# 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

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.

#### 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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