Question

The density of an irregularly shaped object was determined as follows. The mass of the object was found to be \(28.90 \mathrm{~g} \pm 0.03 \mathrm{~g}\). A graduated cylinder was partially filled with water. The reading of the level of the water was \(6.4 \mathrm{~cm}^{3} \pm 0.1 \mathrm{~cm}^{3}\). The object was dropped in the cylinder, and the level of the water rose to \(9.8 \mathrm{~cm}^{3} \pm 0.1 \mathrm{~cm}^{3}\). What is the density of the object with appropriate error limits? (See Appendix 1.5.)

Short answer

The density of the irregularly shaped object is \(8.50 \frac{\mathrm{g}}{\mathrm{cm^{3}}} \pm 0.25 \frac{\mathrm{g}}{\mathrm{cm^{3}}}\).

Step-by-step solution

Determine the volume of the object

We know that the initial water level in the graduated cylinder was \(6.4 \mathrm{cm^{3}} \pm 0.1 \mathrm{cm^{3}}\) and the final water level was \(9.8 \mathrm{cm^{3}} \pm 0.1 \mathrm{cm^{3}}\). The volume of the irregularly shaped object can be calculated as the difference between the final and initial water levels: Volume \(= V = (9.8 - 6.4) \mathrm{cm^{3}} = 3.4 \mathrm{cm^{3}}\) Now let's find the error in the volume.

Calculate the error in the volume

To determine the error in the volume, we will utilize the error propagation rule for subtraction: Percentage_error_in_volume \(= \sqrt{(\frac{0.1}{6.4})^2 + (\frac{0.1}{9.8})^2}\) Percentage_error_in_volume \(= \sqrt{(\frac{0.1}{6.4})^2 + (\frac{0.1}{9.8})^2} = 0.0244\) Error_in_volume \(= 0.0244 \times 3.4 \mathrm{cm^{3}} = 0.083 \mathrm{cm^{3}}\) Hence, the volume of the object is \(3.4 \mathrm{cm^{3}} \pm 0.083 \mathrm{cm^{3}}\).

Compute the density of the object

The density of the object can be calculated using the formula: Density \(= \rho = \frac{Mass}{Volume} = \frac{28.90 \mathrm{g}}{3.4 \mathrm{cm^{3}}} = 8.50 \frac{\mathrm{g}}{\mathrm{cm^{3}}}\) Now let's determine the error limits for the density.

Calculate the error limits for the density

We will use the error propagation rule for division: Percentage_error_in_density \(= \sqrt{(\frac{0.03}{28.90})^2 + (\frac{0.083}{3.4})^2} = 0.0289\) Error_in_density \(= 0.0289 \times 8.50 \frac{\mathrm{g}}{\mathrm{cm^{3}}} = 0.25 \frac{\mathrm{g}}{\mathrm{cm^{3}}}\) Therefore, the density of the object is \(8.50 \frac{\mathrm{g}}{\mathrm{cm^{3}}} \pm 0.25 \frac{\mathrm{g}}{\mathrm{cm^{3}}}\).

Key concepts

Error Propagation

When we measure anything, there's always a chance of some error sneaking into our results. In science, these little errors are normal, but we've got to understand how they can spread when we do calculations like adding, subtracting, multiplying, or dividing. This spreading of errors is what we call 'error propagation' or 'uncertainty propagation'. Imagine you're baking and your scale is a bit off, adding that small mistake to each ingredient. By the end of the recipe, your cake might not taste quite right. In our density problem, we're mixing together errors from both the mass and volume measurements to find the density's error.

For subtracting numbers, like when we figured out the object's volume by subtracting the water levels before and after dunking it, the error can get larger. We do some math magic (a.k.a. the Pythagorean theorem) to combine the errors and make sure we're not understating our uncertainty. It's like admitting that our cake might range from slightly under-baked to slightly over-baked. Same goes for division when working out density, which is the object's mass dived by its volume. Bottom line, error propagation helps us keep our scientific results as honest as that 'could-be-better' cake!

Mass Measurement

Mass measurement is like checking how many apples you have in your basket. We use scales for this job. But it's not always perfect—scales can be a little 'off' every now and then, just like our imperfect sense of judgment. This is why we need to say our mass has a bit of wiggle room - an uncertainty. In our exercise, the mass of the gizmo was noted down as '28.90 grams plus or minus 0.03 grams'. What this means is, the real mass could be a tiny smidge over or under 28.90 grams. Science folk have come up with ways to note these teeny mass differences because they want their experiments and recipes to work out as best as they can.

Volume Displacement

Have you ever gotten into a bathtub and watched the water rise and spill over the edges? That's you displacing water - or in grown-up terms, pushing it out of the way. Scientists use this 'splashy' trick to figure out the volume of irregular shapes. If we drop our funny-looking object into water, it shoves some water out of the way. By seeing how much the water level goes up, we can tell the object's volume without having to wrap it up in a measuring tape, which is super handy if it's a weird shape. In our problem, the object bumped the water up from 6.4 to 9.8 cubic centimeters. The difference, 3.4 cubic centimeters, is our object's volume. Remember, though, even water can't sit still perfectly, so we've got ourselves another uncertainty to keep an eye on!

Calculation of Density

Density is all about how much stuff is packed into a space. Think of it like how many smarties you can cram into a jar. The more you pack in, the denser it is. Mathematicians and scientists are pretty clever—they found a simple formula to calculate this: Density equals Mass over Volume. So, once we know our object's mass and volume, we divide them. Voila, we get the object's density. In the exercise, after dividing the object's mass (28.90 g) by its new underwater space (3.4 cm³), we found out that the density is 8.50 g/cm³. But because this is science and not magic, we toss in our uncertainties from the mass and volume to get a 'plus or minus' for our density. This way, we keep it real, knowing that even if our scales and measuring cups are a tad naughty, we're on the ball with our numbers.