If a body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\) after 1 hour its temperature is \(60^{\circ} \mathrm{C}\), then the additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is

\(\begin{aligned}
& \frac{\mathrm{d} T}{\mathrm{~d} t}=-K(T-10) \\
& \Rightarrow T-10=e^{-k t+C} \\
& \Rightarrow T=10+e^C \cdot e^{-k t}
\end{aligned}\)
For \(t=0\),
\(\begin{aligned}
& T=110 \\
& \Rightarrow e^C=100 \text { i.e. } \\
& T=10+100 \cdot e^{-k t}
\end{aligned}\)
For \(t=1\),
\(\begin{aligned}
& T=60 \\
& \Rightarrow 60=10+100 \cdot e^{-k \times 1} \\
& \Rightarrow-k=\log \frac{1}{2} \\
& \Rightarrow k=\log 2
\end{aligned}\)
\(\Rightarrow T=10+100 \cdot e^{-\left(\log _2\right) t}\)
Putting \(T=30\)
\(\begin{aligned}
& 30=10+100 \cdot e^{-\left(\log _2\right) t} \\
& \Rightarrow \log \left(\frac{1}{5}\right)=-(\log 2) t \\
& \Rightarrow t=\frac{\log 5}{\log 2}
\end{aligned}\)
Additional time \(=t-1=\frac{\log 5}{log2}-1\)