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Chapter 14: Problem 3

You purchase a rectangular piece of metal that has dimensions \(5.0 \times 15.0 \times 30.0 \mathrm{mm}\) and mass 0.0158 \(\mathrm{kg}\) . The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated?

Short Answer

Expert verified
The average density is approximately 7.02 g/cm鲁. Yes, you were cheated; this is not gold.

Step by step solution

01

Calculate the Volume of the Metal

The volume of a rectangular prism (which is the shape of the metal piece) can be calculated using the formula \( V = l \times w \times h \). Here, the length \( l = 30.0 \) mm, the width \( w = 15.0 \) mm, and the height \( h = 5.0 \) mm. Thus, the volume is \( V = 30.0 \times 15.0 \times 5.0 = 2250 \) cubic millimeters.
02

Convert Volume to Cubic Centimeters

The density formula requires the volume in cubic centimeters. We know that \(1\) cubic centimeter equals \(1000\) cubic millimeters. Therefore, the volume in cubic centimeters is \( \frac{2250}{1000} = 2.25 \) cubic centimeters.
03

Calculate the Density

Density \( \rho \) is defined by the formula \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. Here, the mass \( m = 0.0158 \) kg, which converts to \( 15.8 \) g since \(1\) kg = \(1000\) g. Hence, the density is \( \rho = \frac{15.8}{2.25} \approx 7.02 \) g/cm\(^3\).
04

Compare with Gold's Density

The density of gold typically is around \(19.32\) g/cm\(^3\). The calculated density of the metal piece is approximately \(7.02\) g/cm\(^3\), which is much lower than the density of gold.

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Key Concepts

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

Rectangular Prism
Understanding the shape of the object is important not just in geometry but also in real-world applications, such as determining volumes. A rectangular prism is a 3D object with six faces, where each face is a rectangle. It often looks like a box.

To calculate the volume of a rectangular prism, we use the formula:
  • Volume, \( V = l \times w \times h \)
where \( l \) is the length, \( w \) is the width, and \( h \) is the height. This formula allows us to find out how much space the prism occupies.

In our scenario, the rectangular piece of metal has the dimensions \( 5.0 \times 15.0 \times 30.0 \) mm, leading to a volume of 2250 cubic millimeters. This is crucial for further calculations such as density.
Volume Conversion
When solving problems involving density, it is often necessary to convert between different volume units. Many problems will present dimensions in millimeters or other smaller units, but density calculations frequently use cubic centimeters.

This requires a conversion step. Remember:
  • 1 cubic centimeter (cm\(^3\)) = 1000 cubic millimeters (mm\(^3\))

For the given metal piece, the calculated volume is 2250 mm\(^3\). To convert this into cubic centimeters, divide by 1000:
  • \( \text{Volume in cm}^3 = \frac{2250}{1000} = 2.25 \text{ cm}^3 \)
Volume conversion is a fundamental step for further calculations whenever density is involved.
Density of Gold
Density is a key property of materials, indicating how much mass is contained within a given volume. Gold is well known for its high density, typically around 19.32 g/cm\(^3\). This is relatively high compared to many other metals and materials.

Knowing the typical density of gold can help determine the authenticity of an object. In this case, the metal piece had a calculated density of approximately 7.02 g/cm\(^3\), which is significantly lower than gold's characteristic density.

This discrepancy suggests that the piece is not pure gold, demonstrating the importance of understanding material properties like density.
Mass to Volume Ratio
The concept of density is fundamentally about the mass to volume ratio, which tells us how much mass exists in a given space. This is often expressed by the formula:
  • Density, \( \rho = \frac{m}{V} \)
where \( m \) is the mass and \( V \) is the volume. It's crucial to express mass in grams and volume in cubic centimeters to match the units of density often used in material science.

In the given problem, the mass was 0.0158 kg, which converts to 15.8 grams (since 1 kg = 1000 g). With a volume of 2.25 cm\(^3\), the density calculation became:
  • \( \rho = \frac{15.8}{2.25} \approx 7.02 \text{ g/cm}^3 \)
This ratio reflects how densely packed the atoms are in the metal piece, which gives insight into its material identity.

