If the body cools from \(135^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) at room temperature of \(25^{\circ} \mathrm{C}\) in 60 minutes,
then the temperature of body after 2 hours is

Let \(\theta^{\circ} \mathrm{C}\) be the temperature of the body at time \(\mathrm{t}\) min. Room temp is given \(25^{\circ} \mathrm{C}\) Then by Newton's law of cooling, we write
\( \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-25) \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-25) \)
\( \therefore \int \frac{\mathrm{d} \theta}{\theta-25}=\int-\mathrm{kdt} \)
\( \log (\theta-25)=-\mathrm{kt}+\mathrm{c} ...(1) \)
\( \text {Initially when } \mathrm{t}=0, \theta=135 \)
\( \therefore \log (135-25)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 110 \)
\( \log (\theta-25)=-\mathrm{kt}+\log 110 \)
\( \therefore \log \left(\frac{\theta-25}{110}\right)=-\mathrm{kt} ...(2) \)
Now when \(t=60, \theta=80\)
\(\text {w when } t=60, \theta=80 \)
\( \log \left(\frac{55}{110}\right)=-60 k \Rightarrow k=-\frac{1}{60} \log \left(\frac{1}{2}\right)\)
From (2) \(\log \left(\frac{\theta-25}{110}\right)=\frac{t}{60} \log \left(\frac{1}{2}\right)\)
At \(t=120\), we get \( \log \left(\frac{\theta-25}{110}\right)=2 \log \frac{1}{2} \Rightarrow\) \(\log \left(\frac{\theta-25}{110}\right)=\log \left(\frac{1}{4}\right) \) \(\therefore \frac{\theta-25}{110}=\frac{1}{4} \Rightarrow 4 \theta-100=110 \Rightarrow\) \(4 \theta=210 \Rightarrow \theta=52.5^{\circ} \mathrm{C}\)