A liquid cools from 50°C to 45°C in 5 minutes  and from 45 to 41.5°C in the next 5 minutes. The temperature of the surrounding is

According to Newton's law of cooling 
$\mathrm{ms}\left(\frac{{\mathrm{\theta }}_1-{\mathrm{\theta }}_2}{\mathrm{t}}\right)\propto \left[\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right]$
where ${\mathrm{\theta }}_2$= temperature of surroundings.
In first case,
$\mathrm{ms}\left(\frac{50-45}{5}\right)\propto \left[\frac{50+45}{2}-{\mathrm{\theta }}_0\right] .....\left(1\right)$
In second case,
$\mathrm{ms}\left(\frac{45-41\cdot 5}{5}\right)\propto \left[\frac{45+41\cdot 5}{2}-{\mathrm{\theta }}_0\right] ....\left(2\right)$
Solving eqs, (1) and (2), we get
${\mathrm{\theta }}_0=33\cdot 3^{\circ }\mathrm{C}$