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Chapter 14: Problem 73

A hunk of aluminum is completely covered with a gold shell to form an ingot of weight 45.0 \(\mathrm{N}\) . When you suspend the ingot from a spring balance and submerge the ingot in water, the balance reads 39.0 \(\mathrm{N}\) . What is the weight of the gold in the shell?

Short Answer

Expert verified
The weight of the gold in the shell is approximately 29.3 N.

Step by step solution

01

Identify Information and Concepts

First, note the given information: the total weight of the ingot is 45.0 N in air and 39.0 N when submerged in water. The apparent weight loss when submerged (6.0 N) represents the buoyant force, which is equal to the weight of the water displaced by the ingot. We need to find the weight of the gold shell.
02

Calculate Volume Displaced

The buoyant force is given by the difference in weight measurements in air and in the water, which is 6.0 N. Use the relationship: \( F_b = \rho_{water} \times V_{displaced} \times g \) to find the volume displaced, where \( \rho_{water} = 1000 \, \text{kg/m}^3 \). Therefore, \( V_{displaced} = \frac{F_b}{\rho_{water} \times g} = \frac{6.0}{1000 \times 9.8} \).
03

Determine Total Volume of Ingot

Calculate the volume displaced using the formula from the previous step: \( V_{displaced} = \frac{6.0}{1000 \times 9.8} = 6.12 \times 10^{-4} \, \text{m}^3 \). This is the volume of the entire ingot.
04

Calculate Volume of Aluminum

Let \( V_{Al} \) be the volume of aluminum. The density of aluminum is approximately \( 2700 \, \text{kg/m}^3 \). The weight of the aluminum alone would be \( W_{Al} = \rho_{Al} \times V_{Al} \times g \). Since we have the total weight of the ingot, we can express it as 45.0 N = \( \rho_{Al} \times V_{Al} \times g + \rho_{Au} \times (V_{displaced} - V_{Al}) \times g\), where \( \rho_{Au} \) is the density of gold (approximately \( 19300 \, \text{kg/m}^3 \)).
05

Solve for Volume of Gold

Rearrange to express \( V_{Al} \) and \( V_{Au} = V_{displaced} - V_{Al} \). Substitute and solve for \( V_{Au} \). From the equation: \( 45.0 = 2700 \times V_{Al} \times 9.8 + 19300 \times (6.12 \times 10^{-4} - V_{Al}) \times 9.8 \). Solve for \( V_{Au} \), and substituting back gives \( V_{Au} = 1.55 \times 10^{-4} \, \text{m}^3 \).
06

Determine Weight of Gold

Calculate the weight of the gold using its volume: \( W_{Au} = \rho_{Au} \times V_{Au} \times g \). Substituting the appropriate values: \( W_{Au} = 19300 \times 1.55 \times 10^{-4} \times 9.8 \) which gives the weight of the gold.

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Key Concepts

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

Density of Aluminum
Aluminum is a lightweight metal with a density of approximately 2700 kg/m鲁. It is much less dense compared to other metals like gold. Density is a key factor in understanding the overall weight and buoyancy when an object is immersed in a fluid.

For example, when an aluminum piece is submerged in water, its density influences how much it appears to weigh, known as its apparent weight. Aluminum's lower density results in less weight per unit volume, making it ideal for lightweight applications.

In our exercise, the aluminum forms the core of the ingot. Its volume can be calculated given its density and the overall weight equation of the submerged ingot. This understanding helps in determining how much of the ingot's total weight is due to the aluminum inside.
Density of Gold
Gold is a dense and heavy metal with a density around 19300 kg/m鲁, making it significantly denser than many other materials, including aluminum.

This high density means that even a small volume of gold becomes quite heavy, which is why gold is so valuable and sought after.

In this exercise, the gold forms a shell over the aluminum, contributing significantly to the total weight of the ingot in air.
When calculating the weight of the gold shell, understanding gold's density is crucial. This density helps us determine the volume of gold required to reach a particular weight, alongside aluminum, when both are submerged in a fluid like water.
Weight in Fluids
Weight in fluids refers to how an object's weight changes when it is submerged in a fluid, such as water. This apparent weight change is due to the buoyant force exerted by the fluid.

The buoyant force is equal to the weight of the fluid displaced by the object, causing the object to feel "lighter" when submerged.
In this scenario, the ingot weighs 45.0 N in air but only 39.0 N when submerged in water, implying a 6.0 N buoyant force acting on it.

This concept is vital in understanding the relationship between an object's density, its weight in air, and its weight when fully submerged in a fluid. The difference in weight (apparent loss) helps determine the volume of the entire ingot.
Archimedes' Principle
Archimedes' Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that the object displaces.

This principle helps in understanding why objects seem to lose weight in fluids, and it plays a crucial role in calculating the effective weight of submerged objects.
In the exercise, the buoyant force on the ingot of aluminum and gold in water is 6.0 N. This force reflects the weight of the water that the ingot displaces.

By using Archimedes' Principle, we determine the volume of water displaced, and hence the total volume of the ingot. This provides crucial insight into separating the contributions to weight by different layers, like the gold shell in this ingot.

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Most popular questions from this chapter

At one point in a pipeline the water's speed is 3.00 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(5.00 \times 10^{4} \mathrm{Pa}\) . Find the gauge pressure at a second point in the line, 11.0 \(\mathrm{m}\) lower than the first, if the pipe diameter at the second point is twice that at the first.

Black Smokers. Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 \(\mathrm{m}\) to 3200 \(\mathrm{m}\) below the surface. What is the gauge pressure at a \(3200-\mathrm{m}\) deep vent, assuming the density of water does not vary? Express your answer in pascals and atmospheres.

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 hot-air balloon has a volume of 2200 \(\mathrm{m}^{3}\) . The balloon fabric (the envelope) weighs 900 \(\mathrm{N}\) . The basket with gear and full propane tanks weighs 1700 \(\mathrm{N}\) . If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?

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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