(b)
(17 marks)
A tank contains 900L of water with 15kg of dissolved mineral. Water is injected into the tank at a rate of 10L/minute while the solution is mixed thoroughly in the tank. At the same time, the solution ejected from the tank at the same rate as the injection of water. How much mineral is removed from the tank after an hour? (8 marks)

To solve this problem, we need to model the situation using a differential equation that represents the amount of dissolved mineral in the tank over time.

### Step 1: Set up the differential equation

Let \( M(t) \) represent the amount of dissolved mineral (in kg) in the tank at time \( t \) (in minutes). Initially, \( M(0) = 15 \) kg.

- The volume of the solution in the tank remains constant at 900 liters since water is injected and ejected at the same rate.
- The rate at which water is injected into the tank is 10 L/min.
- The concentration of the mineral in the tank at any time \( t \) is \( \frac{M(t)}{900} \) kg/L.
- The amount of mineral leaving the tank per minute is given by \( \text{(rate of outflow)} \times \text{(concentration)} \), which is \( 10 \times \frac{M(t)}{900} \) kg/min.

The rate of change of the mineral in the tank is the difference between the inflow and outflow rates. Since there is no mineral in the inflow (pure water is injected), the differential equation is:

\[
\frac{dM(t)}{dt} = - \frac{10}{900} M(t)
\]

This simplifies to:

\[
\frac{dM(t)}{dt} = -\frac{1}{90} M(t)
\]

### Step 2: Solve the differential equation

This is a first-order linear differential equation. We can solve it using separation of variables.

\[
\frac{dM(t)}{M(t)} = -\frac{1}{90} dt
\]

Integrating both sides:

\[
\ln |M(t)| = -\frac{t}{90} + C
\]

Exponentiate both sides to solve for \( M(t) \):

\[
M(t) = e^{-\frac{t}{90} + C} = e^C e^{-\frac{t}{90}} = M(0) e^{-\frac{t}{90}}
\]

Given the initial condition \( M(0) = 15 \) kg:

\[
M(t) = 15 e^{-\frac{t}{90}}
\]

### Step 3: Determine the amount of mineral removed after one hour

After one hour (i.e., 60 minutes), the amount of mineral in the tank is:

\[
M(60) = 15 e^{-\frac{60}{90}} = 15 e^{-\frac{2}{3}}
\]

Calculate the numerical value:

\[
M(60) \approx 15 \times 0.5134 = 7.701 \text{ kg}
\]

### Step 4: Find the amount of mineral removed

The amount of mineral removed from the tank after one hour is:

\[
\text{Mineral removed} = M(0) - M(60) = 15 - 7.701 \approx 7.299 \text{ kg}
\]

### Final Answer

Approximately **7.299 kg** of mineral is removed from the tank after one hour.

Would you like more details on any part of this solution?

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Here are 5 related questions to expand your understanding:

1. What would happen to the mineral concentration if the rate of water injection increased?
2. How would the differential equation change if the outflow rate was different from the inflow rate?
3. How does the amount of mineral removed change if the process continued for two hours instead of one?
4. Can this problem be solved using another method, like integrating factors?
5. How does the solution change if the initial amount of dissolved mineral was different?

**Tip:** When solving differential equations, always pay close attention to initial conditions, as they are crucial in finding the specific solution to the problem.

Learn how to calculate the amount of mineral removed from a tank over time using a differential equation. This problem involves injecting water into a tank while the solution is mixed and ejected at the same rate, affecting the concentration of dissolved mineral. Detailed solution with step-by-step explanation.

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