A metal Ball sphere , when suspened in a constant temperature enclosure , cools from 10•C To 5•C in 10 min and 2.5 •C in next 10 min , then find the temp of surroundings.
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Explanation :
Let θ₀ be tempecture of the surrounding.
Case (I) : Consider a cooling from 10°C to 5°C in 10 minutes.
- Initial tempecture (θ₁) = 10°C
- Final tempecture (θ₂) = 5°C
- Time taken (t) = 10 min
Apply newton's law of cooling,
dθ/dt = C(θ - θ₀)
.°. θ₁ - θ₂/dt = C(θ₁ + θ₂/dt - θ₀)
=> 10 - 5/10 = C(10 + 5/10 - θ₀)
=> 0.5 = C(1.5 - θ₀) _______(1)
Case (II) : Consider a cooling from 5°C to 2.5°C in 10 minutes.
- Initial tempecture (θ₁) = 5°C
- Final tempecture (θ₂) = 2.5°C
- Time taken (t) = 10 min
Apply newton's law of cooling,
dθ/dt = C(θ - θ₀)
.°. θ₁ - θ₂/dt = C(θ₁ + θ₂/dt - θ₀)
=> 5 - 2.5/10 = C(5 + 2.5/10 - θ₀)
=> 0.25 = C(0.75 - θ₀) _______(2)
Divide eqn. (1) & (2),
0.5/0.25 = C(1.5 - θ₀)/C(0.75 - θ₀)
=> 0.5(0.75 - θ₀) = 0.25(1.5 - θ₀)
=> 0.375 - 0.5θ₀ = 0.375 - 0.25θ₀
=> 0.375 - 0.375 = 0.5θ₀ - 0.25θ₀
=> 0 = -0.25θ₀
=> θ₀ = 0°C
Hence :
The tempecture of the surrounding is 0°C.
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