Question 4.53: A steel bar has a length of 4.0 m at 10°C. What will be the ...
A steel bar has a length of 4.0 m at 10°C. What will be the length of the bar when it is heated to 350°C? If a sphere of diameter 15 cm is made from the same material, what will be the percentage increase in surface area if the sphere is subject to the same initial and final temperatures?
Using \alpha=12 \times 10^{-6} from the above table,
increase in length of the bar is given by:
| Material | Linear expansion coefficient α/°C |
| Invar | 1.5 \times 10^{-6} |
| Glass | 9 \times 10^{-6} |
| Cast iron | 10 \times 10^{-6} |
| Concrete | 11 \times 10^{-6} |
| Steel | 12 \times 10^{-6} |
| Copper | 17 \times 10^{-6} |
| Brass | 19 \times 10^{-6} |
| Aluminium | 24 \times 10^{-6} |
\begin{aligned} x=\alpha l\left(t_2-t_1\right) &=\left(12 \times 10^{-6}\right)(4.0)(350-10) \\&=0.0163 \ m\end{aligned}
This can now be added to the original length to give the final length = 4.0 + 0.0163 = 4.0163 m.
Increase in surface area of the sphere = 2 \alpha A\left(t_2-t_1\right). We first need to find the original surface area which is given by:
A = 4πr² = 4π × (0.075)² = 0.0707 m²
and, from above, the increase in surface area
\begin{aligned}&=(2)\left(12 \times 10^{-6}\right)(0.0707)(340) \\&=5.769 \times 10^{-6} \ m ^2\end{aligned}
Therefore, the percentage increase in area
\begin{aligned}&=\frac{\text { increase in area }}{\text { original area }} \times 100 \\&=\frac{5.769 \times 10^{-4}}{0.0707} \times 100=0.82 \ \% \end{aligned}