Question

The density of an irregularly shaped object was determined as follows. The mass of the object was found to be \(28.90 \mathrm{g}=\) 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 \(8.50 \frac{g}{cm^3} \pm 0.47 \frac{g}{cm^3}\).

Step-by-step solution

Identify the given values

From the problem, we have the following data: - Mass of the object: \(m = 28.90g \pm 0.03g\) - Initial volume of water: \(V_1 = 6.4cm^3 \pm 0.1cm^3\) - Final volume of water (after object drops in): \(V_2 = 9.8cm^3 \pm 0.1cm^3\)

Calculate the volume of the object

To calculate the volume of the object, find the difference between the final and initial water volumes in the graduated cylinder: \(V_{object} = V_2 - V_1\)

Find the error of the volume

Since both initial and final volume readings have the same error of \(0.1cm^3\), the error in the volume of the object is the sum of the errors: \(Error_{volume} = Error_{V1} + Error_{V2}\)

Calculate the density of the object

Now, we can use the mass and volume of the object to find its density using the formula: \(density = \dfrac{mass}{volume}\)

Determine the error limits in density

To estimate the error limits on the calculated density, use the following formula: \(Error_{density} = density \times \sqrt{(\dfrac{Error_{mass}}{mass})^2+(\dfrac{Error_{volume}}{volume})^2}\) Now, by substituting the given values and calculating the results:

Calculate the volume of the object

\(V_{object} = 9.8 cm^3 - 6.4 cm^3 = 3.4 cm^3\)

Find the error of the volume

\(Error_{volume} = 0.1cm^3 + 0.1cm^3 = 0.2cm^3\)

Calculate the density of the object

\(density = \dfrac{28.90g}{3.4 cm^3} = 8.50 \frac{g}{cm^3}\)

Determine the error limits in density

\(Error_{density} = 8.50 \frac{g}{cm^3} \times \sqrt{(\dfrac{0.03g}{28.90g})^2+(\dfrac{0.2cm^3}{3.4cm^3})^2} \approx 0.47 \frac{g}{cm^3}\)

Write the final result with error limits

The density of the object with appropriate error limits is: \(8.50 \frac{g}{cm^3} \pm 0.47 \frac{g}{cm^3}\)

Key concepts

Volume Calculation

When we want to find out the density of an object, one of the crucial steps is calculating its volume. If the object is irregularly shaped, we often use the displacement method with a graduated cylinder. In this method, you measure how much the water level in the cylinder rises after you immerse the object. The initial step is to record the initial volume of water, then add the object, and note the final water level.

• This means the volume of the object is the difference between the final and initial water volume: \[ V_{\text{object}} = V_2 - V_1 \]
• In our example, where the initial water level is 6.4 cm³ and the final level is 9.8 cm³, the volume of the object is: \[ 9.8 \, cm^3 - 6.4 \, cm^3 = 3.4 \, cm^3 \]

Understanding this helps in calculating the density, as density requires both mass and volume as inputs, with the formula \[ \text{density} = \frac{\text{mass}}{\text{volume}} \].

Error Analysis

When performing any measurement, it's crucial to account for potential errors to ensure accurate results. Error analysis helps to understand the precision and reliability of our measurements. In the case of measuring volume and mass, errors can arise from several sources:

• Both initial and final water volume readings come with a possible error. In our case, each has an error of 0.1 cm³.
• When calculating the object's volume from these measurements, the errors from each reading accumulate, giving us a total volume error: \[ \text{Error}_{\text{volume}} = \text{Error}_{V1} + \text{Error}_{V2} \]
• Here, this results in a total error in the object's volume of \[ 0.2 \, cm^3 \]

Additionally, when calculating density, we also incorporate errors from mass. By applying the formula \[ \text{Error}_{\text{density}} = \text{density} \times \sqrt{\left(\frac{\text{Error}_{\text{mass}}}{\text{mass}}\right)^2+\left(\frac{\text{Error}_{\text{volume}}}{\text{volume}}\right)^2} \], one can find the combined uncertainty. This comprehensive view ensures a more accurate representation of the possible variation in our density calculation.

Graduated Cylinder Measurement

A graduated cylinder is a reliable tool for measuring liquid volume. Its transparency allows precise reading of the liquid level, and it's particularly useful for the volume displacement method. However, accuracy in measurement requires careful technique:

• Ensure the cylinder is placed on a flat surface to avoid skewed readings.
• Eye level should match the level of the liquid to accurately read the bottom of the meniscus—the curve seen at the liquid's top surface.
• In the context of our exercise, we used a graduated cylinder both before and after immersing the object to measure how much the water level rises.
• With an initial reading of 6.4 cm³ and a final of 9.8 cm³, these observations provided crucial data for calculating the object's volume.

Thus, understanding how to correctly use a graduated cylinder and account for any observational errors ensures precise measurements, forming the backbone of reliable experimental results.