91Ó°ÊÓ

i) Dimensions of the glass window pane,$20\text{cm}×30\text{cm}$

ii) Initial temperature of glass window pane$T_i={10}^0\text{Cor283K}$

iii) Final temperature of glass window pane$T_f={40}^0\text{Cor313K}$

iv) Coefficient of linear expansion of the glass is.$\mathrm{Î\pm }=(9×{10}^{−6}{/}^0\text{C})$

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. Here, when the window pane is heated, this leads to a change in its length, which is given by the difference between the increased length and the original length.We can use the equation of thermal expansion to write an equation for the area of the glass window pane before and after the expansion. From these equations, we can find the increased area of the glass pane at the required temperature.

Formula:

Change in length due to thermal expansion,$\mathrm{Δ}L=\mathrm{Î\pm }L\mathrm{Δ}T$ …(¾±)

where $\mathrm{Î\pm }$is the coefficient of thermal linear expansion of the substance$\mathrm{Δ}T$, is the temperature difference at both the ends of the body,and $L$is the original length.

Letbe the area at temperature at${10}^0\text{C}$, given by

$A_1=L_1{L}_2$ …(¾±¾±)

Let$A_2$be the area at temperature${40}^{0}C$. The area is foundusing equation (i) because of length expansion as

$\begin{array}{c}{A}_2=(L_1+\mathrm{Δ}{L}_1)(L_2+\mathrm{Δ}{L}_2)\\ =L_1{L}_2+L_1\mathrm{Δ}{L}_2+L_2\mathrm{Δ}{L}_1+\mathrm{Δ}{L}_1\mathrm{Δ}{L}_2\end{array}$ …(¾±¾±¾±)

Here$\mathrm{Δ}{L}_1\mathrm{Δ}{L}_2$can beneglected because it ismuch smaller than the other terms. So, equation (iii) becomes

$A_2=L_1{L}_2+L_1\mathrm{Δ}{L}_2+L_2\mathrm{Δ}{L}_1$ …(¾±±¹)

Therefore, the increasing change in the area of the glass pane is given by subtracting equation (ii) from equation (iv) as

$\begin{array}{c}\mathrm{Δ}A=L_1{L}_2+L_1\mathrm{Δ}{L}_2+L_2\mathrm{Δ}{L}_1−L_1{L}_2\\ =L_1\mathrm{Δ}{L}_2+L_2\mathrm{Δ}{L}_1\\ =L_1\mathrm{Î\pm }{L}_2\mathrm{Δ}T+L_2\mathrm{Î\pm }{L}_1\mathrm{Δ}T\text{(usingequation(i))}\\ =\mathrm{Î\pm }\mathrm{Δ}T(L_1{L}_2+L_1{L}_2)\\ =2\mathrm{Î\pm }{L}_1{L}_2\mathrm{Δ}T\\ =2(9×{10}^{−6}{/}^0\text{C})\left(30\text{cm}\right)\left(20\text{cm}\right)\left({(40−10)}^0\text{C}\right)\text{(}âˆ\mu \text{substitutinggivenvalues)}\\ =0.32{\text{cm}}^{\text{2}}\end{array}$

Therefore, the increased area of the glass window pane at the temperature of ${40}^0\text{C}$is${\text{0.32cm}}^{\text{2}}$.