To solve this problem, we need to analyze how the concentration of salt changes over time as pure water enters the tank and the mixed solution drains out. Here’s a step-by-step approach:

Step 1: Initial Setup

• The initial volume of brine in the tank is half of 600 L, so it's 300 L.
• The initial salt content is 20 kg in 300 L, giving an initial salt concentration of: $\frac{20 \text{ kg}}{300 \text{ L}} = \frac{1}{15} \text{ kg/L}$

Step 2: Rates of Water Inflow and Outflow

• Inflow rate: Pure water enters at $10$ L/min, meaning the volume of liquid in the tank increases by 5 L/min (since 5 L/min also drains out).
• Outflow rate: The thoroughly mixed solution leaves at $5$ L/min.

Step 3: Determine When the Tank Fills

Since the tank fills at a net rate of $5$ L/min (because $10$ L/min in minus $5$ L/min out equals a net of $5$ L/min), we calculate the time required to fill the remaining 300 L: $t = \frac{300 \text{ L}}{5 \text{ L/min}} = 60 \text{ min}$ So, the tank will be completely filled after 60 minutes.

Step 4: Set Up a Differential Equation for Salt Content

Let $S(t)$ represent the amount of salt (in kg) in the tank at time $t$ minutes.

1. Inflow of Salt: Since pure water enters, no salt is added.
2. Outflow of Salt: The rate at which salt leaves is proportional to the concentration of salt in the solution, which is: $\text{Outflow rate of salt} = \frac{S(t)}{\text{Volume of solution}} \times 5 \text{ L/min}$ At time $t$, the volume in the tank is: $\text{Volume}(t) = 300 + 5t$ So, the outflow rate of salt is: $\frac{S(t)}{300 + 5t} \times 5$ Simplifying, we get: $\text{Outflow rate of salt} = \frac{5 S(t)}{300 + 5t} = \frac{S(t)}{60 + t}$

Step 5: Write the Differential Equation

The rate of change of $S(t)$ is given by: $\frac{dS}{dt} = -\frac{S(t)}{60 + t}$ This is a separable differential equation, and we can solve it by integrating.

Step 6: Solve the Differential Equation

Separating variables: $\frac{dS}{S} = -\frac{dt}{60 + t}$ Integrate both sides: $\ln S = -\ln(60 + t) + C$ Exponentiate both sides to solve for $S$: $S = \frac{C}{60 + t}$ Using the initial condition $S(0) = 20$: $20 = \frac{C}{60} \Rightarrow C = 1200$ Thus, the solution is: $S(t) = \frac{1200}{60 + t}$

Step 7: Find the Salt Content at $t = 60$ Minutes

Substitute $t = 60$: $S(60) = \frac{1200}{60 + 60} = \frac{1200}{120} = 10 \text{ kg}$ After 60 minutes, the salt content is $10$ kg.

Step 8: Determine the Concentration Reduction

The original salt content was $20$ kg. At $t = 60$ minutes, the salt content is $10$ kg, which is exactly half of the original salt content. Therefore, the concentration has been reduced to half of its initial value.

Conclusion

Yes, at the moment the tank is filled, the salt content will have been reduced to half of its original concentration.

Would you like a more detailed breakdown of any step or have further questions?

---

Questions for Further Exploration:

1. How would the solution change if the inflow and outflow rates were equal?
2. What if the inflow rate was slower than the outflow rate?
3. How would you analyze the concentration change if the initial salt content was different?
4. How would the model adjust if the tank was initially empty?
5. How could you model this situation if impurities were added along with the water?

Tip: When setting up differential equations for similar problems, always define the variables and the rates carefully, as this ensures accuracy in both the equation and its interpretation.