Q159 / 180 · KCET 2010 · Physics
Hot water cools from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the first $10 \mathrm{~min}$ and to $42^{\circ} \mathrm{C}$ in the next $10 \mathrm{~min}$. Then the temperature of the surroundings is
- A$20^{\circ} \mathrm{C}$
- B$30^{\circ} \mathrm{C}$
- C$15^{\circ} \mathrm{C}$
- Correct answerD$10^{\circ} \mathrm{C}$
Solution
According to Newton's law of cooling
$$
\frac{\theta_{2}-\theta_{1}}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{s}}\right]
$$
where, $\theta_{s}$ is the temperature of the surroundings.
$$
\begin{aligned}
\frac{60-50}{10} &=K\left[\frac{60+50}{2}-\theta_{s}\right] \\
1 &=K\left[55-\theta_{s}\right] \quad \text{...(i)}
\end{aligned}
$$
Similarly, $\quad \frac{50-42}{10}=\mathrm{K}\left(46-\theta_{\mathrm{s}}\right)$
$$
\frac{8}{10}=K\left(46-\theta_{s}\right) \quad \text{...(ii)}
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
&\frac{10}{8}=\frac{K\left(55-\theta_{s}\right)}{K\left(46-\theta_{s}\right)} \\
&\theta_{s}=10^{\circ} \mathrm{C}
\end{aligned}
$$
$$
\Rightarrow \quad \theta_{\mathrm{s}}=10^{\circ} \mathrm{C}
$$
Hence, the correct answer is option: 'D'.