Certain quantity of water cools from 70°C to 60°C
in the first 5 minutes and to 54°C in the next
5 minutes. The temperature of the surroundings is
To=?
[AIPMT-2014]
(1/45°C
(2) 20°C
(3) 42°C
(4) 10°C
Answers
Answer:
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So, the temperature of the surroundings is 45°C.
We know,
According to Newton's law of cooling, rate of cooling is directly proportional to the temperature difference between the body and its surroundings.
If a body in time 't' cools from θ₁ °C to θ₂ °C, then
[(θ₁ - θ₂)/t] = k [(θ₁ + θ₂)/2 - θ₀]
where θ₀ is the temperature of the surroundings.
When water is cooled from 70°C to 60°C
then θ₁ = 70°C and θ₂ = 60°C and t = 5 minutes.
then,
[(70 - 60)/5] = k [(70 + 60)/2 - θ₀]
⇒ k [65 - θ₀] = 2
⇒ k = 2/[65 - θ₀] ...(1)
When water is cooled again from 60°C to 54°C
then θ₁ = 60°C and θ₂ = 54°C and t = 5 minutes.
then,
[(60 - 54)/5] = k [(60 + 54)/2 - θ₀]
⇒ k [57 - θ₀] = 1.2
⇒ k = 1.2/[57 - θ₀] ...(2)
As (1) and (2) are equal,
So, 2/[65 - θ₀] = 1.2/[57 - θ₀]
⇒ 2[57 - θ₀] = 1.2[65 - θ₀]
⇒ 114 - 2θ₀ = 78 - 1.2θ₀
⇒ (2 - 1.2)θ₀ = 114 - 78 = 36
⇒ 0.8 θ₀ = 36
⇒ θ₀ = 36/0.8 = 45 °C