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Chapter 13: Problem 49

A metal object "weighs" \(26.0\) g in air and \(21.48 \mathrm{~g}\) when totally immersed in water. What is the volume of the object? What is its mass density?

Short Answer

Expert verified
Volume: 4.52 cm鲁, Density: 5.75 g/cm鲁

Step by step solution

01

Understand the problem

We have a metal object that weighs 26.0 g in air and 21.48 g when submerged in water. We need to find its volume and mass density.
02

Calculate Buoyant Force

The buoyant force is the difference between the weight of the object in air and the weight in water. Calculate by subtracting 21.48 g from 26.0 g:\[\text{Buoyant Force} = 26.0 \text{ g} - 21.48 \text{ g} = 4.52 \text{ g}\]
03

Convert Buoyant Force to Volume

The buoyant force in grams is equivalent to the volume in cubic centimeters (cm鲁) of the object, since water has a density of 1 g/cm鲁. Therefore, the volume of the object is 4.52 cm鲁.
04

Calculate Mass Density

Mass density is calculated by dividing the mass by the volume. Use the weight in air for mass:\[\text{Density} = \frac{\text{Mass in air}}{\text{Volume}} = \frac{26.0 \text{ g}}{4.52 \text{ cm}^3} \approx 5.75 \text{ g/cm}^3\]

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Key Concepts

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

Buoyant Force
The concept of buoyant force is a fundamental principle in fluid mechanics. It is the upward force that a fluid exerts on a submerged or partially submerged object. This force makes the object feel lighter when it is in the fluid. Archimedes' Principle helps to understand this phenomenon, stating that the buoyant force is equal to the weight of the fluid displaced by the object. This means that an object submerged in water will experience a buoyant force equal to the weight of the water it moves out of its way.
To calculate the buoyant force in our exercise, we take the difference between the object's weight in air and its weight when submerged in water:
  • Original weight in air: 26.0 g
  • Weight when submerged: 21.48 g
  • Buoyant force = 26.0 g - 21.48 g = 4.52 g
This difference of 4.52 grams signifies the weight of the water displaced by the object, demonstrating the magnitude of the buoyant force acting upward on the object.
Mass Density Calculation
Mass density is a measure of how much mass is contained in a given volume. It reflects the compactness of a substance, typically measured in grams per cubic centimeter (g/cm鲁). In our exercise, to find the mass density of the metal object, we use its weight in air as the mass. This ensures accuracy by excluding the effects of buoyancy.
With the volume calculated previously as 4.52 cm鲁 and knowing the mass to be 26.0 g when in air, the formula for density is:
  • Density = Mass / Volume
  • Density = 26.0 g / 4.52 cm鲁
The calculated density is approximately 5.75 g/cm鲁. This value indicates that the object is denser than water, as expected for most metals.
Volume Displacement
Volume displacement is a key concept used in determining the volume of irregularly shaped objects. According to Archimedes' Principle, when an object is immersed in a fluid, it displaces a volume of fluid equal to the volume of the object itself. This technique is especially handy because it removes the need to measure directly with less precise tools.
In this exercise, we use the buoyant force to find the volume. Since water's density is 1 g/cm鲁, the buoyant force in grams equates directly to the object's volume in cubic centimeters. Therefore, the 4.52 g buoyant force indicates a volume of 4.52 cm鲁 for the submerged metal object. This calculation not only shows the elegance of using water displacement to find volume but also highlights the interconnectedness of buoyancy, density, and volume.

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Most popular questions from this chapter

The diameter of the large piston of a hydraulic press is \(20 \mathrm{~cm}\), and the area of the small piston is \(0.50 \mathrm{~cm}^{2}\). If a force of \(400 \mathrm{~N}\) is applied to the small piston, (a) what is the resulting force exerted on the large piston? (b) What is the increase in pressure underneath the small piston? ( \(c\) ) Underneath the large piston?

A solid wooden cube, \(30.0 \mathrm{~cm}\) on each edge, can be totally submerged in water if it is pushed downward with a force of \(54.0\) N. What is the density of the wood?

A beaker contains oil of density \(0.80 \mathrm{~g} / \mathrm{cm}^{3}\). A \(1.6-\mathrm{cm}\) cube of aluminum \(\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) hanging vertically on a thread is submerged in the oil. Find the tension in the thread.

Find the density \(\rho\) of a fluid at a depth \(h\) in terms of its density \(\rho_{0}\) at the surface. If a mass \(m\) of fluid has volume \(V_{0}\) at the surface, then it will have volume \(V_{0}-\Delta V\) at a depth \(h\). The density at depth \(h\) is then which gives $$ \begin{array}{c} \rho=\frac{m}{V_{0}-\Delta V} \quad \text { while } \quad \rho_{0}=\frac{m}{V_{0}} \\ \frac{\rho}{\rho_{0}}=\frac{V_{0}}{V_{0}-\Delta V}=\frac{1}{1-\left(\Delta V / V_{0}\right)} \end{array} $$ However, from Chapter 12, the bulk modulus is \(B=P /\left(\Delta V / V_{0}\right)\) and so \(\Delta V / V_{0}=P / B\). Making this substitution, we obtain $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-P / B} $$ If we assume that \(\rho\) is close to \(\rho 0\), then the pressure at depth \(h\) is approximately \(\rho_{0} g h\), and so $$ \frac{\rho}{\rho_{0}}=\frac{1}{1-\left(\rho_{0} g h / B\right)} $$

A reservoir dam holds an \(8.00-\mathrm{km}^{2}\) lake behind it. Just behind the dam, the lake is \(12.0 \mathrm{~m}\) deep. What is the water pressure \((a)\) at the base of the dam and \((b)\) at a point \(3.0 \mathrm{~m}\) down from the lake's surface? The area of the lake behind the dam has no effect on the pressure against the dam. At any point, \(P=\rho_{w} g h\). (a) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(12.0 \mathrm{~m})=118 \mathrm{kPa}\) (b) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~m})=29 \mathrm{kPa}\)
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