Essay about Surface Area and Cube

Submitted By jessieirons
Words: 897
Pages: 4

Materials and methods
The following materials were used to complete the experiment; two large potatoes, a metric ruler, two 1,000 mL beaker, a sharp knife, a spoon, paper towels and 5% potassium permanganate solution. The two potatoes were used to cut eight cubes; two 10 mm3, two 20 mm3, two 30 mm3, and two 40 mm3 in size. Each potato cube was cut so that there was no skin remaining on either side of the cube. One cube of each size was placed in the beaker so they were not touching each other or the side of the beaker. The timer was set for 30 minutes and started once the beaker was filled with enough potassium permanganate to completely cover the potato cubes. After each 5 minute interval the potato cubes were turned onto a different side using the spoons. Once flipped, it was made certain that no cubes were touching each other or the sides of the beaker. After 30 minutes using a spoon as a barrier to retain the potatoes, the beaker was drained into the other beaker. The cubes were removed using the spoon and placed onto a paper towel. Each cube was blotted dry and then cut in half. Once cut in half the penetrated portion of the potato was measure and recorded. The same process was done for each of the proceeding cubes. Upon completion, the used potassium permanganate was disposed of in a disposal container then the process was repeated with the second set of cubes. Then, the 24 samples, combined from both groups, of all four sizes of penetrated cubes were averaged. The surface area and volume were calculated and were then used to calculate the surface area to volume ratio.
Results
The data gathered from the experiment showed that percentage of absorbed KMnO4 was directly related to the size of the cube (Table 1). In Table 1 it shows that the percentage of KMnO4 absorbed by the 10 mm cube was 51.9%, 34.0% for the 20mm, 26.1% for the 30mm, and 21.4% for the largest cube at 40mm. The results gathered showed that the smaller the cube the more efficient and the larger the cube became, the less efficient.
Table 1

The Effect of Size on the Efficiency of A Cell

Edge Length (mm)
Surface Area Cubed (mm2)
Cube Volume (mm3)
Surface Area: Volume
Average Distance Penetrated (mm) by KMnO4
Edge Length not Penetrated (mm) by KMnO4
Volume not Penetrated (mm3) by KMnO4
Volume Penetrated (mm3) by KMnO4
% Volume Penetrated
10
600
1000
0.6
1.1
7.8
480.7
519.3
51.9
20
2,400
8000
0.3
1.3
17.4
5283.2
2716.8
34.0
30
5,400
27000
0.2
1.4
27.1
19957.6
7042.4
26.1
40
9,600
64000
0.15
1.5
36.9
50311.5
16388.5
21.4

Smaller cells are more efficient than larger cells as it shown in Table 1 and in Figure 1. Figure 1 shows the rapid decrease in efficiency as the cells get larger and Table 1 shows the difference between each cube.
Figure 1

The Relationship between cube size and percentage of KMnO4 absorbed

Conclusion The amount of KMnO4 absorbed my each cube is directly related to the size of the. The data gathered supports the hypothesis that cells are small to be efficient. The cube being smaller means less space to cover and travel. Each cell absorbed about the same amount of KMnO4. The smaller one with less surface area was more efficient with KMnO4 penetrating further than the other sizes. Regardless of size, each cube is penetrated at the same rate making the smallest one more efficient because of there being less space for the KMnO4 to travel. The results clearly