Problem 34 - Calculus: Early Transcendentals

Question

A tank with a capacity of 400 L is full of a mixture of water and chlorine with a concentration of 0.05 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 4 L/s. The mixture is kept stirred and is pumped out at a rate of 10 L/s. Find the amount of chlorine in the tank as a function of time.

Step-by-Step Solution

Answer

\( C(t) = \frac{20}{400^{\frac{5}{3}}} (400 - 6t)^{\frac{5}{3}} \)

Step 1: Understanding the Problem

We have a tank initially filled with 400 L of a water-chlorine mixture at a concentration of 0.05 g/L. Fresh water is added at a rate of 4 L/s, causing a dilution, while the well-mixed solution is pumped out at 10 L/s. Our goal is to find the amount of chlorine as a function of time.

Step 2: Setting Up the Model

Let \( C(t) \) be the amount of chlorine in grams at time \( t \) seconds. The concentration of chlorine at time \( t \) is \( \frac{C(t)}{V(t)} \), where \( V(t) \) is the volume of water in the tank at any given time.

Step 3: Determine the Volume as a Function of Time

The volume of the solution decreases because the rate out (10 L/s) is faster than the rate in (4 L/s). Thus, the volume at time \( t \) is given by: \[ V(t) = 400 - 6t \] as the net flow rate is -6 L/s.

Step 4: Setting Up the Chlorine Balance Equation

Chlorine is being removed at a rate determined by its concentration and the outflow rate. The change in chlorine amount over time is: \[ \frac{dC}{dt} = -10 \cdot \frac{C(t)}{V(t)} \] Substitute \( V(t) = 400 - 6t \): \[ \frac{dC}{dt} = -10 \cdot \frac{C(t)}{400 - 6t} \].

Step 5: Solving the Differential Equation

Rearrange the equation into separable differential form and integrate: \[ \int \frac{dC}{C} = -10 \int \frac{dt}{400 - 6t} \]. Solving these integrals gives: \[ \ln |C| = \frac{10}{6} \ln |400 - 6t| + C_1 \] or equivalently, \( C(t) = K(400 - 6t)^{\frac{5}{3}} \) for some constant \( K \).

Step 6: Determining the Constant of Integration

Initially, at \( t = 0 \), the amount of chlorine \( C(0) = 0.05 imes 400 = 20 \) g. Substituting \( t = 0 \) in \( C(t) = K(400)^{\frac{5}{3}} \), we get: \[ 20 = K(400)^{\frac{5}{3}} \]. Solve for \( K \): \[ K = \frac{20}{400^{\frac{5}{3}}} \].

Step 7: Final Solution

Substitute \( K \) back into the function to get the final expression for \( C(t) \): \[ C(t) = \frac{20}{400^{\frac{5}{3}}} (400 - 6t)^{\frac{5}{3}} \]. This gives the amount of chlorine as a function of time.

Key Concepts

Mixture Problems

Mixture problems involve substances that are combined in varying quantities over time, which often leads to changes in concentration. These kinds of problems are common in fields like chemistry and environmental science. In our specific exercise, we have a tank initially filled with a water-chlorine mixture. Fresh water is entering the tank while the mixed solution exits, leading to a continual change in the mixture's composition.

The essence of solving mixture problems is to track how each component of the mixture changes over time. This requires:

• Identifying the initial conditions: Here, the tank starts with 400 L of a mixture containing 0.05 g of chlorine per liter.
• Understanding the rates of flow: Fresh water enters at 4 L/s and the mixture exits at 10 L/s, making the net flow rate -6 L/s.
• Formulating a mathematical model to describe changes in composition: We express such changes via differential equations, capturing how the chlorine amount decreases over time.

Solving these problems requires both analytical and mathematical insight, especially when dealing with dynamic systems.

Change of Concentration

Change of concentration is a critical aspect in various real-world applications, especially in chemistry and process engineering. This concept describes how the proportion of a particular substance in a mixture varies over time due to shifts in the mixture's constituents. In our example, the concentration of chlorine is affected by both the inflow of fresh water and the outflow of the mixed solution.

Here are the key points to consider:

• The initial concentration of chlorine is 0.05 g/L in a 400 L solution.
• The rate of outflow is greater than the inflow, leading to a dilution of chlorine in the solution.
• The changing volume directly impacts the concentration, expressed as \(\frac{C(t)}{V(t)}\), where \(C(t)\) is the chlorine amount and \(V(t)\) is the current volume.

Understanding change of concentration involves recognizing how external actions, such as adding a solvent or extracting a solution, affect the concentration of a solute over time. By analyzing these changes mathematically, we can use differential equations to find precise substance quantities at any given time.

Calculus Applications

Calculus, especially differential equations, plays a vital role in solving mixture problems like the one tackled here. These mathematical tools are essential because they allow us to quantify how different variables change over time and predict future states of dynamic systems.

In our situation, the differential equation we derived is:\[ \frac{dC}{dt} = -10 \frac{C(t)}{400 - 6t} \]This equation lets us model the rate of change of chlorine concentration in the tank.

Key calculus steps used in solving these kinds of problems include:

• Setting up the differential equation based on the rates of inflow and outflow, and initial concentrations.
• Separation of variables, which transforms the differential equation into a form that is easier to integrate.
• Integration to find a general solution, and adjusting with initial conditions to solve for any constants.
• Substitution of these constants back into the general solution to provide a final, specific solution for the problem at hand.

With these methods, calculus enables us to describe, analyze, and predict the behaviors of changing systems with great precision. This makes it incredibly powerful for real-world applications where systems are in a constant state of flux.