A liquid takes 5 min to cool from 80^oC to 50^oC. How much time will it take to cool from 60^oC to 30^oC? The temperature of the surrounding is 20^oC.

According to Newton's law of cooling

$\begin{array}{l} \left(\frac{{\mathrm{\theta }}_1-{\mathrm{\theta }}_2}{\mathrm{t}}\right)=\mathrm{K}\left[\left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right)\right]\\ \mathrm{so} \mathrm{that} \left(\frac{80-50}{5}\right)=\mathrm{K}\left[\left(\frac{80+50}{2}-20\right)\right]\\ \mathrm{And} \left(\frac{60-30}{\mathrm{t}}\right)=\mathrm{K}\left[\left(\frac{60+30}{2}\right)-20\right]\end{array}$

Solving these for t, we get t = 9 min.