If the surrounding air is kept at $25^{\circ} \mathrm{C}$ and a body cools from $80^{\circ} \mathrm{C}$ to…
If the surrounding air is kept at $25^{\circ} \mathrm{C}$ and a body cools from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 30 minutes, then temperature of the body after one hour will be
$31.72^{\circ} \mathrm{C}$ approximately
$34.74^{\circ} \mathrm{C}$ approximately
$32.36^{\circ} \mathrm{C}$ approximately
$36.36^{\circ}$ approximately
Solution
By Newton's law of cooling, we write
$\begin{aligned}
& \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto\left(\theta-\theta_0\right) \\
& \therefore \frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta_0\right) \Rightarrow \int\left(\frac{\mathrm{d} \theta}{\theta-\theta_0}\right)=\int \mathrm{kdt} \\
& \therefore \log \left|\theta-\theta_0\right|=\mathrm{kt}+\mathrm{c} \\
& \text { When } \mathrm{t}=0, \theta=80 \text { and } \theta_0=25 \\
& \therefore \log |80-25|=0+\mathrm{c} \quad \Rightarrow \mathrm{c}=\log |55| \\
& \text { When } \mathrm{t}=30, \theta=50 \\
& \therefore \log |50-25|=30 \mathrm{k}+\log |55| \\
& \therefore \mathrm{k}=\frac{1}{30} \log \left|\frac{5}{11}\right|
\end{aligned}$
from (1), (2), (3) we write
$\log \left|\theta-\theta_0\right|=\frac{1}{30} \log \left|\frac{5}{11}\right| t+\log |55|$
When $t=60$, we get
$\begin{aligned} & \log \left|\theta-\theta_0\right|=\frac{60}{30} \log \left|\frac{5}{11}\right|+\log |55| \\ & =\log \left(\left|\frac{5}{11}\right|\right)^2+\log |55|=\log \left|\frac{25}{121} \times 55\right|=\log \left|\frac{125}{11}\right| \\ & \therefore \theta-25=\frac{125}{11} \Rightarrow \theta=36.36^{\circ} \mathrm{C}\end{aligned}$