A body cools according to Newton's law from $100^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 20…
A body cools according to Newton's law from $100^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 20 minutes. The
temperature of the surrounding being $20^{\circ} \mathrm{C}$ then the temperature of the body after
one hour is
$15^{\circ} \mathrm{C}$
$30^{\circ} \mathrm{C}$
$40^{\circ} \mathrm{C}$
$20^{\circ} \mathrm{C}$
Solution
Let $\theta^{\prime} \mathrm{C}$ be the temperature of the body at time t. The temperature of surrounding is $20^{\circ} \mathrm{C}$.
According to Newton's law of cooling
$\frac{d \theta}{d t} \propto(\theta-20) \Rightarrow \frac{d \theta}{d t}=-K(\theta-20), \quad$ where $K>0$
$\therefore \int \frac{d \theta}{\theta-20}=\int-K d t \Rightarrow \log |\theta-20|=-K t+c$
We have $\theta=100$ and $t=0$
$\therefore \log |100-20|=0+c \Rightarrow c=\log 80$
$\therefore \log |\theta-20|=-\mathrm{Kt}+\log 80$
$\therefore \log \left|\frac{\theta-20}{80}\right|=-\mathrm{Kt}$
When $t=20, \theta=60$
$\therefore K=\frac{-1}{20} \log \left(\frac{1}{2}\right)$
Thus $\log \left(\frac{\theta-20}{80}\right)=\frac{\mathrm{t}}{20} \log \left(\frac{1}{2}\right)$
When $t=60$
$\log \left(\frac{\theta-20}{80}\right)=3 \log \left(\frac{1}{2}\right)=\log \left(\frac{1}{8}\right) \Rightarrow \frac{\theta-20}{80}=\frac{1}{8} \Rightarrow \theta=30^{\circ}$