Explanation

(b) Using Newton’s law of cooling,
log$\ \frac{\theta_{2} - \theta_{0}}{\theta_{1} - \theta_{0}} = - Kt$
Log $\frac{40 - \theta_{0}}{50 - \theta_{0}} = \ - K \times 5$ …..(i)
Log $\frac{33.33 - \theta_{0}}{40 - \theta_{0}}\ = \ - K \times 5$ …..(ii)
From Eqs.(i) and (ii),
$\frac{40 - \theta_{0}}{50 - \theta_{0}\ }\ = \ \frac{33.33 - \theta_{0}}{40 - \theta_{0}}$
On solving, we get
$\theta_{0} = 19.95℃\ \approx 20℃$