# Thread: losing moisture at a rate proportional to moisture content

1. ## losing moisture at a rate proportional to moisture content

A wet towel hung from a clothesline to dry loses moisture through evaporation at a rate proportional to its moisture content. If after hours the towel has lost % of its original moisture content, after how long will it have lost %?

t=____ hours

2. Originally Posted by ewkimchi
A wet towel hung from a clothesline to dry loses moisture through evaporation at a rate proportional to its moisture content. If after hours the towel has lost % of its original moisture content, after how long will it have lost %?

t=____ hours
$\frac{dy}{dt} = k.y$

where k is the rate, and y is the proportion of the moisture.

so,

$\frac{dy}{y} = k. dt$

Integrate both sides:

$\int \frac{dy}{y} = \int k. dt$

You will get:

$log(y) = kt + C$..................(1)

Use this equation to find out what your question is asking. First start with time $t=0$ to find out $C$. At time $t=0$, the towel

has not lost any moisture. That means $y=100$

So,

$log(100) = k \times 0 + C \rightarrow 2 = 0+C$

$\therefore C=2$

Now find k. To find k, use the first part of your question.

At time $t=5 \text{hours}$, the moisture lost is 10%. So the remaining moisture in the towel is $90$.

So, $log(90) = (k \times 5) + 2$

$\therefore k = -0.00915$

Now, that you have $k$ and $C$, find $t$.

Let $t$ be the time when 70% moisture is lost, that is, the remaining moisture is 30. So, use equation (1) in a similar manner to find the time(t).

3. ## THanks

but I got t= and it's wrong; what did you get?

4. Originally Posted by ewkimchi
but I got t= and it's wrong; what did you get?
I am sorry. I posted the wrong value of $k$.

$k = \frac{log(90)-2}{5} = -0.00915$

With this, and 70% moisture gone, the time should be

$log(30) = (-0.00915) \times t + 2$

which gives $t = 57.14$ hours