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Most popular questions from this chapter

The earth does not have a uniform density; it is most dense at its center and least dense at its surface. An approximation of its density is \(\rho(r)=A-B r,\) where \(A=12,700 \mathrm{kg} / \mathrm{m}^{3}\) and \(B=\) \(1.50 \times 10^{-3} \mathrm{kg} / \mathrm{m}^{4}\) . Use \(R=6.37 \times 10^{6} \mathrm{m}\) for the radius of the earth approximated as a sphere. (a) Geological evidence indicates that the densities are \(13,100 \mathrm{kg} / \mathrm{m}^{3}\) and \(2,400 \mathrm{kg} / \mathrm{m}^{3}\) at the earth's center and surface, respectively. What values does the linear approximation model give for the densities at these two locations? (b) Imagine dividing the earth into concentric, spherical shells. Each shell has radius \(r\) , thickness \(d r\) , volume \(d V=4 \pi r^{2} d r,\) and mass \(d m=\rho(r) d V .\) By integrating from \(r=0\) to \(r=R,\) show that the mass of the earth in this model is \(M=\frac{4}{3} \pi R^{3}\left(A-\frac{3}{4} B R\right)\) (c) Show that the given values of \(A\) and \(B\) give the correct mass of the earth to within 0.4\(\%\) (d) We saw in Section 12.6 that a uniform spherical shell gives no contribution to \(g\) inside it. Show that \(g(r)=\frac{4}{3} \pi G r\left(A-\frac{3}{4} B r\right)\) inside the earth in this model. (e) Verify that the expression of part (d) gives \(g=0\) at the center of the earth and \(g=9.85 \mathrm{m} / \mathrm{s}^{2}\) at the surface. (f) Show that in this model \(g\) does not decrease uniformly with depth but rather has a maximum of \(4 \pi G A^{2} / 9 B=10.01 \mathrm{m} / \mathrm{s}^{2}\) at \(r=2 A / 3 B=5640 \mathrm{km} .\)

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is 15.0 \(\mathrm{cm}\) (Fig. 14.37 . (a) What is the gauge pressure at the water-mercury interface? (b) Calculate the vertical distance \(h\) from the top of the mercury in the right-hand arm of the tube to the top of the water in the left-hand arm.

Oceans on Mars. Scientists have found evidence that Mars may once have had an ocean 0.500 \(\mathrm{km}\) deep. The acceleration due to gravity on Mars is 3.71 \(\mathrm{m} / \mathrm{s}^{2}\) , (a) What would be the gauge pressure at the bottom of such an ocean, assuming it was freshwater? (b) To what depth would you need to go in the earth's ocean to experience the same gauge pressure?

Lift on an Airplane. Air streams horizontally past a small airplane's wings such that the speed is 70.0 \(\mathrm{m} / \mathrm{s}\) over the top surface and 60.0 \(\mathrm{m} / \mathrm{s}\) past the bottom surface. If the plane has a wing area of 16.2 \(\mathrm{m}^{2}\) on the top and on the bottom, what is the net vertical force that the air exerts on the airplane? The density of the air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) .

Assume that crude oil from a supertanker has density 750 \(\mathrm{kg} / \mathrm{m}^{3}\) . The tanker runs aground on a sandbar. To refloat the tanker, its oil cargo is pumped out into steel barrels, each of which has a mass of 15.0 \(\mathrm{kg}\) when empty and holds 0.120 \(\mathrm{m}^{3}\) of oil. You can ignore the volume occupied by the steel from which the barrel is made. (a) If a salvage worker accidentally drops a filled, sealed barrel overboard, will it float or sink in the seawater? (b) If the barrel floats, what fraction of its volume will be above the water surface? If it sinks, what minimum tension would have to be exerted by a rope to haul the barrel up from the ocean floor? (c) Repeat parts (a) and (b) if the density of the oil is 910 \(\mathrm{kg} / \mathrm{m}^{3}\) and the mass of each empty barrel is 32.0 \(\mathrm{kg}\) .
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