If a body cools from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the room temperature of…
If a body cools from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in the room temperature of $25^{\circ} \mathrm{C}$ in 30 minutes, then the temperature of the body after 1 hour is
$31.36^{\circ} \mathrm{C}$
$32.25^{\circ} \mathrm{C}$
$36.36^{\circ} \mathrm{C}$
$33.25^{\circ} \mathrm{C}$
Solution
Let $\theta$ be the temperature of the body at any time $t$.
$\begin{aligned}
\therefore \quad & \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-25) \\
& \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-25), \mathrm{k}>0
\end{aligned}$
Integrating on both sides, we get
$\begin{aligned}
& \log |\theta-25|=-\mathrm{kt}+\mathrm{c} \\
& \text { When } \mathrm{t}=0, \theta=80^{\circ} \\
\therefore \quad & \log 55=0+\mathrm{c} \\
\Rightarrow \quad \mathrm{c}=\log 55 & \log |\theta-25|=-\mathrm{kt}+\log 55 \\
\therefore \quad & \text { When } \mathrm{t}=30, \theta=50^{\circ} \\
\therefore \quad & \log 25=-30 \mathrm{k}+\log 55 \\
\Rightarrow & \mathrm{k}=-\frac{1}{30} \log \frac{5}{11} \\
\therefore \quad & \log |\theta-25|=\frac{\mathrm{t}}{30} \log \frac{5}{11}+\log 55
\end{aligned}$
When $\mathrm{t}=1$ hour $=60$ minutes, ... [From (i)]
$\begin{aligned}
& \log |\theta-25|=\frac{60}{30} \log \frac{5}{11}+\log 55 \\
& \Rightarrow \log \left(\frac{\theta-25}{55}\right)=2 \log \left(\frac{5}{11}\right) \\
& \Rightarrow \frac{\theta-25}{55}=\left(\frac{5}{11}\right)^2 \\
& \Rightarrow \frac{\theta-25}{55}=\frac{25}{121} \\
& \Rightarrow \theta=25+\frac{125}{11}=25+11.36=36.36^{\circ} \mathrm{C}
\end{aligned}$