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
$\begin{aligned}
& \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-25) \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-25) \\
\therefore & \int \frac{\mathrm{d} \theta}{\theta-25}=\int-\mathrm{kdt} \\
& \log (\theta-25)=-\mathrm{kt}+\mathrm{c} ...(1) \\
\text { Initially when } \mathrm{t}=0, \theta=135 \\
\therefore & \log (135-25)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 110 \\
& \log (\theta-25)=-\mathrm{kt}+\log 110 \\
\therefore & \log \left(\frac{\theta-25}{110}\right)=-\mathrm{kt} ...(2)
\end{aligned}$
Now when $t=60, \theta=80$
$\begin{array}{l}
\text { w when } t=60, \theta=80 \\
\log \left(\frac{55}{110}\right)=-60 k \Rightarrow k=-\frac{1}{60} \log \left(\frac{1}{2}\right)
\end{array}$
From (2) $\log \left(\frac{\theta-25}{110}\right)=\frac{t}{60} \log \left(\frac{1}{2}\right)$
At $t=120$, we get $$ \log \left(\frac{\theta-25}{110}\right)=2 \log \frac{1}{2} \Rightarrow \log \left(\frac{\theta-25}{110}\right)=\log \left(\frac{1}{4}\right) $$ $\therefore \frac{\theta-25}{110}=\frac{1}{4} \Rightarrow 4 \theta-100=110 \Rightarrow 4 \theta=210 \Rightarrow \theta=52.5^{\circ} \mathrm{C}$