Question

100 Ib of NaCl are initially dissolved in 600 gal of a NaCl solution in a container. Water is continuously added to the container at a rate of 6 gal/min and the container content, which is kept uniform by mixing, is also withdrawn out of the container at 6 gal/min. Determine how much NaCl will be in the tank at the end of 3 hours.

Answer 1

16.52 lb NaCl will be in the tank at the end of 3 hours.

Step-by-step explanation:

Let s(t) = gal of NaCl in a container at time t (in minutes).

s'(t) = The rate at which the amount of gal of NaCl in the container is changing.

s'(t) = Rate of NaCl going in - Rate of NaCl going out

`s'(t)=0-(6)/(600)s`

`(ds)/(dt)=-(1)/(100)s`

By variable separable method we get

`(ds)/(s)=-(1)/(100)dt`

Integrate both sides.

`\int (ds)/(s)=\int -(1)/(100)dt`

`\ln (s)=-(t)/(100)+ln C`

`e^(\ln (s))=e^{-(t)/(100)+ln C}`

`s=e^{-(t)/(100)}* e^(ln C)`

`s=Ce^{-(t)/(100)}`
`[\because e^(\ln x)=x]`

100 Ib of NaCl are initially dissolved in 600 gal of a NaCl solution in a container. it means s(0)=100.

`100=Ce^{-((0))/(100)}`

`100=C`

Therefore the required function is

`s=100e^{-(t)/(100)}` .... (1)

We need to find the amount of NaCl will be in the tank at the end of 3 hours.

3 hours = 180 minutes.

Substitute t=180 in equation (1).

`s=100e^{-((180))/(100)}`

`s=16.52988`

`s\approx 16.52`

Therefore, 16.52 lb NaCl will be in the tank at the end of 3 hours.