- #1
Scalper
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Homework Statement
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?
Homework Equations
Let x= g chlorine in the tank
let t= time in seconds
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 I am sure is not right. what did i do wrong? lol