If a body cools from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in the room temperature of \(25^{\circ} \mathrm{C}\) in 30 minutes, then the temperature of the body after 1 hour is

Let \(\theta\) be the temperature of the body at any time \(t\).
\(\therefore \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-25) \)
\( \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-25), \mathrm{k}>0\)
Integrating on both sides, we get
\(\log |\theta-25|=-\mathrm{kt}+\mathrm{c} \)
\( \text {When } \mathrm{t}=0, \theta=80^{\circ} \)
\( \therefore \log 55=0+\mathrm{c} \)
\( \Rightarrow \mathrm{c}=\log 55 \log |\theta-25|=-\mathrm{kt}+\log 55 \)
\( \therefore \text { When } \mathrm{t}=30, \theta=50^{\circ} \)
\( \therefore \log 25=-30 \mathrm{k}+\log 55 \)
\( \Rightarrow \mathrm{k}=-\frac{1}{30} \log \frac{5}{11} \)
\( \therefore \log |\theta-25|=\frac{\mathrm{t}}{30} \log \frac{5}{11}+\log 55\)
When \(\mathrm{t}=1\) hour \(=60\) minutes, ... [From (i)]
\(\log |\theta-25|=\frac{60}{30} \log \frac{5}{11}+\log 55 \)
\( \Rightarrow \log \left(\frac{\theta-25}{55}\right)=2 \log \left(\frac{5}{11}\right) \)
\( \Rightarrow \frac{\theta-25}{55}=\left(\frac{5}{11}\right)^2 \)
\( \Rightarrow \frac{\theta-25}{55}=\frac{25}{121} \)
\( \Rightarrow \theta=25+\frac{125}{11}=25+11.36=36.36^{\circ} \mathrm{C}\)