Question

A body cools from $50{ }^{\circ} C$ to $40{ }^{\circ} C$ in 5 minutes. Its temperature comes down to $33.33$ { }^{\circ} C$ in the next $5$ minutes. The temperature of the surroundings is

• A. $15^\circ C$
• B. $20{ }^{\circ} C$
• C. $25 \quad{ }^{\circ} C$
• D. $10^\circ C$

Answer 1

Answer: (B)

Let $\theta _{0}$ = temperature of surrounding,
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
$\theta_{0}=19.95^{\circ} C \quad \approx 20{ }^{\circ} C$