The rate at which the metal cools in moving air is proportional to the difference of tempratures between the…
The rate at which the metal cools in moving air is proportional to the difference of tempratures between the metal and air. If the air temperature is $290 \mathrm{~K}$ and the metal temperature drops from $370 \mathrm{~K}$ to $330 \mathrm{~K}$ in 10 minutes, then the time required to drop the temperature upto $295 \mathrm{~K}$ is
40min
20min
35min
30min
Solution
We know that $\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{Tm})$, where
$\mathrm{k}$ is proportionality constant, $\mathrm{T}=$ Temperature of body, $\mathrm{Tm}=$ Temperature of surrounding.
$\therefore \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-290) \Rightarrow \int \frac{\mathrm{dT}}{\mathrm{T}-290}=\int-\mathrm{k} \mathrm{dt}$
$\log |\mathrm{T}-290|=-\mathrm{kt}+\mathrm{C}$
Initially, $\mathrm{t}=0, \mathrm{~T}=370$
$\therefore \log |370-290|=0+C \Rightarrow C=\log |80|$
$\therefore \log |\mathrm{T}-290|=-\mathrm{kt}+\log |80|$
When $\mathrm{t}=10, \mathrm{~T}=330$
$\therefore \log |330-290|=-10 k+\log |80|$
$\therefore \log \left[\left[\frac{40}{80}\right]\right]=-10 \mathrm{k} \Rightarrow \mathrm{k}=\frac{1}{10} \log 2$
$\therefore \log [|T-290|]=-\frac{\log 2}{10} t+\log |80|$
When $\mathrm{T}=295$, we write
$\log [295-290]]=-\frac{\log 2}{10} t+\log 80$
$\therefore \frac{(\log 5-\log 80)}{\log 2} (-10)=1$
$\frac{\log \left(\frac{1}{10}\right)}{\log 2} (-10)=1$
$\therefore t=\frac{-4 \log 2}{\log 2} (-10)$
$=40$