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Chapter 1: Problem 79

A 40 -lb container of peat moss measures \(14 \times 20 \times 30\) in. A 40 -lb container of topsoil has a volume of \(1.9 \mathrm{gal}\). (a) Calculate the average densities of peat moss and topsoil in units of \(\mathrm{g} / \mathrm{cm}^{3}\). Would it be correct to say that peat moss is "lighter" than topsoil? Explain. (b) How many bags of peat moss are needed to cover an area measuring \(15.0 \mathrm{ft} \times 20.0 \mathrm{ft}\) to a depth of \(3.0\) in.?

Short Answer

Expert verified
The average density of peat moss is approximately 0.1317 g/cm鲁, while the average density of topsoil is approximately 2.5256 g/cm鲁. Therefore, peat moss is lighter than topsoil. To cover an area measuring 15.0 ft 脳 20.0 ft to a depth of 3.0 in, 16 bags of peat moss are needed.

Step by step solution

01

(a) Calculate the average densities of peat moss and topsoil

First, we need to convert the volumes and masses from different units to the same units. Volume of peat moss container: \(14 \times 20 \times 30 = 8400 \ \mathrm{in}^3\) Convert \(\mathrm{in}^3\) to \(\mathrm{cm}^3\): 1 in = 2.54 cm So, \(1 \mathrm{in}^3 = (2.54 \ \mathrm{cm})^3 = 16.3871 \ \mathrm{cm}^3\) Thus, volume of peat moss container = \(8400 \ \mathrm{in}^3 \times 16.3871 \frac{\mathrm{cm}^3}{\mathrm{in}^3} \approx 137653.34 \ \mathrm{cm}^3\) Convert 40 pounds to grams: 1 lb = 453.592 g Weight of peat moss container = \(40 \ \mathrm{lb} \times 453.592 \frac{\mathrm{g}}{\mathrm{lb}} = 18143.68 \ \mathrm{g}\) Now we can calculate the density of peat moss: Density = \(\frac{mass}{volume}\) Density of peat moss = \(\frac{18143.68 \ \mathrm{g}}{137653.34 \ \mathrm{cm}^3} \approx 0.1317 \ \mathrm{g/cm}^3\) Topsoil container volume is given in gallons. We need to convert it to cm鲁. 1 gallon = 3.78541 L 1 L = 1000 cm鲁 So, volume of topsoil container = \(1.9 \ \mathrm{gal} \times 3.78541 \frac{\mathrm{L}}{\mathrm{gal}} \times 1000 \frac{\mathrm{cm}^3}{\mathrm{L}} \approx 7184.279 \ \mathrm{cm}^3\) Since the weight of topsoil container is also 40 lb, we can now calculate the density of topsoil: Density of topsoil = \(\frac{18143.68 \ \mathrm{g}}{7184.279 \ \mathrm{cm}^3} \approx 2.5256 \ \mathrm{g/cm}^3\) Comparing the densities, peat moss is lighter, with an average density of approximately 0.1317 g/cm鲁, whereas topsoil has an average density of approximately 2.5256 g/cm鲁.
02

(b) Calculate the number of bags of peat moss needed to cover an area

We are given an area of \(15.0 \ \mathrm{ft} \times 20.0 \ \mathrm{ft}\) and a depth of \(3.0\) in. Our goal is to find out how many bags of peat moss are needed to cover this area. First, let's convert the length, width, and height dimensions to inches: Length = \(15\ \mathrm{ft} \times 12\ \mathrm{in/ft} = 180\ \mathrm{in}\) Width = \(20\ \mathrm{ft} \times 12\ \mathrm{in/ft} = 240\ \mathrm{in}\) Now we can calculate the volume of peat moss needed to cover the area to the given depth: Volume = Length x Width x Depth Volume = \(180\ \mathrm{in} \times 240\ \mathrm{in} \times 3\ \mathrm{in} = 129600\ \mathrm{in}^3\) Next, we need to find out how many bags of peat moss are needed: One bag of peat moss has a volume of \(14\times 20 \times 30= 8400\ \mathrm{in}^3\) So, \(\frac{129600\ \mathrm{in}^3}{8400\ \mathrm{in}^3} = 15.428\) Since we can't have a fraction of a bag, it is necessary to round up to the nearest whole number to fully cover the area. Therefore, 16 bags of peat moss are needed to cover the area measuring 15.0 ft 脳 20.0 ft to a depth of 3.0 in.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion
Understanding unit conversion is crucial for meaningful scientific calculation and comparison. It allows us to express quantities in different units, which is essential when working with measurements in multiple systems.

