A spherical metal ball at \(80^{\circ} \mathrm{C}\) cools in 5 minutes to \(60^{\circ} \mathrm{C}\), in surrounding temperature of \(20^{\circ} \mathrm{C}\), then the temperature of the ball after 20 minutes is approximately

Let \(\theta\) be the temperature of ball at any time ' \(t\) '.
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto(\theta-20) \\
& \quad \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-20), \mathrm{k}\gt0
\end{aligned}\)
Integrating on both sides, we get
\(\log |\theta-20|=-\mathrm{kt}+\mathrm{c}\)
when \(\mathrm{t}=0, \theta=80^{\circ}\)
\(\begin{array}{ll}
\therefore & c=\log 60 \\
\therefore & \log |\theta-20|=-k t+\log 60...(i)
\end{array}\)
\(\begin{aligned} & \quad \text { When } \mathrm{t}=5, \theta=60^{\circ} \\ & \therefore \quad \log 40=-5 \mathrm{k}+\log 60 \\ & \Rightarrow 5 \mathrm{k}=\log 60-\log 40 \\ & \Rightarrow 5 \mathrm{k}=\log \left(\frac{3}{2}\right) \\ & \Rightarrow \mathrm{k}=\frac{1}{5} \log \left(\frac{3}{2}\right)\end{aligned}\)
\(\therefore \quad \log |\theta-20|=\frac{-1}{5} \log \left(\frac{3}{2}\right) \mathrm{t}+\log 60\)
...[From (i)]
when \(\mathrm{t}=20\),
\(\begin{aligned}
& \Rightarrow \log |\theta-20|=\frac{-1}{5} \log \left(\frac{3}{2}\right) 20+\log 60 \\
& \Rightarrow \log |\theta-20|=-4 \log \left(\frac{3}{2}\right)+\log 60 \\
& \Rightarrow \log |\theta-20|=\log \left(\frac{2}{3}\right)^4+\log 60 \\
& \Rightarrow \log |\theta-20|=\log \left(\frac{16 \times 60}{81}\right) \\
& \Rightarrow \log |\theta-20|=\log (11.85) \\
& \Rightarrow \theta-20=11.85 \\
& \Rightarrow \theta=11.85+20 \\
& \Rightarrow \theta=31.85
\end{aligned}\)
The temperature of ball after 20 minutes is \(31.85^{\circ} \mathrm{C}\)