A body is heated to $110^{\circ} \mathrm{C}$ and placed in air at $10^{\circ} \mathrm{C}$. After 1 hour its…
A body is heated to $110^{\circ} \mathrm{C}$ and placed in air at $10^{\circ} \mathrm{C}$. After 1 hour its temperature is
$60^{\circ} \mathrm{C}$. The additional time required for it to cool to $30^{\circ} \mathrm{C}$ is
$\left(\frac{\log 2}{\log 5}+1\right)$ hours
$\left(\frac{\log 5}{\log 2}\right)$ hours
$\left(\frac{\log 5}{\log 2}-1\right)$ hours
$\left(\frac{\log 2}{\log 5}\right)$ hours
Solution
Let $\theta$ be the temperature of body at time $t$. Temperature of air is given to be $10^{\circ} \mathrm{C}$. $=\theta_{0}$ (say).
$\begin{aligned}
& \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto\left(\theta-\theta_{0}\right) \\
& \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}\left(\theta-\theta_{0}\right), \mathrm{k}>0 \\
\therefore & \int \frac{\mathrm{d} \theta}{\theta-\theta_{0}}=\int-\mathrm{kt} \quad \Rightarrow \log \left|\theta-\theta_{0}\right|=-\mathrm{kt}+\log \mathrm{c}...(1)
\end{aligned}$
$\therefore\left(\frac{\theta-\theta_{0}}{\mathrm{c}}\right)=\mathrm{e}^{-\mathrm{kt}}$ $\therefore \theta=\theta_{0}+\mathrm{ce}^{-\mathrm{kt}}$ When $\mathrm{t}=0, \theta=110$ and $\theta_{0}=10, \quad \therefore \theta=10+100 \mathrm{e}^{-\mathrm{kt}}$ $\quad 110=10+\mathrm{c} \Rightarrow \mathrm{c}=100 \quad \therefore \quad \mathrm{c}$ When $\theta=60^{\circ} \mathrm{C}, \mathrm{t}=1$ Substituting value in equation (1), $\therefore \quad \log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1$
$\quad \mathrm{k}=-\log \left(\frac{1}{2}\right)$
$\therefore \quad \log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \cdot \log \left(\frac{1}{2}\right)$
When $\theta=30^{\circ} \mathrm{C}$, then
$\log \left(\frac{30-10}{100}\right)=\mathrm{t} \cdot \log \left(\frac{1}{2}\right)$
$\log \left(\frac{20}{100}\right)=t \cdot \log \frac{1}{2}$
$-\log 5=-t \log 2 \Rightarrow t=\frac{\log 5}{\log 2}$
$\therefore$ Additional time required is $\frac{\log 5}{\log 2}-1$