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.
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Answer 1

Initial temperature, T1 = 27.0°C

Diameter of the hole at T1, d1 = 4.24 cm

Final temperature, T2 = 227°C

Diameter of the hole at T2 = d2

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

For co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:

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

[ (πd22/ 4) - (πd12 / 4) ] / (πd11 / 4) = ∆A / A

∴ ∆A / A = (d22 - d12) / d12

But β = 2α

∴ (d22 - d12) / d12 = 2α∆T

(d22 / d12) - 1 = 2α(T2 - T1)

d22 / 4.242 = 2 × 1.7 × 10-5 (227 - 27) +1

d22 = 17.98 × 1.0068 = 18.1

∴ d2 = 4.2544 cm

Change in diameter = d2 – d1 = 4.2544 – 4.24 = 0.0144 cm

Hence, the diameter increases by 1.44 × 10–2 cm.

Answer 2

Answer:

Explanation:

Here, diameter of hole ( D1) = 4.24 cm

So, initial area of hole (Ao)= πr² = 22/7 ( 4.24/2)²

= 4.494π cm²

intial temperature ( T1) = 27°C = 27+ 273 = 300 K

Final temperature ( T2) = 227°C = 227+ 273 = 500K

coefficient of linear expansion (a) = 1.7 × 10^-5/°C

coefficient of superficial expansion (b) = 2× linear expansion

= 2 × 1.7 × 10^-5/°C

= 3.4 × 10^-5 /°C

Use formula,

A = Ao( 1 + b∆T)

A = 4.494π [1 + 3.4 × 10^-5 × (500-300)]

= 4.494π[ 1 + 3.4 × 10^-5 × 200]

= 4.494π [ 1 + 6.8 × 10^-3 ]

= 4.494π [ 1 + 0.0068]

= 4.494π × 1.00068

= 4.525π cm² = πD2²/4

D2² = 4.525 × 4

D2 = 4.2544 cm

Change in diameter (∆D) = D2 - D1

= 4.2544 - 4.24

= 0.0144 cm