Hot water cools from ${60}^0 C$ to ${50}^0$C in the first 10 minutes and to ${42}^0$C in the next 10 minutes. The temperature of the surrounding is

According to Newton’s law of cooling$\frac{{\theta }_1-{\theta }_2}{t}=K\left[\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0\right]$
In the first case$\frac{(60-50)}{10}=\mathrm{K}\left[\frac{60+50}{2}-{\mathrm{\theta }}_0\right]$
In the second case$\frac{(50-42)}{10}=K\left[\frac{50+42}{2}-{\theta }_0\right]$
Solving the above,$\frac{1}{0.8}=\frac{55-\mathrm{\theta }}{46-\mathrm{\theta }}\Rightarrow {\mathrm{\theta }}_0={10}^{\circ }\mathrm{C}$