At ${\mathbf{20}}^{\mathbf{°}}\mathit{C}$, a brass cube has an edge length of $\mathbf{30}\mathbf{}\mathit{c}\mathit{m}$.What is the increase in the cube’s surface area when it is heated from${\mathbf{20}}^{\mathbf{°}}\mathit{C}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{75}\mathbf{°}\mathit{C}$?

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

The linear expansion of length is given by:

$∆L=L{\alpha }_b∆T$ …… (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" $∆L^2\approx 0$. Hence,

$∆A=12L∆L$

Therefore, the change in area is determined as:

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

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

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

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