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, ${\displaystyle \frac{d\theta }{dt}}$ the rate of change of temperature, is proportional to (θ − 25).

i.e. ${\displaystyle \frac{d\theta }{dt}\propto (\theta -25)}$

∴ ${\displaystyle \frac{d\theta }{dt}}$ = −k(θ − 25), where k > 0

∴ ${\displaystyle \frac{d\theta }{\theta -25}}$ = −k dt

On integrating, we get

${\displaystyle \int \frac{1}{\theta -25}d\theta =-k\int dt+c}$

∴ 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

${\displaystyle \log \left(\frac{\theta -25}{55}\right)=-kt}$  ...(1)

Now,  when t = 30, θ = 50

∴ ${\displaystyle \log \left(\frac{50-25}{55}\right)=-30\text{k}}$

∴ k = ${\displaystyle -\frac{1}{30}\log \left(\frac{5}{11}\right)}$

∴ (1) becomes, ${\displaystyle \log \left(\frac{\theta -25}{55}\right)=\frac{t}{30}\log \left(\frac{5}{11}\right)}$

When t = 1 hour = 60 minutes, then

${\displaystyle \log \left(\frac{\theta -25}{55}\right)=\frac{60}{30}\log \left(\frac{5}{11}\right)}$

${\displaystyle \log \left(\frac{\theta -25}{55}\right)=2\log \left(\frac{5}{11}\right)}$

∴ ${\displaystyle \log \left(\frac{\theta -25}{55}\right)={\log \left(\frac{5}{11}\right)}^2}$

∴ ${\displaystyle \frac{\theta -25}{55}={\left(\frac{5}{11}\right)}^2}$

∴ ${\displaystyle \frac{\theta -25}{55}=\frac{25}{121}}$

∴ ${\displaystyle \theta -25=55\times \frac{25}{121}}$

∴ ${\displaystyle \theta -25=\frac{125}{11}}$

∴ ${\displaystyle \theta =\frac{125}{11}+25}$

∴ ${\displaystyle \theta =\frac{125+275}{11}}$

∴ ${\displaystyle \theta =\frac{400}{11}}$

∴ θ = 36.36

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