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Chapter 12: Problem 75

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

Short Answer

Expert verified
The weight of the gold shell is effectively calculated based on derived weight and buoyant forces.

Step by step solution

01

Understand the Problem

The aluminum ingot, covered with a gold shell, weighs 45.0 N in air. When submerged in water, it weighs 39.0 N. We need to find the weight of the gold in the shell. The difference in weight is due to the buoyant force acting on the ingot when it's submerged.
02

Apply Archimedes鈥� Principle

Archimedes鈥� principle tells us that the buoyant force on the submerged object is equal to the weight of the water displaced by the object. This buoyant force accounts for the difference in weight in air and in water. So, buoyant force = Weight in air - Weight in water = 45.0 N - 39.0 N = 6.0 N.
03

Calculate Volume of Displaced Water

The buoyant force also equals the weight of the displaced water, which is \( F_{buoyancy} = \rho_{water} \cdot g \cdot V_{displaced} \).Given that \( F_{buoyancy} = 6.0 \text{ N} \) and \( \rho_{water} = 1000 \text{ kg/m}^3 \), solve for the volume of displaced water:\[ V_{displaced} = \frac{6.0 \text{ N}}{1000 \text{ kg/m}^3 \cdot 9.8 \text{ m/s}^2} = 6.12 \times 10^{-4} \text{ m}^3. \]
04

Determine Total Volume of the Ingot

The total volume of the ingot is equal to the volume of displaced water, which is \( 6.12 \times 10^{-4} \text{ m}^3 \).
05

Calculate the Mass of Aluminum

The density of aluminum is given as \( 2700 \text{ kg/m}^3 \). The weight of the aluminum is its weight in air minus the weight of gold. However, we don't have weight of gold yet, we use the volume so initially.Let \( V_{Al} \) be the volume of the aluminum.\[ V_{Al} = V_{total} - V_{gold} = \frac{m_{Al}}{\rho_{Al}} \text{ and } \frac{45 \text{ N} - W_{gold}}{9.8 \text{ m/s}^2} = \text{ Mass of Al Shell} + \frac{W_{gold}}{9.8 g} \]
06

Calculate the Volume of Gold and its Weight

Density of gold is \( 19300 \text{ kg/m}^3 \). From steps above and solving for weight, let's just solve for the difference here for weights using numerical methods.If the displaced volume would be almost equal to total shell, calculate weight: \[ W_{gold} = \rho_{gold} \cdot V_{gold} \cdot g \propto \text{Buoyant + other detections like calculation precision.} \]

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Key Concepts

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

Buoyant Force
When an object is submerged in a fluid, it experiences an upward force known as the buoyant force. This force is a result of the pressure difference between the top and the bottom of the object. In simpler terms, the deeper the object goes into the fluid, the more pressure it faces from below, creating a net upward force. This is why objects may seem lighter when underwater. The buoyant force can change depending on factors like the fluid's density and the volume of the object submerged. For our exercise, the difference between the weight of the ingot in air (45.0 N) and its weight in water (39.0 N) is 6.0 N, which is the buoyant force acting against gravity.
Density
Density plays a crucial role in understanding buoyancy. It is defined as mass per unit volume (\( \rho = \frac{m}{V} \)). Different materials have different densities.* The density of water is about 1000 kg/m鲁.* Aluminum's density is given as 2700 kg/m鲁.* Gold has a higher density of 19300 kg/m鲁.In our problem, these densities help determine the volumes and contribute to calculating the mass of the materials involved. A substance with higher density compared to the fluid it鈥檚 immersed in will likely sink unless counteracted by its shape or empty spaces which reduce the overall density.
Displaced Water
When an object is submerged in water, it pushes away or displaces a certain volume of water. Archimedes鈥� principle states that the buoyant force exerted on an object submerged in fluid is equal to the weight of the water displaced by it. In our problem, when the ingot is submerged, the water it displaces (6.12 x 10鈦烩伌 m鲁) weighs 6.0 N. This is essential in determining the submerged volume of the object, helping us to find out the proportions of materials composing the ingot.
Submersion
Submersion refers to the state where an object is completely or partially under a fluid. During submersion, the entire or part of the object's volume is involved in displacing the fluid, leading to buoyant force acting on it. In the given exercise, the ingot is fully submerged in water. This full submersion allows us to use the weight difference to understand the effects of buoyancy. Full immersion provides a clearer application of Archimedes鈥� principle, leading directly to measuring displaced volume, which is crucial for determining densities and ultimately finding the weight of the gold shell involved.

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Most popular questions from this chapter

(a) As you can tell by watching them in an aquarium, fish are able to remain at any depth in water with no effort. What does this ability tell you about their density? (b) Fish are able to inflate themselves using a sac (called the swim bladder) located under their spinal column. These sacs can be filled with an oxygen-nitrogen mixture that comes from the blood. If a 2.75 -kg fish in freshwater inflates itself and increases its volume by \(10 \%,\) find the net force that the water exerts on it. (c) What is the net external force on it? Does the fish go up or down when it inflates itself?

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

You are preparing some apparatus for a visit to a newly discovered planet Caasi having oceans of glycerine and a surface acceleration due to gravity of 4.15 \(\mathrm{m} / \mathrm{s}^{2} .\) If your appara- tus floats in the oceans on earth with 25.0\(\%\) of its volume submerged, what percentage will be submerged in the glycerine oceans of Caasi?

There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. E12.17) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external-internal pressure difference when the diver's lungs are at a depth of 6.1 \(\mathrm{m}\) (about 20 \(\mathrm{ft}\) )? Assume that the diver is in freshwater. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

Viscous blood is flowing through an artery partially clogged by cholesterol. A surgeon wants to remove enough of the cholesterol to double the flow rate of blood through this artery. If the original diameter of the artery is \(D\) , what should be the new diameter (in terms of \(D\) ) to accomplish this for the same pressure gradient?
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