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
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
$\displaystyle \frac{dy}{dt} = k.y$
where k is the rate, and y is the proportion of the moisture.
so,
$\displaystyle \frac{dy}{y} = k. dt$
Integrate both sides:
$\displaystyle \int \frac{dy}{y} = \int k. dt$
You will get:
$\displaystyle log(y) = kt + C$..................(1)
Use this equation to find out what your question is asking. First start with time $\displaystyle t=0$ to find out $\displaystyle C$. At time $\displaystyle t=0$, the towel
has not lost any moisture. That means $\displaystyle y=100$
So,
$\displaystyle log(100) = k \times 0 + C \rightarrow 2 = 0+C$
$\displaystyle \therefore C=2$
Now find k. To find k, use the first part of your question.
At time $\displaystyle t=5 \text{hours}$, the moisture lost is 10%. So the remaining moisture in the towel is $\displaystyle 90$.
So, $\displaystyle log(90) = (k \times 5) + 2 $
$\displaystyle \therefore k = -0.00915$
Now, that you have $\displaystyle k $ and $\displaystyle C$, find $\displaystyle t$.
Let $\displaystyle t$ 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).