Question

A body cools from $60^{\circ} C$ to $50^{\circ} C$ in $10$ min. If the room temperature is $25^{\circ} C$ and assuming Newton law of cooling to hold good, the temperature of the body at the end of the next $10$ min will be

• A. $45^{\circ} C$
• B. $42.85^{\circ} C$
• C. $40^{\circ} C$
• D. $38.5^{\circ} C$

Answer 1

Answer: (B)

As the body cools down, its rate of cooling slows down.
From Newtons law of cooling when a hot body is cooled in air, the rate of loss of heat by the body is proportional to the temperature difference between the body and its surroundings.
Given, $\theta_{1}=60{ }^{\circ} C, \theta_{2}=50{ }^{\circ} C, \theta=25{ }^{\circ} C$.
$\therefore $ Rate of loss of heat $=K$
(Mean temp.-Atmosphere temp.)
where $K$ is coefficient of thermal conductivity.
$\frac{\theta_{1}-\theta_{2}}{t}=K\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta\right)$
$\frac{60-50}{10}=K\left(\frac{60+50}{2}-25\right)$
$\Rightarrow K=\frac{1}{30}$
Also putting the value of $K$, we have
$\frac{50-\theta_{2}}{10}=\frac{1}{30}\left(\frac{50+\theta_{2}}{2}-25\right)$
$\Rightarrow \theta_{3}=42.85^{\circ} C$
Note: For Newtons law of cooling to hold good, temperature difference between the body and its surroundings should not be large.