Question

A body takes $10\min$ to cool from $60^{\circ }C$ to $50^{\circ }C$ . If the temperature of the surroundings is $25^{\circ }C$ and $527^{\circ }C$ respectively. The final temperature of the body is

• A. $48^{\circ }C$
• B. $46^{\circ }C$
• C. $49^{\circ }C$
• D. $42.85^{\circ }C$

Answer 1

Answer: (D)

According to Newton's law
$\frac{{\theta }_1-{\theta }_2}{t}=K\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0$
$\therefore \frac{60-50}{10}=K\frac{60+50}{2}-25\dots$ (i)
Let $\theta$ be the temperature after another $10\min$
$\therefore \frac{50-\theta }{10}=K\frac{\theta +50}{2}-25\dots \text{..(ii) }$
Dividing Eq.(i) by Eq. (ii), we get
$\frac{10}{50-\theta }=\frac{30\times 2}{\theta }$
$\therefore \theta =42.85^{\circ }C$