text{2}}-1.4{\text{\hspace{0.17em}m}}^{\text{2}}\\ =8.0\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Hence, the value of the changed area is$8.0\times {10}^{-3}{\text{m}}^{\text{2}}$.