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

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.
\(\frac{60-50}{10} =K\left[\frac{60+50}{2}-\theta_{s}\right] \)
\(1 =K\left[55-\theta_{s}\right] \text{...(i)}\)
Similarly, \(\frac{50-42}{10}=\mathrm{K}\left(46-\theta_{\mathrm{s}}\right)\)
\(\frac{8}{10}=K\left(46-\theta_{s}\right) \text{...(ii)}\)
Dividing Eq. (i) by Eq. (ii), we get
\(\begin{aligned}&\frac{10}{8}=\frac{K\left(55-\theta_{s}\right)}{K\left(46-\theta_{s}\right)} \\&\theta_{s}=10^{\circ} \mathrm{C}\end{aligned}\)
\(\Rightarrow \theta_{\mathrm{s}}=10^{\circ} \mathrm{C}\)