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Chapter 1: Problem 22

Gold, which has a density of \(19.32 \mathrm{~g} / \mathrm{cm}^{3}\), is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold with a mass of \(29.34 \mathrm{~g}\) is pressed into a leaf of \(1.000 \mu \mathrm{m}\) thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius \(2.500\) \(\mu \mathrm{m}\), what is the length of the fiber?

Short Answer

Expert verified
(a) The area of the leaf is approximately 15190 cm虏. (b) The length of the fiber is approximately 774400.31 cm.

Step by step solution

01

Identify Known Variables

We have the mass of gold, the density, the thickness for the leaf, and the radius for the fiber. The known values are: \( \text{mass} = 29.34 \, \text{g} \) \( \text{density} = 19.32 \, \text{g/cm}^3 \) \( \text{thickness of the leaf} = 1.000 \, \mu\text{m} = 0.0001 \, \text{cm} \) \( \text{radius of the fiber} = 2.500 \, \mu\text{m} = 0.00025 \, \text{cm} \)
02

Calculate Volume of Gold

To find the volume \( V \) of the gold, use the formula for density: \( \text{density} = \frac{\text{mass}}{\text{volume}} \). Rearrange to solve for volume:\[ V = \frac{\text{mass}}{\text{density}} = \frac{29.34 \, \text{g}}{19.32 \, \text{g/cm}^3} = 1.519 \, \text{cm}^3 \]
03

Calculate Area of the Gold Leaf

With the thickness of the leaf known, the area \( A \) can be found using the formula for volume of a rectangular prism: \( V = A \times \text{thickness} \). Rearrange to solve for area:\[ A = \frac{V}{\text{thickness}} = \frac{1.519 \, \text{cm}^3}{0.0001 \, \text{cm}} = 15190 \, \text{cm}^2 \]
04

Calculate Length of the Gold Fiber

For a cylindrical shape, use the formula \( V = \pi r^2 h \) where \( h \) is the height (or length) of the cylinder. Rearrange to find \( h \):\[ h = \frac{V}{\pi r^2} = \frac{1.519 \, \text{cm}^3}{\pi \times (0.00025 \, \text{cm})^2} \approx 774400.31 \, \text{cm} \]

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Key Concepts

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

Density
Density is a fundamental property of materials, describing how much mass is contained within a certain volume. It's a critical concept in physics and chemistry, helping us understand how substances behave in different conditions.
The formula for density is simple:
  • Density = \( \frac{\text{Mass}}{\text{Volume}} \)
This equation allows us to calculate the density of an object when both its mass and volume are known.
In our case, we are given the density of gold as \(19.32 \mathrm{~g} / \mathrm{cm}^3 \). This means that each cubic centimeter of gold has a mass of 19.32 grams. Understanding this allows us to find other gold properties, such as its volume, when we know its mass.
Metal Properties
Metals have unique properties, including ductility, which is the ability to be drawn into wires or pressed into thin sheets. Gold is known as the most ductile metal, making it perfect for applications requiring thin sheets or long wires.
Additionally, metals generally have high density and are excellent conductors of electricity and heat, besides being malleable and possessing a shiny luster. This combination enables metals to be used in a variety of applications, from electronics to jewelry.
  • Ductility: This property makes it possible to create gold leaves as thin as even a few micrometers.
  • Conductivity: A highly conductive metal, gold isn't just decorative; it's crucial in electronic components.
These properties don't just affect the way metals feel or look; they play a crucial role in determining how metals are used across industries.
Cylinder Volume Calculation
When calculating the volume of a cylinder, it's vital to understand the geometric shape involved. A cylinder has two circular bases and a certain height, and its volume can be evaluated using:
  • Volume = \( \pi r^2 h \)
where \( r \) is the radius of the circular base and \( h \) is the height (or length in some cases).
For example, in the problem, gold is drawn into a cylindrical fiber with a specific radius. Calculating its volume involves knowing the fiber length and assuming no material loss in conversion.
Area Calculation
Area calculation is often necessary in theoretical and practical applications. In this scenario, we calculate the area of a thin gold leaf.
When a material is turned into a thin sheet, its volume (found from its mass and density) can be used to determine its area, given the thickness of the leaf:
  • Area = \( \frac{V}{\text{thickness}} \)
Using this formula, when the thickness of the gold leaf is known, it allows the calculation of the leaf's area. This principle ensures that every atom of gold is accounted for, providing a comprehensive understanding of spatial distribution in material science.
Understanding how to calculate area helps in planning and resource allocation in various industrial and scientific applications.

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Most popular questions from this chapter

At the end of a year, a motor car company announces that sales of pickup trucks are down by \(43.0 \%\) for the year. If sales continue to decrease by \(43.0 \%\) in each succeeding year, how long will it take for sales to fall below \(10.0 \%\) of the original number?

A vertical container with base area measuring \(14.0 \mathrm{~cm}\) by \(17.0 \mathrm{~cm}\) is being filled with identical pieces of candy, each with a volume of \(50.0 \mathrm{~mm}^{3}\) and a mass of \(0.0200 \mathrm{~g}\). Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of \(0.250\) \(\mathrm{cm} / \mathrm{s}\), at what rate (kilograms per minute) does the mass of the candies in the container increase?

A fortnight is a charming English measure of time equal to \(2.0\) weeks (the word is a contraction of "fourteen nights"). That is a nice amount of time in pleasant company but perhaps a painful string of microseconds in unpleasant company. How many microseconds are in a fortnight?

Einstein's mass-energy equation relates mass \(m\) to energy \(E\) as \(E=m c^{2}\), where \(c\) is speed of light in vacuum. The energy at nuclear level is usually measured in MeV, where \(1 \mathrm{MeV}=\) \(1.60218 \times 10^{-13} \mathrm{~J}\); the masses are measured in unified atomic mass unit \((\mathrm{u})\), where \(1 \mathrm{u}=1.66054 \times 10^{-27} \mathrm{~kg}\). Prove that the energy equivalent of \(1 \mathrm{u}\) is \(931.5 \mathrm{MeV}\).

Because Earth's rotation is gradually slowing, the length of each day increases: The day at the end of \(1.0\) century is \(1.0 \mathrm{~ms}\) longer than the day at the start of the century. In 30 centuries, what is the total of the daily increases in time?
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