Problem 4 - Calculus for Biology and Medicine

Question

Suppose that a tank holds 1000 liters of water, and \(2 \mathrm{~kg}\) of salt is poured into the tank. (a) Compute the concentration of salt in \(\mathrm{g}\) liter \(^{-1}\). (b) Assume now that you want to reduce the salt concentration. One method would be to remove a certain amount of the salt water from the tank and then replace it by pure water. How much salt water do you have to replace by pure water to obtain a salt concentration of \(1 \mathrm{~g}\) liter \(^{-1} ?\) (c) Another method for reducing the salt concentration would be to hook up an overflow pipe and pump pure water into the tank. That way, the salt concentration would be gradually reduced. Assume that you have the choice of two pumps, one that pumps water at a rate of 1 liter \(\mathrm{s}^{-1}\), the other at a rate of 2 liter \(\mathrm{s}^{-1}\). For each pump, find out how long it would take to reduce the salt concentration from the original concentration to 1 gliter \(^{-1}\). (Note that the rate at which water enters the tank is equal to the rate at which water leaves the tank.) (d) Show that, whichever pump you use in part (c), you need more pure water if you use the pump method than if you follow the method in (b). Can you explain why?

Step-by-Step Solution

Answer

(a) 2 g/liter; (b) Replace 500 liters; (c) 1st pump: 693 s, 2nd pump: 346 s; (d) More water needed for the overflow method.

Step 1: Calculate Initial Salt Concentration

To find the initial concentration of salt in the tank, we have the mass of salt as 2 kg, which is equivalent to 2000 g. The volume of water in the tank is 1000 liters. The concentration is given by \( \text{concentration} = \frac{\text{mass of salt}}{\text{volume of water}} \). Hence, the concentration is \( \frac{2000}{1000} = 2 \text{ g/liter} \).

Step 2: Determine Volume to Replace for Desired Concentration

To reduce the salt concentration to 1 g/liter, we want the final concentration to become \( 1 \text{ g/liter} \). Using the initial concentration of 2 g/liter, consider removing \( x \) liters of salty water, meaning you are also removing \( 2x \) grams of salt. Replacing with pure water results in \( 2000 - 2x \) grams of salt remaining in 1000 liters. Set \( \frac{2000 - 2x}{1000} = 1 \). Solving this equation, \( 2000 - 2x = 1000 \), results in \( x = 500 \) liters. So, you must replace 500 liters of salt water with pure water.

Step 3: Evaluate Time for Overflow Method Using First Pump

Using the first pump that pumps at 1 liter/s, both inflow and outflow rates are the same, keeping the tank volume constant. Model this using the differential equation \( \frac{dS}{dt} = -\frac{S}{1000} \), where \( S(t) \) is the amount of salt at time \( t \). Solve it to get \( S(t) = S(0)e^{-t/1000} \). With initial \( S(0) = 2000 \) g and to reach \( 1 \text{ g/liter} \), we solve \( 1000e^{-t/1000} = 1000 \). The equation simplifies to \( e^{-t/1000} = \frac{1}{2} \), giving \( t = 1000 \ln(2) \approx 693 \) seconds.

Step 4: Evaluate Time for Overflow Method Using Second Pump

With the second pump operating at 2 liter/s, similar to before, form \( \frac{dS}{dt} = -\frac{2S}{1000} \). Solving gives \( S(t) = S(0)e^{-2t/1000} \). With the same initial condition and desired concentration, set \( 1000e^{-2t/1000} = 1000 \). Simplify to \( e^{-2t/1000} = \frac{1}{2} \), leading to \( 2t = 1000 \ln(2) \), and thus \( t = 500 \ln(2) \approx 346 \) seconds.

Step 5: Compare Methods

In method (b), 500 liters of pure water is added. For method (c), considering the first pump, 693 liters of pure water are added, and for the second pump, 692 liters, because these amounts represent the flow rate times the time calculated in steps 3 and 4, respectively. The overflow method requires more pure water because it doesn't selectively remove the salt, unlike the replacement method that directly adjusts the concentration by reduction.

Key Concepts

Concentration Calculations

Concentration calculations play a critical role in determining how much solute is present within a given volume of solvent. In our example, we have a tank with 1000 liters of water into which 2 kg of salt has been added. To find the concentration of salt in grams per liter, we start by converting kilograms to grams. Since 1 kg equals 1000 grams, 2 kg of salt is equivalent to 2000 grams.

• Step 1: Determine the mass of salt, which is 2000 grams.
• Step 2: Identify the volume of water in the tank, which is 1000 liters.
• Step 3: Calculate the concentration using the formula:

\[ \text{Concentration} = \frac{\text{Mass of Salt (grams)}}{\text{Volume of Water (liters)}} \]Substituting the known values:\[ \text{Concentration} = \frac{2000}{1000} = 2 \text{ g/L} \]This means that initially, the tank water has a salt concentration of 2 grams per liter.

Pump Rate

The pump rate refers to how quickly water is being added or removed from the tank. When considering pumps with different rates, we must analyze how these rates affect the reduction in salt concentration over time. Consider two pumps with different capacities:

• Pump 1 moves water at 1 liter per second.
• Pump 2 is more efficient at 2 liters per second.

Maintaining a steady pump rate is crucial when using an overflow method for diluting the salt concentration. In this method, as pure water is pumped into the tank, an equal amount of the salt solution overflows out. For both pumps, the crucial understanding is that the net volume of water remains constant due to equal inflow and outflow. The variable here is how quickly the concentration changes, which is directly tied to the pump rate. Faster pump rates lead to faster dilution:

• Pump 1: Takes approximately 693 seconds to reduce concentration to 1 g/L.
• Pump 2: Accomplishes this task in about 346 seconds.

Salt Concentration Reduction

Reducing salt concentration can be approached in different ways, each influencing efficiency in terms of time and resources. Let’s explore two main methods. Replacement Method: Involves removing a portion of the saltwater and replacing it with pure water. For example, by calculating the amount to be replaced to achieve a desired lower concentration, we find:

• Initial concentration is 2 g/L and desired concentration is 1 g/L.
• Calculations show replacing 500 liters of salt water with pure water achieves this.

Overflow Method: Here, pure water is continuously added while letting an equal volume of salt solution overflow. The concentration decreases as:

• Use of Pump 1 results in adding about 693 liters to achieve the target concentration.
• Using Pump 2 requires approximately 692 liters.

Comparatively, overflow requires more water than the replacement method because it gradually dilutes the salt without directly removing it. It highlights the selective efficiency of the replacement method, which targets decreasing the salt ratio directly, whereas overflow involves a complete volume-based swapping of water content.