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Homework Help: Change in concentration over time

  1. Jul 26, 2011 #1
    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. jcsd
  3. Jul 27, 2011 #2

    ehild

    User Avatar
    Homework Helper

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

    ehild
     
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