Question

A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when will it have lost 95% moisture, weather conditions remaining the same.

Answer 1

Let the original amount of moisture in the porous substance be N and the amount of moisture in the porous substance at any time t be P.

\[\text{ Given: }\frac{dP}{dt}\alpha P\]

\[ \Rightarrow \frac{dP}{dt} = - aP\]

\[ \Rightarrow \frac{dP}{P} = - a dt\]

\[ \Rightarrow \log \left| P \right| = - at + C . . . . . \left( 1 \right)\]

Now, P = N at t = 0

Putting P = N and t = 0 in (1), we get

\[\log \left| N \right| = C\]

\[\text{ Putting }C = \log \left| N \right|\text{ in }\left( 1 \right), \text{ we get }\]

\[\log \left| P \right| = - \text{ at }+ \log \left| N \right|\]

\[ \Rightarrow \log \left| \frac{P}{N} \right| = - \text{ at }. . . . . \left( 2 \right)\]

According to the question,

\[P = \frac{1}{2}N\text{ at }t = 1\]

\[\log \left| \frac{N}{2N} \right| = - a\]

\[ \Rightarrow a = \log \left| 2 \right|\]

\[\text{ Putting }a = \log \left| 2 \right|\text{ in }\left( 2 \right),\text{ we get }\]

\[\log \left| \frac{P}{N} \right| = - t \log\left| 2 \right|\]

To find the time when it will loss 95 % moisture, we have

\[P = \left( 1 - 95\%  \right)N = \frac{5}{100}N\]

\[ \therefore \log \left| \frac{5N}{100N} \right| = - t \log \left| 2 \right|\]

\[ \Rightarrow \log 20 = t \log 2\]

\[ \Rightarrow t = \frac{\log 20}{\log 2}\]