Calculating Surface Area to Volume Ratio for Emulsified Droplets

What is the surface area to volume ratio of the droplet?Answer: 6.0Surface Area / Volume = 4 x (3.14) x .108^2 / (4/3) x 3.14 x .108^3 = 6.04 or 6.0B. Now, suppose that this sphere were emulsified into 100 essentially equal-sized droplets. What is the average surface area to volume ratio of each droplet?Answer: 60.0Surface Area / Volume = 4 x (3.14) x .108^2 / (4/3) x 3.14 x (.108/100)^3 = 60.
  • #1
Searchme
14
0

Homework Statement



Suppose a lipid droplet is essentially a sphere with a diameter of 1 cm.

For A – D below, the area of a sphere is 4π r2, and volume is 4/3πr3.

For π, use a value of 3.14.

Show your calculations neatly in the space provided to receive credit for your answer.

Round all answers to the nearest tenth.B. Now, suppose that this sphere were emulsified into 100 essentially equal-sized droplets. What is the average surface area to volume ratio of each droplet?C. How much greater is the total surface area of these 100 droplets compared to the original single droplet?

D. How much did the total volume change as a result of emulsification?

Homework Equations



Surface Area / Volume

The Attempt at a Solution



What is the surface area to volume ratio of the droplet?

Answer: 6.0

Surface Area / Volume = 4 x (3.14) x .5^2 / (4/3) x 3.14 x .5^3 = 6.04 or 6.0I figured out A but am stuck on part B through D. Can you help? Thank you! Thank you!

I don't know how the equation evolves when the sphere is emulsified into 100 equal droplets. I know that the surface area
must increase but I cannot get it to work :(
 
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  • #2
When the original droplet splits into 100 equal size droplets, each one of these smaller droplets has an volume that is 1/100 of the original droplet. Use this and the equation for the volume of a sphere to calculate the radius of the smaller droplet, and then knowing their radius you can calculate their surface area.
 
  • #3
You have already posted this question in the Calculus forum. Don't multi-post.
 
  • #4
... and in the Topology & Geometry forum...
 
  • #5
LCKurtz said:
You have already posted this question in the Calculus forum. Don't multi-post.

Sorry I am stupid when it comes to math so i posted at three places since i really didnt know where to put :-(
 
  • #6
gerben said:
When the original droplet splits into 100 equal size droplets, each one of these smaller droplets has an volume that is 1/100 of the original droplet. Use this and the equation for the volume of a sphere to calculate the radius of the smaller droplet, and then knowing their radius you can calculate their surface area.

I don't understand. I thought that radius is already understood given that we were given the diameter of 1 cm. I was under the impression that radius is .5 cm / 100 droplets which equals 0.005 cm. In other words, can I plug in "0.005" whenever "r" in the original surface area / volume equation? Maybe, I'm I doing something wrong?
 
  • #7
No, you cannot just divide the radius by 100. If you break a sphere into two and make two smaller sphere's from the halves their radius will not be halve of the radius the original sphere. However their volume will be halve the volume.
 
  • #8
Does breaking up the droplet into 100 smaller droplets change the mass or the density one iota? What does this mean with respect to question D? What does that answer mean with respect to questions B and C?

BTW, what was question A?
 
  • #9
Searchme said:
Sorry I am stupid when it comes to math so i posted at three places since i really didnt know where to put :-(

(Multiple threads merged. We trust this won't happen again.)
 
  • #10
D H said:
Does breaking up the droplet into 100 smaller droplets change the mass or the density one iota? What does this mean with respect to question D? What does that answer mean with respect to questions B and C?

BTW, what was question A?

Question A was What is the surface area to volume ratio of the droplet?

Answer: 6.0

Surface Area / Volume = 4 x (3.14) x .5^2 / (4/3) x 3.14 x .5^3 = 6.04 or 6.0
 
  • #11
berkeman said:
(Multiple threads merged. We trust this won't happen again.)

I'm sorry for the trouble again.
 
