We are dealing with a classic mixing problem where salt enters a tank and mixes with water, while some of the solution flows out. The concentration of salt changes over time due to this inflow and outflow. Let's break this down step by step to derive the equation for the mass of salt in the tank after $t$ minutes and determine when the concentration of salt reaches $0.1 \, \text{kg/L}$.

Step 1: Define Variables

Let:

• $V(t)$ be the volume of the solution in the tank after $t$ minutes.
• $A(t)$ be the mass of salt (in kg) in the tank after $t$ minutes.

Step 2: Volume of Solution in the Tank

Initially, the tank contains 100 L of pure water, and brine flows in at 6 L/min while the solution flows out at 5 L/min. Therefore, the net rate of change in the volume of the solution is:

$\frac{dV}{dt} = 6 \, \text{L/min} - 5 \, \text{L/min} = 1 \, \text{L/min}.$

So, after $t$ minutes, the volume of the solution is:

$V(t) = 100 + t \, \text{L}.$

Step 3: Differential Equation for Salt Mass $A(t)$

The mass of salt in the tank changes due to:

• Inflow of salt: $0.5 \, \text{kg/L} \times 6 \, \text{L/min} = 3 \, \text{kg/min}$ entering the tank.
• Outflow of salt: The concentration of salt in the tank at any time $t$ is $\frac{A(t)}{V(t)}$, and the outflow rate is $5 \, \text{L/min}$. So, the rate of salt flowing out is $5 \times \frac{A(t)}{V(t)}$.

Thus, the differential equation for the rate of change of salt mass is:

$\frac{dA}{dt} = 3 - 5 \frac{A(t)}{V(t)}.$

Substitute $V(t) = 100 + t$ into this equation:

$\frac{dA}{dt} = 3 - 5 \frac{A(t)}{100 + t}.$

Step 4: Solve the Differential Equation

This is a linear first-order differential equation. We can solve it using an integrating factor.

1. Write the equation in standard form:

$\frac{dA}{dt} + \frac{5}{100 + t} A(t) = 3.$

2. The integrating factor $\mu(t)$ is:

$\mu(t) = e^{\int \frac{5}{100 + t} dt} = e^{5 \ln(100 + t)} = (100 + t)^5.$

3. Multiply the differential equation by the integrating factor:

$(100 + t)^5 \frac{dA}{dt} + 5(100 + t)^4 A(t) = 3(100 + t)^5.$

The left-hand side is the derivative of $(100 + t)^5 A(t)$, so we can rewrite it as:

$\frac{d}{dt} \left( (100 + t)^5 A(t) \right) = 3(100 + t)^5.$

4. Integrate both sides with respect to $t$:

$(100 + t)^5 A(t) = \int 3(100 + t)^5 dt = \frac{3(100 + t)^6}{6} + C.$

Simplifying:

$A(t) = \frac{(100 + t)}{2} + \frac{C}{(100 + t)^5}.$

Step 5: Apply Initial Condition

Initially, the tank contains pure water, so $A(0) = 0$. Substituting $t = 0$ and $A(0) = 0$ into the solution:

$0 = \frac{100}{2} + \frac{C}{100^5},$ $C = -50 \times 100^5.$

Now the solution for $A(t)$ becomes:

$A(t) = \frac{100 + t}{2} - \frac{50 \times 100^5}{(100 + t)^5}.$

Step 6: Find When Concentration Reaches $0.1 \, \text{kg/L}$

The concentration of salt in the tank is $\frac{A(t)}{V(t)}$. We want to find $t$ when the concentration is $0.1 \, \text{kg/L}$, i.e.,

$\frac{A(t)}{100 + t} = 0.1.$

Substitute the expression for $A(t)$:

$\frac{\frac{100 + t}{2} - \frac{50 \times 100^5}{(100 + t)^5}}{100 + t} = 0.1.$

This equation can now be solved numerically for $t$.

Would you like the exact solution for $t$, further explanation, or help with the final computation?