A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the open air loses half its moisture during the first hour, then the time t , in which \(99 \%\) of the moisture will be lost, is

Let \(y\) be the amount of moisture at time t .
\(\begin{aligned}
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{dt}}=-\alpha y \\
& \Rightarrow \frac{\mathrm{~d} y}{y}=-\alpha \mathrm{dt}
\end{aligned}\)
Integrating on both sides, we get
\(\int \frac{\mathrm{d} y}{y}=-\alpha \int \mathrm{dt}\)
\(\therefore \quad \log y=-\alpha \mathrm{t}+\mathrm{c}\)
when \(\mathrm{t}=0, y=1\)
\(\therefore \quad\) From (i), we get
\(\mathrm{c}=0\)...(i)
when \(\mathrm{t}=1, y=0.5\)
\(\therefore \quad\) From (i) and (ii), we get
\(\begin{array}{ll}
& \log (0.5)=-\alpha \\
\therefore \quad & \alpha=\log 2
\end{array}\)
\(\therefore \quad\) From (i), (ii) and (iii), we get
\(\log y=-(\log 2) t\)
\(\therefore \quad\) When \(99 \%\) of the moisture will be lost,
\(\begin{array}{ll}
& y=0.01 \\
\therefore \quad & \log (0.01)=-(\log 2) t \\
\therefore \quad & t=\frac{2 \log 10}{\log 2}
\end{array}\)