Question

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

• A. $15^{\circ }C$
• B. $20{}^{\circ }C$
• C. $25{}^{\circ }C$
• 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
${\theta }_0=19.95^{\circ }C\approx 20{}^{\circ }C$