The volume of the container can be calculated by
V = h*d*l
with h for height, d for depth and l for length.
The volume of the cylinder can be determined using the formula
V = h*π*((d2)/4)
with h for height and d for diameter.
After the lengths have been measured and the volumes calculated, the volumes will be measured by filling the containers with water and pouring the water into a measurement cylinder. We will then compare calculated and measured volume.
The volume of one drop will be measured by using an eye dropper and filling a graduated cylinder to the 1.00 mL mark and counting the drops; in a second round, the cylinder will be filled the same way from 1.00 mL to 2.00 mL. The number of drops will be averaged.
As an additional experiment, the volume of water in a plastic spoon will be determined by pouring water from such a spoon into the graduated cylinder; the average of two measurements will be determined.
Here, the average mass of five pennies will be determined to the milligram unit, using an electronic balance. In addition, the total mass of these five pennies on the balance will be determined by placing them all at once onto the balance.
3. Results and Discussion
A. Length and volume
The straight edged box has the following dimensions:
l = 12.5 cm; h = 14.0 cm; d = 19.5 cm.
Therefore, the volume of the box is
V(box) = 14.0 cm * 19.5 cm * 12.5 cm = 3,412.5 cm3
The drinking mug as cylinder yields the following measurements:
h = 9.5 cm; d = 7.7 cm.
Thus, the volume of the cylinder is
V(cylinder) = 9.5 cm * π * (7.7 cm 2)/4 = 442.4 cm3
Filling the containers with water and pouring it into a graduated cylinder yields the following volumina:
Vmeasured(box) = 3.391 L
Vmeasured(cylinder) = 0.420 L.
The measurements are smaller than the calculations, as not all water is completely transferred out of one into the other container, and it is difficult to fill the original box and mug completely, so they automatically have a smaller volume than calculated.
Using the eye dropper, it took 43 drops to fill the graduated cylinder to the 1.00 mL mark and 39 drops to fill it from 1.00 mL to 2.00 mL. Therefore, the average number of drops in one mL is
n = ½ * (43 + 39) = 41; the volume per drop is therefore
V(drop) = 1.00 mL / n * 0.02 mL.
This is in line with a general rule of thumb that assumes that a drop measures ca. 25 *L.
The volume in a plastic tea spoon was determined to be
V1 = 4.51 mL and V2 = 4.32 mL; the average value is therefore V average = ½ * (V1 + V2) or
V average = (4.51 mL + 4.32 mL) * 0.5 * 4.42 mL.
The five pennies were measured to be
2.176 g, 2.396 g, 2.424 g, 2.346 g, 2.188 g with an average of
m = 2.306 g.
If we place the pennies all at once onto the balance, we get
m = 11.550 g, which is almost the sum of all five pennies together; the actual sum would be 11.530 g, which means we have a deviation of 0.2%.
In all cases, the results met the expectation - by and large, with some unexpected findings. For example, when comparing measured to calculated volume, the measurements were distinctly smaller than calculated - this was surprising at first, yet made good sense, as we can't fill all devices completely. Measurement cylinders and other devices have actual graduations that do not aim at completely filling the device, so we always have to account for some 'dead' volume when filling a container. In that way, the experiments added a practical angle to our understanding of mass, length and volume. If one were to improve the experiments, one could combine mass and volume measurements by measuring the density of various liquids - alcohol, for example, will have a lower mass than water for the same volume.
1. The pencil is 8.70 cm long.
2A. Conversion factor: m/dm = 0.1 and dm/m = 10.
Therefore, 105 dm = x => x = 105 dm * m/dm = 105 * 0.1 m = 10.5 m
105 dm = 10.5 m.
2B. Conversion factors: g/mg = 0.001 and mg/g = 1,000.
Thus, 2,455 mg = x => x = 2,455 mg * g/mg = 2,455 * g * 0.001
2,455 mg = 0.002455 g.
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