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

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 \(t=60\)
\(\log \left(\frac{\theta-20}{80}\right)=3 \log \left(\frac{1}{2}\right)=\log \left(\frac{1}{8}\right) \Rightarrow \frac{\theta-20}{80}=\frac{1}{8} \Rightarrow\) \(\theta=30^{\circ}\)