Last edited:
  • #12
gerben said:
No, you cannot just divide the radius by 100. If you break a sphere into two and make two smaller sphere's from the halves their radius will not be halve of the radius the original sphere. However their volume will be halve the volume.

Thanks for the clarification. I will find the volume right now and will post the latter part of my answer in a few hours, after my class. Volume = 4/3πr3

Volume = (4/3) x 3.14 x 0.005^3 = 5.23333333334e-7Or in the alternative:

It could be
Volume = 4/3πr3 Volume = (4/3) x 3.14 x 0.5^3 = .5233333333; then divided by 100 = .00523333333333I wonder which among the two is correct?
 
Last edited:
  • #13
Second answer is correct. The questions is about relation between lengths, surface area and volume. Think about how these relate. Try a simpler shape than the sphere to get some feel for it first perhaps.

Take a cube for example:
a cube with sides of length 2 has volume 2x2x2=8

Cut it in half so that you get two cubes of each 2x2x1, so each has volume 4.

This not the same as just halving the lengths of the sides and then calculating
1x1x1=1.
 
  • #14
gerben said:
Second answer is correct. The questions is about relation between lengths, surface area and volume. Think about how these relate. Try a simpler shape than the sphere to get some feel for it first perhaps.

Take a cube for example:
a cube with sides of length 2 has volume 2x2x2=8

Cut it in half so that you get two cubes of each 2x2x1, so each has volume 4.

This not the same as just halving the lengths of the sides and then calculating
1x1x1=1.

Gerben... I worked out the equation based on your thoughts and some research. I hope this looks right or kinda right. Can you take a look B, C, and D to see if its interpreted correctly? Thank you so much. You are a life saver!


.00523333333333 = (4/3) x 3.14 x r^3
Multiply both sides by 3 to get rid of the 3 from (4/3)
.00523333333333 x (3) = 4 x 3.14 x r^3
.0157 = 4 x 3.14 x r^3
Divide (4 x 3.14) against .0157 to get rid of it in the right side
.00125 = r^3
Cube it!
³√x.00125 = radius = .108
Substitute .108 in the Surface Area Formula SA= 4 x (3.14) x (r ^2)
4 x (3.14) x (.108^2)
12.56 x (0.011664) =0 .1465 = Surface Area
Plug in the numbers in the equation Surface Area / Volume
0.1465 / .00523333333333 = 27.99
B. Now, suppose that this sphere were emulsified into 100 essentially equal-sized droplets. What is the average surface area to volume ratio of each droplet?
27.99 = (28.0)
C. How much greater is the total surface area of these 100 droplets compared to the original single droplet? 466.7 Greater
28.0 x 100 = 2800
2800/ 6 = 466.7
D. How much did the total volume change as a result of emulsification? Total volume of 100 droplet is the same as the BIG droplet ( before being emulsified)
 
  • #15
Also in case you are wondering- I got the 6.0 from my answer to question "A"

What is the surface area to volume ratio of the droplet?

Answer: 6.0

Surface Area / Volume = 4 x (3.14) x .5^2 / (4/3) x 3.14 x .5^3 = 6.04 or 6.0
 
  • #16
B and D seem okay. When you say "Cube it" you mean the inverse namely "take cube root". Cube it means take it to the 3rd power.

You could also first write it out in symbols so you do not have to round off in between like this:

Volume of large droplet:
[tex] V = \frac{4}{3}\pi(\frac{1}{2})^3 = \frac{4}{3}\pi\frac{1}{8} [/tex]

Volume of small droplet:
[tex] V_s = (\frac{4}{3}\pi\frac{1}{8})/100 = \frac{4}{3}\pi\frac{1}{800} [/tex]

Get radius of of small droplet from equation:
[tex] V_s = \frac{4}{3}\pi r_s^3 [/tex]
[tex] \frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi\frac{1}{800} [/tex]
[tex] r_s^3 = \frac{1}{800} [/tex]
[tex] r_s = \sqrt[3]{\frac{1}{800}}[/tex]

