# Homework Help: Change in concentration over time

1. Jul 26, 2011

### Scalper

1. The problem statement, all variables and given/known data
A tank with a capacity of 400L is full of a mixture of water and chlorine with a concentration of 0.05g of chlorine per liter. The chlorine concentration is to be reduced by pumping in fresh water at the rate of 4L/s. The mixture is kept stirred and pumped out at the rate of 10L/s. What is the concentration of chlorine in the tank 15 seconds later?

2. Relevant equations
Let x= g chlorine in the tank
let t= time in seconds
3. The attempt at a solution
dx/dt=(rate in)-(rate out)

dx/dt= [(4L/s)(0g/l)]-[(10L/s)(x/(400-6t))]

dx/dt=0-10x/(400-6t)

dx/dt=-5x/(200-3t)

(-1/5)(dx/x)=dt/(200-3t)

integrating both sides:

(-1/5)*ln(|x|)=(-1/3)*ln(|200-3t|)+C

3*ln(|x|)=5*ln(|200-3t|)+C

ln(|x|)=(5/3)*ln(|200-3t|)+C

|x|=e^[(5/3)*ln(|200-3t|)]+C

|x|=e^[ln(|200-3t|)]^(5/3)+C

e^ln cancels

|x|=(|200-3t|)^(5/3/)+C

solving for C, When t=0, x=400*.05=20

|20|=(|200-3*0|)^(5/3)+C

20-200^(5/3)=C

then solving for x when t=15

x=(|200-3*15|)^(5/3)+20-200^(5/3)

this gets a -2300ish, which im sure is not right. what did i do wrong? lol

2. Jul 27, 2011

### ehild

"]" is at the wrong place: |x|=e^[(5/3)*ln(|200-3t|)+C]

ehild