(Solved): The flow system shown in the figure is activated at time t=0. Let Q.(t) denote the amount of so ...

The flow system shown in the figure is activated at time $$t=0$$. Let $$Q_{.}(t)$$ 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 $$0.8\mathrm{kg}/\mathrm{L}$$, and the concentration of solute in the inflow to Tank 2 (from a source other than Tank 1) is $$0\mathrm{kg}/\mathrm{L}$$. Assume each tank is mixed perfectly. a. Set up a system of first-order differential equations that models this situation. $$\left[\begin{array}{l}{Q}_1^{?}\\ Q_2^{?}\end{array}\right]=[]\left[\begin{array}{l}{Q}_1\\ Q_2\end{array}\right]+[]$$ b. If $$Q_1(0)=10\mathrm{kg}$$ and $$Q_2(0)=0\mathrm{kg}$$, find the amount of solute in each tank after $$t$$ minutes. $$\begin{array}{l}{Q}_1(t)=\\ Q_2(t)=\end{array}$$ c. As $$t??$$, how much solute is in each tank? In the long run, Tank 1 will have $$\mathrm{kg}$$ of solute. In the long run, Tank 2 will have $$\mathrm{kg}$$ of solute. (Reread the question and think about why this answer makes sense.)