Surface area of small droplet:
[tex] A_s = 4\pi r_s^2 = 4\pi \left(\sqrt[3]{\frac{1}{800}} \right)^2 \approx 0,1458[/tex]

Surface area to volume ratio of small droplet:
[tex] A_s/V_s =\left. 4\pi \left(\sqrt[3]{\frac{1}{800}} \right)^2 \right/
\frac{4}{3}\pi\frac{1}{800} =
3\cdot 800 \cdot \left(\sqrt[3]{\frac{1}{800}} \right)^2 \approx 27.8495[/tex]

For C, you need to calculate: "Surface area all small droplets relative to surface area large droplet"
but you used "surface area to volume ratio" you should just divide the surface area of all small droplets by the surface area of the large droplet.
 
  • #17
For C, the professor's question was "How much greater is the total surface area of these 100 droplets compared to the original single droplet?"
Should I subtract instead of dividing? It seems like his question can be interpreted in multiple ways!

Total Surface Area of 100 Small Droplets = 0.1465 x 100 droplets = 14.65

14.65 (The Total SA of 100 Droplets) - 3.14 (SA of the Original Surface Area) = 11.48

Surface area is greater by 11.48.

Or if I go by your recommendation:

14.65 (The Total SA of 100 Droplets) / 3.14 (SA of the Original Surface Area) = 4.67 or 4.7. Total Surface Area of the 100 droplets is 4.7 times larger.

What units can use to express surface area and volume given that the original droplet had a 1 centimeter diameter. Millimeters maybe?

You are heaven sent gerben! I'm very grateful for taking the time to answer my questions!
 
  • #18
Searchme said:
For C, the professor's question was "How much greater is the total surface area of these 100 droplets compared to the original single droplet?"

Should I subtract instead of dividing? It seems like his question can be interpreted in multiple ways!

"How much greater" seems to indicate "how many times as large", so I guess he wants you to divide the two.

Otherwise he should have asked "how much more surface area do the small droplets have than the large droplet".

Searchme said:
What units can use to express surface area and volume given that the original droplet had a 1 centimeter diameter. Millimeters maybe?

Since you used centimeters for the radius (0.5 cm) the surface area is in cm2 and the volume in cm3 .
 
  • #19
Thank you Thank you Gerben!:biggrin:
 

Related to Calculating Surface Area to Volume Ratio for Emulsified Droplets

What is the surface area to volume ratio?

The surface area to volume ratio is a mathematical concept that compares the amount of surface area of an object to its volume. It is expressed as a ratio and is used to understand the relationship between the size and shape of an object and its ability to exchange materials with its environment.

Why is the surface area to volume ratio important?

The surface area to volume ratio is important because it affects the efficiency of various processes within an object. For example, in cells, a larger surface area to volume ratio allows for more efficient exchange of nutrients and waste products. In chemical reactions, a higher surface area to volume ratio can increase the rate of reaction by providing more contact points for reactants.

How is the surface area to volume ratio calculated?

The surface area to volume ratio is calculated by dividing the surface area of an object by its volume. The surface area can be calculated by adding the area of each face of the object, while the volume is calculated by multiplying the length, width, and height of the object. The resulting ratio is a dimensionless number.

How does the surface area to volume ratio change with the size of an object?

The surface area to volume ratio changes as the size of an object changes. As the size of an object increases, its volume increases at a faster rate than its surface area. This results in a decrease in the surface area to volume ratio. For example, a cube with sides of 1cm will have a higher surface area to volume ratio compared to a cube with sides of 10cm.

What is the significance of a high or low surface area to volume ratio?

A high surface area to volume ratio indicates that an object has a large surface area in proportion to its volume. This can be beneficial in processes that require efficient exchange of materials, such as in cells. A low surface area to volume ratio can be beneficial in situations where insulation is needed, as a smaller surface area reduces heat loss. However, it can also limit the efficiency of processes that rely on surface area, such as chemical reactions.

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