Water at $100^{\circ} \mathrm{C}$ cools in 15 minutes to $75^{\circ} \mathrm{C}$ in a room temperature of…
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
$\left(\frac{400}{9}\right)^{\circ} \mathrm{C}$
$\left(\frac{526}{9}\right)^{\circ} \mathrm{C}$
$\left(\frac{335}{9}\right)^{\circ} \mathrm{C}$
$\left(\frac{175}{3}\right)^{\circ} \mathrm{C}$
Solution
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 $t=15, \theta=75^{\circ}$
$\begin{aligned}
& \log \left(\frac{50}{75}\right)=-15 \mathrm{~K} \Rightarrow \mathrm{K}=\frac{-1}{15} \log \left(\frac{2}{3}\right) \\
\therefore & \log \left(\frac{\theta-25}{75}\right)=\frac{\mathrm{t}}{15} \log \left(\frac{2}{3}\right) ...[from 2] \\
\text { When } \mathrm{t}=30^{\circ} & \log \left(\frac{\theta-25}{75}\right)=\frac{30}{15} \log \left(\frac{2}{3}\right) \\
& \log \left(\frac{\theta-25}{75}\right)=2 \log \left(\frac{2}{3}\right)=\log \left(\frac{2}{3}\right)^{2}=\log \left(\frac{4}{9}\right) \\
\therefore & \frac{\theta-25}{75}=\frac{4}{9} \Rightarrow \theta=\frac{525}{9}=\left(\frac{175}{3}\right)^{\circ} \mathrm{C}
\end{aligned}$