Let ${\theta }_0$ = temperature of surrounding,

Using Newton's law of cooling,

log$\frac{{\mathrm{\theta }}_2-{\mathrm{\theta }}_0}{{\mathrm{\theta }}_1-{\mathrm{\theta }}_0}=-Kt$

log $\frac{40-{\mathrm{\theta }}_0}{50-{\mathrm{\theta }}_0}=-K\times 5$ .....(i)

log $\frac{33.33-{\mathrm{\theta }}_0}{40-{\mathrm{\theta }}_0} =-K\times 5$ .....(ii)

From Eqs.(i) and (ii),

$\frac{40-{\mathrm{\theta }}_0}{50-{\mathrm{\theta }}_0\mathrm{ }} = \frac{33.33-{\mathrm{\theta }}_0}{40-{\mathrm{\theta }}_0}$

On solving, we get

${\theta }_0=19.95°\mathrm{C} \approx 20 °\mathrm{C}$