Answer to Question #183882 in Differential Equations for josephine

Question

If a wet sheet in a dryer losses its moisture at a rate proportional to its moisture content, and losses half of its moisture during the first 10 minutes, when it will be practically dry? Say when will it have lost 99% of its moisture.

Answer 1

$\dfrac{dV}{dt}∝ V$

Where V is Volume of moisture in towel

$\dfrac{dV}{dt }= kV$

Where k is proportionality constant

On Integrating,

$\ln V = kt +C$

At t = 0s, V=V_i m³

$C =\ln V_i$

At t=600s, V= 0.5V_i m³

$\ln(0.5V_i) =600k+ \ln V_i\\ \ln(0.5V_i) - \ln V_i = 600k\\$

$\dfrac{\ln(\frac{V_i}{2Vi})}{600 }= k$

$k = \dfrac{\ln(\frac{1}{2})}{600}$

$k = \dfrac{\ln(1) - \ln(2)}{600}$

$k= \dfrac{0 -ln(2)}{600 }= \dfrac{\ln(2)}{600}=\dfrac{ -0.693}{600}$

$k = -1.1552 × 10^{{-3}}$

According to the question,

$V= V_i - \dfrac{99V_i}{100} = \dfrac{Vi}{100}$

$\ln\dfrac{V_i}{100}= -1.1552×10^{-3} t +\ lnV_i$

$\dfrac{\ln V_i - ln\frac{V_i}{100}}{1.1552×10^{-3}}= t$

$t = \dfrac{\ln(100)}{1.1552×10^{-3}}=\dfrac{4.6051}{1.1552×10^{-3}}$

$t = 3986 s= 66.43\ min = 1.107\ hour$

Note: You can also use mass or moles instead of volume in the calculations because it will still cancel out.