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 \ldots $ (i)
Let $\theta$ be the temperature after another $10 \,\min$
$\therefore \frac{50-\theta}{10}=K \frac{\theta+50}{2}-25 \ldots \text {..(ii) }$
Dividing Eq.(i) by Eq. (ii), we get
$\frac{10}{50-\theta}=\frac{30 \times 2}{\theta} $
$\therefore \theta=42.85^{\circ} C$