Question

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × $10^{-5} K^{-1}.$

Answer 1

Initial temperature, $T_1$ = 27.0°C
Diameter of the hole at $T_1,d_1$= 4.24 cm
Final temperature, $T_2$ = 227°C
Diameter of the hole at $T_2= d_2$
Co-efficient of linear expansion of copper, α$_{Cu}$= 1.70 × 10$^{-5}K^{-1}$
For co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:
$\frac{Change in Area (\triangle A)}{Original Area(A)}$ = β△T

$\frac{\bigg(\pi\frac{d^2_2}{4}-\pi\frac{d^2_1}{4}\bigg)}{\bigg(\pi\frac{d^2_1}{4}\bigg)}$ = $\frac{\triangle A}{A}$
∴ $\frac{\triangle A}{A}$ = $\frac{d^2_2-d^2_1}{d^2_1}$

But β = 2α

∴$\frac{d^2_2-d^2_1}{d^2_1}$= 2α△T

$\frac{d^2_2}{d^2_1}-1$ = 2α(T2 - T1)

$\frac{d^2_2}{(4.24)^2}$ = 2 x 1.7 x 10$^{-5}$ x (227-27) +1

${d^2_2}$ = 17.98 x 1.0068 = 18.1

∴ $d_2$ = 4.2544 cm
Change in diameter = $d_2-d_1$ = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 × 10$^{-2}$ cm.