91Ó°ÊÓ

• i) Edge length of the brass cube$at20°CisL=30cmor30×{10}^{-2}m$ is
• ii) Initial temperature $=20°C$
• iii) Final temperature$=75°C$

The linear expansion of length is given by:

$∆L=L{\mathrm{Î\pm }}_b∆T$ …… (i)

The total surface area of cube is given by:

$A=6{\left(L\right)}^2$ ……. (ii)

The brass cube has six faces so that it has six-time linear expansion. So, the change in area due to expansion using equation (ii) is given as:
$$\begin{gathered} ∆A=6(L+∆L)(L+∆L)-6L^2(Since,newexpandedlengthis(L+∆L) \\ =6{(L+∆L)}^2-6L^2 \\ =6(L^2+∆L^2+2L∆L)-6L^2 \end{gathered}$$

Since,$∆L$ is very small,localid="1661433128129" $∆L^2≈0$. Hence,

$∆A=12L∆L$

Therefore, the change in area is determined as:

localid="1661432593498" $$\begin{gathered} ∆A=12(L\mathrm{Î\pm }∆T) \\ =12L^2{\mathrm{Î\pm }}_b∆T \\ =12×{(30×{10}^{-2})}^2×19×{10}^{-6}×(75-20) \end{gathered}$$

(Since the linear coefficient of expansion for brass is$a_b=19×{10}^{-8}\frac{1}{°C}$)

$$\begin{gathered} ∆A=12×900×{10}^{-4}×19×{10}^{-6}×55 \\ =0.0011m^2 \\ =11c{m}^2 \end{gathered}$$

Hence, the increase in the total surface area of the cube is$11c{m}^2$.