If a body cools from \(80^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in the room temperature of \(30^{\circ} \mathrm{C}\) in 30 min , then the temperature of a body after one hour is

Let \(\theta\) be the temperature of the body at any time ' \(t\) '.
\(\begin{array}{ll}
\therefore & \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-30) \\
\therefore & \frac{\mathrm{d} \theta}{\mathrm{dt}}=k(\theta-30)
\end{array}\)
Integrating on both sides, we get
\(\begin{array}{ll}
& \log (\theta-30)=k t+C \\
& \text { when } \mathrm{t}=0, \theta=80^{\circ} \mathrm{C} \\
\therefore \quad & \log (80-30)=\mathrm{k}(0)+\mathrm{C} \\
\Rightarrow & C=\log 50
\end{array}\)
\(\begin{array}{ll}
\therefore \quad & \log (\theta-30)=\mathrm{kt}+\log 50 ...(i)\\
& \text { When } \mathrm{t}=30, \theta=60 \\
\therefore \quad & \log 30=30 \mathrm{k}+\log 50 \\
& \Rightarrow \log 30-\log 50=30 \mathrm{k} \\
& \Rightarrow \mathrm{k}=\frac{1}{30} \log \left(\frac{3}{5}\right)
\end{array}\)
Equation (i) becomes,
\(\log (\theta-30)=\frac{1}{30} \log \left(\frac{3}{5}\right) t+\log 50\)
when \(t=60\) minutes, we have
\(\begin{aligned}
& \log (\theta-30)=\frac{1}{30} \log \left(\frac{3}{5}\right) \times 60+\log 50 \\
& \log (\theta-30)=2 \log \left(\frac{3}{5}\right)+\log 50 \\
& \log (\theta-30)=\log \left(\frac{9}{25} \times 50\right) \\
& \Rightarrow \theta-30=18 \\
& \Rightarrow \theta=48^{\circ} \mathrm{C}
\end{aligned}\)