Let's solve the problem step by step: --- **Step 1: Find the steady-state concentration in Tank 1 as \( t \to \infty \).** - Tank 1 receives inflow with a concentration of 0.3 kg/L, at a rate of 10 L/min. - The total volume in Tank 1 is constant at 1000 L. - Outflow from Tank 1 is also 10 L/min (since the volume is constant). Let \( Q_1(t) \) be the amount of solute in Tank 1 at time \( t \). The concentration in Tank 1 is \( \frac{Q_1(t)}{1000} \) kg/L. **At steady state (\( t \to \infty \)), the rate of solute in = rate of solute out:** \[ \text{Inflow rate} = 10 \text{ L/min} \times 0.3 \text{ kg/L} = 3 \text{ kg/min} \] \[ \text{Outflow rate} = 10 \text{ L/min} \times \left(\frac{Q_1}{1000}\right) \text{ kg/L} \] Set inflow rate = outflow rate: \[ 3 = 10 \times \left(\frac{Q_1}{1000}\right) \implies Q_1 = 300 \] **Explanation:** As \( t \to \infty \), Tank 1 will have **300 kg of solute**. --- **Step 2: Find the steady-state concentration in Tank 2 as \( t \to \infty \).** - Tank 2 receives all its inflow from Tank 1 (since inflow from other sources has 0 solute). - The inflow rate is 10 L/min, and the concentration entering Tank 2 is the same as the outflow concentration of Tank 1, which is \( \frac{Q_1}{1000} = 0.3 \) kg/L. - The outflow rate from Tank 2 is also 10 L/min. Let \( Q_2 \) be the amount of solute in Tank 2 at steady state. \[ \text{Inflow rate} = 10 \text{ L/min} \times 0.3 \text{ kg/L} = 3 \text{ kg/min} \] \[ \text{Outflow rate} = 10 \text{ L/min} \times \left(\frac{Q_2}{1000}\right) \text{ kg/L} \] Set inflow = outflow: \[ 3 = 10 \times \left(\frac{Q_2}{1000}\right) \implies Q_2 = 300 \] **Explanation:** As \( t \to \infty \), Tank 2 will have **300 kg of solute**. --- ### Final Answer - **In the long run, Tank 1 will have 300 kg of solute.** - **In the long run, Tank 2 will have 300 kg of solute.**