For example, in the solved problem, the volume initially measures in inches cubed is converted to centimeters cubed. Recognizing that 1 inch equals 2.54 centimeters, and using the formula that \(1 \text{in}^3 = (2.54 \text{cm})^3\text{or} 16.3871 \text{cm}^3\), enables accurate volume conversion.

Similarly, mass conversion from pounds to grams uses the fact that 1 pound is equal to 453.592 grams. These conversions provide the necessary step to compare densities and to determine the amount of materials in correct units, which is critical for scientific, engineering, and everyday applications.

A simple breakdown of unit conversion involves:
  • Knowing the conversion factors between units.
  • Using multiplication or division to convert quantities.
  • Remembering to convert all dimensions in the case of converting units of volume.
Mass to Volume Ratio
The mass to volume ratio, commonly known as density, is fundamental to understanding the physical properties of materials. It's defined as the mass of a substance divided by its volume.

In our textbook solution, we calculated the density of two different substances鈥攑eat moss and topsoil鈥攗sing this ratio. The formula used is \(\text{Density} = \frac{\text{mass}}{\text{volume}}\). By dividing the mass (after converting it to grams) by the volume (after converting it to cubic centimeters), we obtain the average density of peat moss and topsoil.

This ratio allows us to compare different substances and answer questions about their properties. Peat moss appears 'lighter' because it has a lower density, despite the same mass as topsoil. This concept answers practical questions too, such as how much substance is needed to fill or cover a certain volume.

Key takeaways for mass to volume ratio:
  • It is also known as density.
  • Crucial for comparing different materials.
  • Calculated by the formula \(\text{Density} = \frac{\text{mass}}{\text{volume}}\).
Density of Materials
The provided textbook solution demonstrates the density of materials, which is a characteristic property of every physical object. It's indicative of how compact a substance is.

In the context of the exercise, peat moss has a lower density compared to topsoil, indicating that for the same mass, peat moss occupies more volume. This property can influence decisions in various applications, such as agriculture, construction, and manufacturing.

When discussing 'lightness' or 'heaviness' of a material, it's essential to consider density and not just the weight. For instance, peat moss cannot be deemed 'lighter' than topsoil based simply on its mass, but rather on its lower density. It is this intrinsic property that often determines usage such as for insulation, where a lower density material might be chosen.

Essential aspects of material density include:
  • Varies between different materials.
  • Important for understanding buoyancy, material science, and product design.
  • Dependent on both the atomic/molecular structure and how close together these particles are.

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Most popular questions from this chapter

Identify each of the following as measurements of length, area, volume, mass, density, time, or temperature: (a) \(25 \mathrm{ps}\), (b) \(374.2 \mathrm{mg}\), (c) \(77 \mathrm{~K}\), (d) \(100,000 \mathrm{~km}^{2}\), (e) \(1.06 \mu \mathrm{m}\), (f) \(16 \mathrm{~nm}^{2}\), (g) \(-788^{\circ} \mathrm{C}\), (h) \(2.56 \mathrm{~g} / \mathrm{cm}^{3}\), (i) \(28 \mathrm{~cm}^{3}\). [Section 1.4]

In 2009, a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is \(0.20 \mathrm{~g} / \mathrm{cm}^{3}\), and its surface area is \(1242 \mathrm{~m}^{2}\) per gram of material. (a) Calculate the volume of a 10.0-mg sample of this material. (b) Calculate the surface area for a \(10.0\)-mg sample of this material. (c) A \(10.0\)-mL sample of contaminated water had \(7.748 \mathrm{mg}\) of mercury in it. After treatment with \(10.0 \mathrm{mg}\) of the new spongy material, \(0.001 \mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?

The density of air at ordinary atmospheric pressure and \(25^{\circ} \mathrm{C}\) is \(1.19 \mathrm{~g} / \mathrm{L}\). What is the mass, in kilograms, of the air in a room that measures \(14.5 \mathrm{ft} \times 16.5 \mathrm{ft} \times 8.0 \mathrm{ft}\) ?

The inside dimension of a box that is cubic is \(24.8 \mathrm{~cm}\) on each edge with an uncertainty of \(0.2 \mathrm{~cm}\). What is the volume of the box? What do you estimate to be the uncertainty in the calculated volume?

Automobile batteries contain sulfuric acid, which is commonly referred to as "battery acid." Calculate the number of grams of sulfuric acid in \(1.00\) gal of battery acid if the solution has a density of \(1.28 \mathrm{~g} / \mathrm{mL}\) and is \(38.1 \%\) sulfuric acid by mass.
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