Question

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​

Answer 1

Answer:

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Answer 2

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