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If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour. - Mathematics and Statistics

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Sum

If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.

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Solution

Let θ° C be the temperature of the body at time t minutes. Room temperature is given to be 25°C.

Then by Newton’s law of cooling, `("d"theta)/"dt"` the rate of change of temperature, is proportional to (θ - 25).

i.e. `("d"theta)/"dt" ∝  (θ - 25)`

∴ `("d"theta)/"dt"` = - k(θ - 25) , where k > 0

∴ `("d"theta)/(theta - 25)` = - k dt

On integrating, we get

`int "dP"/"P" = "k" int "dt" + "c"`

∴ log P = kt + c

Initially, i.e. when t = 0, let P = P0

∴ log P0 = k × 0 + c

∴ c = log P0 

∴ log P = kt + log P0 

∴ log P - log P0 = kt

∴ `log ("P"/"P"_0)`= kt    ...(1)

Since the population doubles in 60 hours, i.e. when t = 60, P = 2P0 

∴ `log ((2"P"_0)/"P"_0)` = 60k

∴ k = `1/60` log 2

∴ (1) becomes, `log ("P"/"P"_0) = "t"/60` log 2

When population becomes triple, i.e. when P = 3P0 , we get

`log ((3"P"_0)/"P"_0) = "t"/60` log 2

∴ `log 3 = ("t"/60)` log 2

∴ t = `60 ((log 3)/(log 2)) = 60 (1.0986/0.6912)`

= 60 × 1.5894 = 95.364 ≈ 95.4 years

∴ the population becomes triple in 95.4 years (approximately).

Concept: Application of Differential Equations
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