Help me solving this problem with reasons. I`m unable to solve this....
Preeti , 7 Years ago
Grade 11
Thermal Physics> Help me solving this problem with reasons...
1 AnswersArun
Last Activity: 7 Years ago
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 × 10–2cm.
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 × 10–2cm.
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