Question

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.

Answer 1

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, `(dθ)/dt` the rate of change of temperature, is proportional to (θ − 25).

i.e. `(dθ)/dt ∝ (θ - 25)`

∴ `(dθ)/dt` = −k(θ − 25), where k > 0

∴ `(dθ)/(θ - 25)` = −k dt

On integrating, we get

`int 1/(θ - 25)dθ = -k int dt + c`

∴ log (θ − 25) = −kt + c

Initially, i.e. when t = 0, θ = 80

∴ log (80 − 25) = −k × 0 + c  ...(∴ c = log 55)

∴ log (θ − 25) = −kt + log 55

∴ log (θ − 25) − log 55 = −kt

`log ((θ - 25)/55) = -kt`  ...(1)

Now,  when t = 30, θ = 50

∴ `log ((50 - 25)/55) = - 30"k"`

∴ k = `- 1/30 log (5/11)`

∴ (1) becomes, `log ((θ - 25)/55) = t/30 log (5/11)`

When t = 1 hour = 60 minutes, then

`log ((θ - 25)/55) = 60/30 log (5/11)`

`log ((θ - 25)/55) = 2 log (5/11)`

∴ `log ((θ - 25)/55) = log (5/11)^2`

∴ `(θ - 25)/55 = (5/11)^2`

∴ `(θ - 25)/55 = 25/121`

∴ `θ - 25 = 55 xx 25/121`

∴ `θ - 25 = 125/11`

∴ `θ = 125/11 + 25`

∴ `θ = (125 + 275)/11`

∴ `θ = 400/11`

∴ θ = 36.36

∴ θ the temperature of the body will be 36.36°C after 1 hour.