Q8 of 26 Page 294

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.

Given,

Initial temperature, T1 = 27.0°C = 27 + 273.15 = 300.15 K


Diameter of the hole at T1, d1 = 4.24 cm


Final temperature, T2 = 227°C = 227 + 273.15 = 500.15 K


Co-efficient of linear expansion of copper, αCu = 1.70 × 10-5 K-1


Let the diameter of the hole at T2 be d2.


If the co-efficient of superficial expansion is β, and the change in temperature is ΔT, then


Change in area (∆A)/Original area (A) = β∆T


∆A/A = [(πd22/4) - (πd12/4)]/(πd12/4)]


⇒ ∆A/A = (d22 - d12)/d12


But β = 2α


∴ (d22 - d12)/d12 = 2α∆T


⇒ (d22/d12) -1 = 2α(T2 – T1)


⇒ d22/(4.24 cm)2 – 1 = 2 × (1.7 × 10-5 K-1) (500.15 K – 300.15 K)


⇒ d22/(17.98 cm2) = 68 × 10-4 + 1


⇒ d22 = 17.98 cm2 × 1.0068


⇒ d22 = 18.1022 cm2


⇒ d2 = 4.2546 cm


Change in diameter = d2– d1 = 4.2546 cm – 4.24 cm =0.0146 cm Hence, the diameter increases by 0.0146 cm.


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