Water at \(100^{\circ} \mathrm{C}\) cools in 15 minutes to \(75^{\circ} \mathrm{C}\) in a room temperature of \(25^{\circ} \mathrm{C}\). Then the
temperature of water after 30 minutes is

Let \(\theta^{\circ} \mathrm{C}\) be the temperature of water at time \(\mathrm{t}\) min. Room temperature is given \(25^{\circ} \mathrm{C}\). then by Newton's law of cooling \(\frac{-d \theta}{d t} \propto(\theta-25)\)
\(\therefore \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{K}(\theta-25)\) where \(\mathrm{K}>0\)
\(\therefore \frac{\mathrm{d} \theta}{\theta-25}=-\mathrm{K} \mathrm{dt} \Rightarrow \int \frac{\mathrm{d} \theta}{\theta-25}=-\mathrm{K} \int \mathrm{dt}\)
\(\log (\theta-25)=-\mathrm{Kt}+\mathrm{c}\) \(\ldots(1)\)
When \(\mathrm{t}=0, \theta=100^{\circ}\)
\(\therefore \log 75=\mathrm{c}\)
\(\therefore\) from (1) \(\log (\theta-25)=-\mathrm{K} t+\log 75\)
\(\therefore \log \left(\frac{\theta-25}{75}\right)=-\mathrm{Kt}\) \(\ldots(2)\)
When \(t=15, \theta=75^{\circ}\)
\(\log \left(\frac{50}{75}\right)=-15 \mathrm{~K} \Rightarrow \mathrm{K}=\frac{-1}{15} \log \left(\frac{2}{3}\right) \)
\( \therefore \log \left(\frac{\theta-25}{75}\right)=\frac{\mathrm{t}}{15} \log \left(\frac{2}{3}\right) ...[from 2] \)
\( \text { When } \mathrm{t}=30^{\circ} \log \left(\frac{\theta-25}{75}\right)=\frac{30}{15} \log \left(\frac{2}{3}\right) \)
\( \log \left(\frac{\theta-25}{75}\right)=2 \log \left(\frac{2}{3}\right)=\log \left(\frac{2}{3}\right)^{2}=\log \left(\frac{4}{9}\right) \)
\( \therefore \frac{\theta-25}{75}=\frac{4}{9} \Rightarrow \theta=\frac{525}{9}=\left(\frac{175}{3}\right)^{\circ} \mathrm{C}\)