Question

A wet porus 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 its moisture during the first
hour, then find the time when it has lost 95% moisture provided the weather
conditions remain the same.

Answer 1

Let the moisture content be m.Thus, rate of moisture loss is given by $-dm/dt$ .

Given; $-{dm/dt} \propto m \implies dm/dt = -\lambda m$ , where $\lambda$ is a constant.

$\int dm/m =-\lambda \int dt$

$\implies m=m_0e^{- \lambda t}$ ; where $m_0$ is the initial moisture content at $t=0$ .

Also given, $m_0/2=m_0e^{-\lambda} \implies \lambda =(ln2 )hr^{-1}$

If 95% moisture is lost at time t; then $m=5\%(m_0)=0.05m_0$

Thus, $0.05 m_0=m_0e^{-t*ln2} \implies t=log_2(20)=4.322hr.=259.3min.$

Hence, the time when it has lost 95% moisture provided the weather conditions remain the same is after 4.322 hours.