Mass Transfer Through Kool-Aid
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Diffusion still applies to the Kool-Aid example. It would not require stirring for the powder Kool-Aid to travel through the water to a less concentrated area. The diffusion would create movement of the Kool-Aid powder and sugar to the top of the container until all of the water was red and sugary. This process would take a long time and would be faster if stirring were used, therefore we stir Kool-Aid when we make it. Now that we have learned about mass transfer and the equation that describes using Fick’s Law, we are ready to apply all of our knowledge to finding how mass transfer affects us when we make Kool-Aid. We asked ourselves what would happen if we never stirred the Kool-Aid when we made it. We decided that it would slowly stir itself because diffusion would move the sugar particles throughout the water. Diffusion, remember, is the movement of things from high concentration to low concentration. Let’s do an example to determine how long it would take for the Kool-Aid to mix completely if it is never stirred. |
Now that we have learned about mass transfer and the equation that describes Fick’s Law and we have seen an example using Fick’s Law, we are ready to apply all of our knowledge to finding how mass transfer affects us when we make Kool-Aid. We asked ourselves what would happen if we never stirred the Kool-Aid when we made it. We decided that it would slowly stir itself because diffusion would move the sugar particles throughout the water. Diffusion, remember, is the movement of things from high concentration to low concentration. Let’s do an example to determine how long it would take for the Kool-Aid to mix completely if it is never stirred.
Question:
Suppose you make Kool-Aid in a container that is about 30 cm tall. It is a cylindrical pitcher with a radius of 5 cm. First you put the cup of sugar and Kool-Aid mix in the bottom of the container. Then you pour the 8 cups of water into the container. Without stirring, how long will it take for the sugar to be evenly distributed throughout the water? The density of sugar is 1.5620 g/cm3. The diffusion coefficient for sugar in water is 0.52x10-5 cm2/s.
Answer:
To answer this question, we need to make a couple of assumptions. First we need to assume that the Kool-Aid powder has the same exact properties as the sugar. Since there is much more sugar than Kool-Aid powder, the sugar is more important, so this is not a bad assumption. The next assumption we will make is that the sugar we will be using is glucose, which has a molecular weight of 180 g/mol. Glucose is a common type of sugar, so this is not a bad assumption either. Our final assumption will be that the sugar is initially saturated in the water. What this means is that there is as much sugar in the water as the water can hold at that temperature. This amount can be looked up in many reference books. It was found to be 5.6x10-3 mol/cm3.
So let’s think about what is going on. Initially, there is no sugar at the top of the container, but we have a pile of sugar at the bottom of the container that is moving upward to the area of lower concentration. This is going to continually happen until the amount of sugar at the top of the container is equal to that at the bottom. What we want to find is the amount of time that it will take for this to happen. If we can find the flux, which is in units of mol/area*time, then we can find time by knowing the moles and area of the container. So, let’s first solve for the flux, J.
J = -D * D C/D x
where: D = diffusion coefficient = 0.52 x 10-5 cm2/s, D C = concentration gradient = Csat - 0 = 5.3 x 10-3 mol/cm3, and D x = height of the container = 30 cm
therefore,
J = -D * D C/D x
J = - (0.52 x 10-5 cm2/s) * (5.3 x 10-3 mol/cm3) / 30cm
J = -9.2 x 10-10 mol/(cm2/s)
Now that we have the flux of sugar through the water, we can analyze the units to figure out how to get time:
flux = mol / (cm2*s)
flux * area = mol / time
time = mol / (flux * area)
So, to find time, we need to find the area and the moles of sugar.
Area:
The area of the container can be found by the common equation A = P r2. Since the radius of the container is 5 cm, the area can be calculated:
A = P r2
A = P *(5cm)2
A = 78.5 cm2
Moles:
To find the moles of sugar at the top of the container, we need to assume that the sugar is distributed evenly. Therefore, the amount of sugar at the top of the container will be the same as the amount of sugar at the bottom of the container. This uniform solution is made up of 1 cup of sugar in 8 cups of water. Using the density of the sugar (1.5620 g/cm3), we can convert 1 cup of sugar to grams of sugar. (236.6 cm3 = 1 cup)
1 cup sugar *(236.6 cm3 / 1 cup) * (1.5620 g/
cm3) = 359.6 g sugar
The grams of sugar can then be converted to moles of sugar by dividing by the molecular weight.
359.6 g sugar * (1 mol sugar / 180 g sugar) = 2 mol sugar
We can now return to the equation we developed that said:
time = mol / (flux * area)
time = (2 mol) / ( 9.2 x 10-10 mol/[s*cm2] * 78.5 cm2)
time = 2.77 x 107 sec
We can convert this to a more understandable time frame by:
2.77 x 107 sec * (1 hour / 3600 sec)
* (1 day / 24 hours) = 321 days
As you can see, it would take nearly a year for the Kool-Aid to mix by itself. Since this is not a realistic time to wait to have a drink of Kool-Aid, we stir it. When the Kool-Aid is stirred, it takes about 10 to 30 seconds to mix the sugar. The speed of mixing is due to the forced convection that we have from the spoon. So, I think we have learned to never wait for the Kool-Aid to "stir itself" because you will die of thirst before you get to drink it.
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