Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first 10 minutes and to…
Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. The temperature of the surroundings is:
$25^{\circ} \mathrm{C}$
$10^{\circ} \mathrm{C}$
$15^{\circ} \mathrm{C}$
$20^{\circ} \mathrm{C}$
Solution
By Newton's law of cooling
$
\frac{\theta_1-\theta_2}{\mathrm{t}}=-\mathrm{K}\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right]
$
where $\theta_0$ is the temperature of surrounding. Now, hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes,
$
\frac{60-50}{10}=-K\left[\frac{60+50}{2}-\theta_0\right]
$
Again, it cools from $50^{\circ} \mathrm{C}$ to $42^{\circ} \mathrm{C}$ in next 10 minutes.
$
\frac{50-42}{10}=-\mathrm{K}\left[\frac{50+42}{2}-\theta_0\right]
$
Dividing equations (i) by (ii) we get
$
\begin{aligned}
&\frac{1}{0.8}=\frac{55-\theta_0}{46-\theta_0} \\
&\frac{10}{8}=\frac{55-\theta_0}{46-\theta_0} \\
&460-10 \theta_0=440-8 \theta_0 \\
&2 \theta_0=20 \\
&\theta_0=10^{\circ} \mathrm{c}
\end{aligned}
$