Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surrounding is

According to Newton’s law of cooling
$\frac{{\mathrm{\theta }}_1-{\mathrm{\theta }}_2}{\mathrm{t}}=\mathrm{K}\left[\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right]$
In the first case,
$\frac{(60-50)}{10}=\mathrm{K}\left[\frac{60+50}{2}-{\mathrm{\theta }}_0\right]\Rightarrow 1=\mathrm{K}(55-\mathrm{\theta })\dots \dots \left(\mathrm{i}\right)$
In the second case,
$\frac{(50-42)}{10}=K\left[\frac{50+42}{2}-{\theta }_0\right]\Rightarrow 0.8=k\left(46-{\theta }_0\right)\dots ..\left(\text{ ii }\right)$
Dividing (i) by (ii), we get
$\frac{1}{0.8}=\frac{55-\mathrm{\theta }}{46-\mathrm{\theta }}\Rightarrow 46-{\mathrm{\theta }}_0=44-0.8\mathrm{\theta }\Rightarrow {\mathrm{\theta }}_0={10}^0\mathrm{C}$