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

(D)
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
\(\log [295-290]]=-\frac{\log 2}{10} t+\log 80\)
\(\therefore \frac{(\log 5-\log 80)}{\log 2} \backslash(-10)=1\)
\(\frac{\log \left(\frac{1}{10}\right)}{\log 2} \backslash(-10)=1\)
\(\therefore t=\frac{-4 \log 2}{\log 2} \backslash(-10)\)
\(=40\)