Problem 14 - 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 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 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 \(1 \mathrm{~g}\) liter \(^{-1}\) and how much pure water is needed in each case. (Note that the rate at which water enters the tank is equal to the rate at which water leaves the tank.) Compare the amount of water needed using the pumps with the amount of water needed in part (b).

Step-by-Step Solution

Answer

(a) 2 g L\(^{-1}\). (b) Replace 1000 L. (c) 1000 s for 1 L/s pump, 500 s for 2 L/s pump, using 1000 L in each case.

Step 1: Compute initial concentration

First, calculate the initial concentration of salt in grams per liter. The tank holds 1000 liters of water, and it has 2 kg of salt, which is equivalent to 2000 grams. Therefore, the concentration is given by \( \frac{2000 \text{ g}}{1000 \text{ L}} = 2 \text{ g L}^{-1} \).

Step 2: Determine volume to replace for target concentration

To achieve a target concentration of 1 g L\(^{-1}\), we need to decrease the salt amount to 1000 g (since 1 g L\(^{-1}\) \(\times\) 1000 L = 1000 g). Therefore, we have to replace some water to remove excess salt. Each liter replaced reduces the salt by \(2 \text{ g/L} - 1 \text{ g/L} = 1 \text{ g/L}\). To remove 1000 g (from 2000 g to 1000 g), we need to replace 1000 liters of salt water with pure water.

Step 3: Calculate time using first pump

The first pump operates at 1 L/s. Since we need to replace 1000 liters of water, it will take \( \frac{1000 \text{ L}}{1 \text{ L/s}} = 1000 \text{ s} \). The total amount of pure water needed is 1000 liters.

Step 4: Calculate time using second pump

The second pump operates at 2 L/s. Replacing 1000 liters would thus take \( \frac{1000 \text{ L}}{2 \text{ L/s}} = 500 \text{ s} \). Like before, the total amount of pure water needed is 1000 liters, as the input equals the output.

Step 5: Compare the methods

In both the method of replacing a specific volume of water (step 2) and using either pump (step 3 and 4), the same amount of pure water (1000 liters) is used to achieve a concentration of 1 g L\(^{-1}\). However, the pumps provide flexibility with time efficiency: the second pump is faster than the first. Exact replacement, as in part b, instantly achieves the desired concentration without waiting.

Key Concepts

Concentration Calculation

When tackling problems involving salt concentration, the first step is always to determine the initial concentration of the solution. In this exercise, we start with a tank that contains 1000 liters of water and 2 kg of salt. To find the concentration in grams per liter, we need to convert the salt's mass into grams: 2 kg equals 2000 grams. The formula for concentration is:\[\text{Concentration} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (L)}}\]
Substituting the values, we have:\[\frac{2000 \text{ g}}{1000 \text{ L}} = 2 \text{ g/L}\]
This simple equation provides the concentration in terms of grams of salt per liter of water, which is crucial for understanding how much salt is present initially in the tank.

Replacement Method

To reduce the salt concentration using the replacement method, we have a target to achieve—a concentration of 1 g/L. Initially, the concentration is 2 g/L, which means the total salt in the tank is 2000 grams. Our goal is to decrease this amount of salt to 1000 grams to meet the 1 g/L target. Here's how you can perform this replacement method:

• Identify the required final amount of salt; here, it's 1000 grams (since 1 g/L \(\times\) 1000 L = 1000 g).
• Calculate the excess salt to remove, which is 1000 grams (2000 grams initial - 1000 grams desired).
• Recognize that every liter of replaced water decreases the salt concentration by 1 g (2 g/L initial - 1 g/L desired). As a consequence, to reduce 1000 grams, you must replace 1000 liters of salt water with pure water.

This method achieves the desired concentration level quickly, effectively using the exact calculation from the concentration change.

Pumping Method

Another effective way to alter the salt concentration is through the pumping method. This involves using overflow pipes to continuously introduce pure water while maintaining a steady outflow. Two pumps are considered, one pumping at 1 liter/second and another at 2 liters/second. The effectiveness of these pumps can be illustrated by the time each takes to achieve a 1 g/L concentration.

Pumping Method Calculation:

• First Pump: It operates at 1 L/s. To replace 1000 liters of water, it would need 1000 seconds, as calculated by \(\frac{1000 \text{ L}}{1 \text{ L/s}} = 1000 \text{ s}\).
• Second Pump: It operates at 2 L/s. It requires 500 seconds to replace 1000 liters, calculated by \(\frac{1000 \text{ L}}{2 \text{ L/s}} = 500 \text{ s}\).

While both methods require 1000 liters of water, the second pump cuts the operation time in half, illustrating the advantage in time savings over less efficient methods. This makes the pumping method very practical, especially when time efficiency is a factor.