If the surrounding air is kept at \(25^{\circ} \mathrm{C}\) and a body cools from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 30 minutes, then temperature of the body after one hour will be

By Newton's law of cooling, we write
\(
\begin{aligned}
& \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto\left(\theta-\theta_0\right) \\
& \therefore \frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta_0\right) \Rightarrow \int\left(\frac{\mathrm{d} \theta}{\theta-\theta_0}\right)=\int \mathrm{kdt} \\
& \therefore \log \left|\theta-\theta_0\right|=\mathrm{kt}+\mathrm{c} \\
& \text { When } \mathrm{t}=0, \theta=80 \text { and } \theta_0=25 \\
& \therefore \log |80-25|=0+\mathrm{c} \quad \Rightarrow \mathrm{c}=\log |55| \\
& \text { When } \mathrm{t}=30, \theta=50 \\
& \therefore \log |50-25|=30 \mathrm{k}+\log |55| \\
& \therefore \mathrm{k}=\frac{1}{30} \log \left|\frac{5}{11}\right|
\end{aligned}
\)
From (1), (2), (3) we write
\(
\log \left|\theta-\theta_0\right|=\frac{1}{30} \log \left|\frac{5}{11}\right| t+\log |55|
\)
When \(t=60\), we get
\( \log \left|\theta-\theta_0\right|=\frac{60}{30} \log \left|\frac{5}{11}\right|+\log |55| \)
\( =\log \left(\left|\frac{5}{11}\right|\right)^2+\log |55|=\log\) \(\left|\frac{25}{121} \times 55\right|=\log \left|\frac{125}{11}\right| \)
\( \therefore \theta-25=\frac{125}{11} \Rightarrow \theta=36.36^{\circ} \mathrm{C}\)