The flow system shown in the figure is activated at time $t=0$. Let $Q_i(t)$ denote the amount of solute present in the $i$th tank at time $t$. Assume that all the flow rates are a constant $10 \ \mathrm{L/min}$. It follows that the volume of solution in each tank remains constant; assume this volume to be $1000 \ \mathrm{L}$. The concentration of solute in the inflow to Tank 1 (from a source other than Tank 2) is $0.3 \ \mathrm{kg / L}$, and the concentration of solute in the inflow to Tank 2 (from a source other than Tank 1) is $0 \ \mathrm{kg / L}$. Assume each tank is mixed perfectly.

1. Set up a system of first-order differential equations that models this situation.
   
   

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $Q_1^{\,\prime}$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$ $Q_2^{\,\prime}$

   =

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $Q_1$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$ $Q_2$

   +

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

   
   
   
2. If $Q_1(0) = 20 \ \mathrm{kg}$ and $Q_2(0) = 0 \ \mathrm{kg}$, find the amount of solute in each tank after $t$ minutes.
   $Q_1(t) =$ kg
   $Q_2(t) =$ kg
   
   
3. As $t \to \infty$, how much solute is in each tank?
   In the long run, Tank 1 will have kg of solute.
   In the long run, Tank 2 will have kg of solute.
   (Reread the question and think about why this answer makes sense.)

You can earn partial credit on this problem.