A body cools according to Newton's law of cooling from \(100^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 15 minutes. If the temperature of the surrounding is \(20^{\circ} \mathrm{C}\), then the temperature of the body after cooling down for one hour is

Let \(\theta\) be the temperature of the body at any time \(t\).
\(\begin{aligned}
\therefore \quad & \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-20) \\
& \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-20), \mathrm{k}\gt0
\end{aligned}\)
Integrating on both sides, we get
\(\begin{aligned}
& \log |\theta-20|=-\mathrm{kt}+\mathrm{c} \\
& \text {When } \mathrm{t}=0, \theta=100^{\circ} \\
& \therefore \quad \log 80=-\mathrm{k}(0)+\mathrm{c} \\
& \Rightarrow \mathrm{c}=\log 80
\end{aligned}\)
\(\begin{array}{ll}
\therefore \quad & \log |\theta-20|=-\mathrm{kt}+\log 80 \\
& \text {When } \mathrm{t}=15, \theta=60^{\circ} \\
\therefore \quad & \log 40=-15 \mathrm{k}+\log 80 \\
\Rightarrow & \mathrm{k}=\frac{-1}{15} \log \frac{1}{2}
\end{array}\)
\(\begin{aligned} \therefore \quad & \log |\theta-20|=\frac{\mathrm{t}}{15} \log \frac{1}{2}+\log 80 \ldots[\text { From }(\mathrm{i})] \\ & \text { When } \mathrm{t}=1 \text { hour }=60 \text { minutes, } \\ & \log |\theta-20|=\frac{60}{15} \log \frac{1}{2}+\log 80 \\ & \Rightarrow \log \left(\frac{\theta-20}{80}\right)=4 \log \frac{1}{2} \\ & \Rightarrow \frac{\theta-20}{80}=\left(\frac{1}{2}\right)^4 \\ & \Rightarrow \theta=5+20=25^{\circ} \mathrm{C}\end{aligned}\)