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
$$
\begin{aligned}
& \log \left|\theta-\theta_0\right|=\frac{60}{30} \log \left|\frac{5}{11}\right|+\log |55| \\
& =\log \left(\left|\frac{5}{11}\right|\right)^2+\log |55|=\log \left|\frac{25}{121} \times 55\right|=\log \left|\frac{125}{11}\right| \\
& \therefore \theta-25=\frac{125}{11} \Rightarrow \theta=36.36^{\circ} \mathrm{C}
\end{aligned}
$$