# Thread: Finding how much chemical is left over

1. ## Finding how much chemical is left over

A 500 liter tank originally contains 300 liters of liquid in which there's dissolved 50 grams of some chemical. Liquid containing 30 grams per liter of the dissolved chemical flows into the tank at the rate of 4 liters/min. The mixture is uniform and it flows out at a rate of 2 liters/min. How much of the chem is inside at the time it overflows?

Here's my work:

There is 300+2t liters of liquid in the tank. I'm going to simplify liters with L.

$\displaystyle Q(t) = $$\displaystyle 4 \displaystyle \frac{L}{min} x \displaystyle 30 \displaystyle \frac{grams}{L} , \displaystyle Q(0) = 50 \displaystyle Q(t) = \displaystyle 2 \frac{liters}{min} x \displaystyle \frac{Q(t)}{300+2t} \displaystyle Q'(t) = 120 - \frac{2Q(t)}{300+2t} \displaystyle = 120 - \frac{2Q(t)}{2(150+t)} \displaystyle = 120 - \frac{Q(t)}{150+t} \displaystyle Q'(t) + Q(t)\frac{1}{150+t} = 120 Linear Equation. \displaystyle e^{\int\frac{1}{150+t}dt \displaystyle = e^{ln|150+t|} \displaystyle = 150+t \displaystyle Q'(t)(150+t) + \frac{(150+t)Q(t)}{150+t} = 120(150+t) \displaystyle Q'(t)(150+t)+ Q(t) = 18000+120t \displaystyle \int [Q(t)(150+t)]'dt = \int 18000 dt + \int 120t dt \displaystyle Q(t)(150+t) = 18000t + 60t^2 + C \displaystyle Q(t) = \frac{18000t}{150+t}+\frac{60t^2}{150+t}+\frac{C}{ 150+t} Use initial condition \displaystyle Q(0) = 50. \displaystyle 50 = \frac{C}{150} \displaystyle 7500 = C \displaystyle Q(t) = \frac{18000t}{150+t}+\frac{60t^2}{150+t}+\frac{750 0}{150+t} Determining the time of overflow: \displaystyle 300+2t = 500 \displaystyle 2t = 200 \displaystyle t = 100 Sub in for t. \displaystyle Q(t) = \frac{18000(100)+60(100)^2+7500}{150+(100)} \displaystyle = 9,630 \displaystyle \frac{grams}. 2. Originally Posted by A Beautiful Mind A 500 liter tank originally contains 300 liters of liquid in which there's dissolved 50 grams of some chemical. Liquid containing 30 grams per liter of the dissolved chemical flows into the tank at the rate of 4 liters/min. The mixture is uniform and it flows out at a rate of 2 liters/min. How much of the chem is inside at the time it overflows? Here's my work: There is 300+2t liters of liquid in the tank. I'm going to simplify liters with L. \displaystyle Q(t) =$$\displaystyle 4$ $\displaystyle \frac{L}{min}$ x $\displaystyle 30$ $\displaystyle \frac{grams}{L}$ , $\displaystyle Q(0) = 50$

$\displaystyle Q(t) =$ $\displaystyle 2 \frac{liters}{min}$ x $\displaystyle \frac{Q(t)}{300+2t}$

$\displaystyle Q'(t) = 120 - \frac{2Q(t)}{300+2t}$

$\displaystyle = 120 - \frac{2Q(t)}{2(150+t)}$

$\displaystyle = 120 - \frac{Q(t)}{150+t}$

$\displaystyle Q'(t) + Q(t)\frac{1}{150+t} = 120$ Linear Equation.

$\displaystyle e^{\int\frac{1}{150+t}dt$

$\displaystyle = e^{ln|150+t|}$

$\displaystyle = 150+t$
$\displaystyle Q'(t)(150+t) + \frac{(150+t)Q(t)}{150+t} = 120(150+t)$

$\displaystyle Q'(t)(150+t)+ Q(t) = 18000+120t$

$\displaystyle \int [Q(t)(150+t)]'dt = \int 18000 dt + \int 120t dt$

$\displaystyle Q(t)(150+t) = 18000t + 60t^2 + C$

$\displaystyle Q(t) = \frac{18000t}{150+t}+\frac{60t^2}{150+t}+\frac{C}{ 150+t}$

Use initial condition $\displaystyle Q(0) = 50$.

$\displaystyle 50 = \frac{C}{150}$

$\displaystyle 7500 = C$

$\displaystyle Q(t) = \frac{18000t}{150+t}+\frac{60t^2}{150+t}+\frac{750 0}{150+t}$

Determining the time of overflow:

$\displaystyle 300+2t = 500$
$\displaystyle 2t = 200$
$\displaystyle t = 100$

Sub in for t.

$\displaystyle Q(t) = \frac{18000(100)+60(100)^2+7500}{150+(100)}$

$\displaystyle = 9,630$ $\displaystyle \frac{grams}$.
Looks OK.

You can check answers here: solve dx&#47;dt &#43; x&#47;&#40;150 &#43; t&#41; &#61; 120 and x&#40;0&#41; &#61; 50 - Wolfram|Alpha