Lab 2: Aluminum Foil Lab
Purpose
The purpose of the experiment was to give students a firmer grasp of the concept of density and its relationship to mass and volume. To accomplish this purpose, students were told to calculate the thickness of a piece of aluminum foil given that the density of aluminum is 2.70 g/(cm^3).
Procedure
PART II of the experiment was a (roughly) four-step process.
Step 1: We measured the length and width of a piece of aluminum foil.
Step 2: Next we massed the piece of aluminum foil using an electronic balance.
Step 3: Next, using the formula for density, we solved for volume. We rearranged the the formula to solve for volume, then substituted the values we'd found for mass as well as the given density of aluminum to solve for the volume of the sheet.
Step 4: Finally, we substituted the values we'd found in Step 1 into the formula for volume and set it equal to the value for volume found in Step 3. Afterwards it was a matter of solving for the height.
Data
The measurements we took were a length of 12.27 cm and a width of 12.07 cm. The piece of aluminum was also found to have mass of 0.6 g.
Density=Mass/Volume
Therefore,
Density*Volume=Mass
Volume= Mass/Density
Thus,
Volume=Mass/Density
= 0.6 g/ 2.70 g/(cm^3)
= 0.2 cm^3
Knowing the volume of the sheet of aluminum, we can then substitute the length and width into the formula for volume in order to find the height (or thickness) of the aluminum sheet.
Rearranging the equation to solve for height,
Volume= Length*Width*Height
Volume/(Length*Width)=Height
Height=Volume/(Length*Width)
Thus, substituting the known values into the equation,
Height= 0.2 cm^3/ (12.27 cm*12.07 cm)
=0.2 cm^3/(148. 0989 cm^2)
=0.001 cm (one sig fig)
Converting to milligrams,
I cm= 10 mm
0.001 cm *(10 mg/ 1 cm)= 0.01 mm
Therefore, the height of the sheet of aluminum foil is 0.01 millimeters.
Conclusion
We encountered some difficulty in the calculations. One reason for such difficulty, we believe, is due to the uncertainty of our measurements. The piece of aluminum did not have straight edges, and we believe that this may have affected the accuracy of our measurements.
Another reason for difficulty could have also been due to the balance. The electronic balance only displayed values up to the tenth digit, meaning that our sig fig count after finding the volume was only one. However, if we had stuck to using one sig fig throughout the entire experiment, we would have ended up with an answer of 0.0. As a result, we decided to include sig fig digits up to the thousandths place in order to have a nonzero number in our answer.
A surprising aspect of this lab, to me, was the problem, the prompt; I did not expect to be asked to calculate the thickness of a sheet of aluminum, nor did I expect a process as complex as this, having to solve for volume then proceeding to solve for a certain unknown dimension; I was pleasantly surprised by this.
The purpose of the experiment was to give students a firmer grasp of the concept of density and its relationship to mass and volume. To accomplish this purpose, students were told to calculate the thickness of a piece of aluminum foil given that the density of aluminum is 2.70 g/(cm^3).
Procedure
PART II of the experiment was a (roughly) four-step process.
Step 1: We measured the length and width of a piece of aluminum foil.
Step 2: Next we massed the piece of aluminum foil using an electronic balance.
Step 3: Next, using the formula for density, we solved for volume. We rearranged the the formula to solve for volume, then substituted the values we'd found for mass as well as the given density of aluminum to solve for the volume of the sheet.
Step 4: Finally, we substituted the values we'd found in Step 1 into the formula for volume and set it equal to the value for volume found in Step 3. Afterwards it was a matter of solving for the height.
Data
The measurements we took were a length of 12.27 cm and a width of 12.07 cm. The piece of aluminum was also found to have mass of 0.6 g.
Density=Mass/Volume
Therefore,
Density*Volume=Mass
Volume= Mass/Density
Thus,
Volume=Mass/Density
= 0.6 g/ 2.70 g/(cm^3)
= 0.2 cm^3
Knowing the volume of the sheet of aluminum, we can then substitute the length and width into the formula for volume in order to find the height (or thickness) of the aluminum sheet.
Rearranging the equation to solve for height,Volume= Length*Width*Height
Volume/(Length*Width)=Height
Height=Volume/(Length*Width)
Thus, substituting the known values into the equation,
Height= 0.2 cm^3/ (12.27 cm*12.07 cm)
=0.2 cm^3/(148. 0989 cm^2)
=0.001 cm (one sig fig)
Converting to milligrams,
I cm= 10 mm
0.001 cm *(10 mg/ 1 cm)= 0.01 mm
Therefore, the height of the sheet of aluminum foil is 0.01 millimeters.
Conclusion
We encountered some difficulty in the calculations. One reason for such difficulty, we believe, is due to the uncertainty of our measurements. The piece of aluminum did not have straight edges, and we believe that this may have affected the accuracy of our measurements.
Another reason for difficulty could have also been due to the balance. The electronic balance only displayed values up to the tenth digit, meaning that our sig fig count after finding the volume was only one. However, if we had stuck to using one sig fig throughout the entire experiment, we would have ended up with an answer of 0.0. As a result, we decided to include sig fig digits up to the thousandths place in order to have a nonzero number in our answer.
A surprising aspect of this lab, to me, was the problem, the prompt; I did not expect to be asked to calculate the thickness of a sheet of aluminum, nor did I expect a process as complex as this, having to solve for volume then proceeding to solve for a certain unknown dimension; I was pleasantly surprised by this.
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