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.