Hot water cools from 60^ C to 50^ C in the first 10 minutes and to...

By Newton's law of cooling \frac{\theta_1-\theta_2}{\mathrm{t}}=-\mathrm{K}\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right] where $\theta_0$ is the temperature of surrounding. Now, hot water cools from $60^{{}^{\circ}} \mathrm{C} to 50^{{}^{\circ}} \mathrm{C}$ in 10 minutes, $\frac{60-50}{10}=-$K\left[\frac{60+50}{2}-\theta_0\right] Again, it cools from 50^{{}^{\circ}} \mathrm{C} to 42^{{}^{\circ}} \mathrm{C}$in next 10 minutes.$\frac{50-42}{10}=-\mathrm{K}\left[\frac{50+42}{2}-\theta_0\right] Dividing equations (i) by (ii) we get$\begin{aligned}&\frac{1}{0.8}=\frac{55-\theta_0}{46-\theta_0}\\ &\frac{10}{8}=\frac{55-\theta_0}{46-\theta_0}\\ &460-10 \theta_0=440-8 \theta_0 \\ &2 \theta_0=20 \\ &\theta_0=10^{{}^{\circ}} \mathrm{c} \end{aligned}$$