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Water and salt problem

  1. Jun 24, 2017 #1
    1. The problem statement, all variables and given/known data
    A tank contains 800 gal of water in which 200 lb of salt is dissolved. Two gallons of fresh water runs in per minute, and 2 gal of the mixture in the tank, kept uniform by stirring, runs out per minute. How much salt is left in the tank after 5 hours ?

    2. Relevant equations



    3. The attempt at a solution

    Let ##y## be the amount of salt present in the tank at any given time.

    ##y' \propto -y ##

    ##y' = -Cy ##

    ## \frac {dy}{dt} = -Cy ##

    ##\int \frac {1}{y} dy = -\int Cdt ##

    ## y = ke^{-Ct} ##

    At ## t=0 , y_{0} = 200 = k##

    ## y = 200e^{-Ct} ##

    I don't know how to find the value of ##C##.
     
  2. jcsd
  3. Jun 24, 2017 #2

    Orodruin

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    What does this tell you?
     
  4. Jun 25, 2017 #3
    In the two gallons per minute, how much of it is salt and how much is water ?
     
  5. Jun 25, 2017 #4

    Orodruin

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    The salt is not notably going to change the volume of the water. (Note that salt water is denser than fresh water!) what does the statement "kept uniform by stirring" tell you?
     
  6. Jun 25, 2017 #5
    It has the same concentration of salt throughout the mixture?
     
  7. Jun 25, 2017 #6

    Orodruin

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    Yes. So what is the rate at which salt is removed?
     
  8. Jun 25, 2017 #7
    That is directly proportional to the amount of salt present in the tank at that time... according to my first equation.
     
  9. Jun 25, 2017 #8

    Orodruin

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    Yes, but you need the numbers and you can get them from this line of argumentation.
     
  10. Jun 25, 2017 #9
    At t = 0, taking y' = 0.5 lb/min as, in 2 gal/min of the mixture flowing out, there will be 0.5 lb of salt.

    ## y' = -200Ce^{-c(0)} ##
    ##0.5 = -200Ce^{-c(0)} ##
    ## C = -0.0025 ##

    at t = 5 hours = 300 min

    ## y = 200e^{-0.0025(300)} ##

    ## y = 94.4733 lb ##.

    Thanks for your help.
     
    Last edited: Jun 25, 2017
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