/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A piece of pure gold \(\left(\rh... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 22

A piece of pure gold \(\left(\rho=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is suspected to have a hollow center. It has a mass of \(38.25 \mathrm{~g}\) when measured in air and \(36.22 \mathrm{~g}\) in water. What is the volume of the central hole in the gold? Remember that you go from a density in \(\mathrm{g} / \mathrm{cm}^{3}\) to \(\mathrm{kg} / \mathrm{m}^{3}\) by multiplying by 1000 . From \(\bar{\rho}=m / V\), ctual volume of \(38.25 \mathrm{~g}\) of gold \(=\frac{0.03825 \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}}=1.982 \times 10^{-6} \mathrm{~m}^{3}\) $$ \begin{array}{c} \text { Volume of displaced water }=\frac{(38.25-36.22) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2.030 \times 10^{-6} \mathrm{~m}^{3} \\ \text { Volume of hole }=(2.030-1.982) \mathrm{cm}^{3}=0.048 \mathrm{~cm}^{3} \end{array} $$

Short Answer

Expert verified
The volume of the central hole in the gold is 0.048 cm鲁.

Step by step solution

01

Calculate the Apparent Weight Loss in Water

The difference in weight between the gold in air and water gives the buoyant force, which equals the weight of the displaced water. This can be calculated as: \( \text{Apparent Weight Loss} = 38.25 \text{ g} - 36.22 \text{ g} = 2.03 \text{ g} \).
02

Convert Weight Loss to Volume of Displaced Water

Using the density of water (1 \( \text{g/cm}^3 \)), convert the weight loss due to buoyancy into volume: \( \text{Volume of Displaced Water} = 2.03 \text{ g} = 2.03 \text{ cm}^3 \).
03

Calculate Volume of Total Gold Using Density

Convert the density of gold from \( \text{g/cm}^3 \) to \( \text{kg/m}^3 \). The volume of the entire piece of gold is given by: \( \text{Volume of Gold (no hole)} = \frac{38.25 \text{ g}}{19.3 \text{ g/cm}^3} = 1.982 \text{ cm}^3 \).
04

Calculate Volume of the Hole

The actual volume of the gold piece is greater than calculated due to the hollow, shown by the greater displaced water volume. The volume of the hole is: \( \text{Volume of Hole} = \text{Volume of Displaced Water} - \text{Volume of Gold} = 2.03 \text{ cm}^3 - 1.982 \text{ cm}^3 = 0.048 \text{ cm}^3 \).

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Key Concepts

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

Density Conversion
Understanding density conversion is crucial when working with different measurement units. Density is defined as mass per unit volume, commonly expressed in units like grams per cubic centimeter (g/cm鲁) or kilograms per cubic meter (kg/m鲁). To convert between these units, use the conversion factor where 1 g/cm鲁 equals 1000 kg/m鲁.

For example, if the density of gold is given as 19.3 g/cm鲁, converting this to kg/m鲁 involves multiplying by 1000, resulting in 19,300 kg/m鲁. This conversion allows for consistency when performing calculations that involve different measurement systems. Always remember:
  • Multiply by 1000 to convert from g/cm鲁 to kg/m鲁.
  • Divide by 1000 to convert from kg/m鲁 to g/cm鲁.
These conversions are essential for precise calculations, especially in scientific and engineering contexts.
Volume Calculation
Calculating volume often involves understanding the relationship between mass, density, and volume. The basic formula is:\[ V = \frac{m}{\rho} \]where \( V \) is volume, \( m \) is mass, and \( \rho \) is density.

In the context of our gold piece, we use the given mass and density to find the volume. For instance, with a mass of 38.25 g and a density of 19.3 g/cm鲁, we calculate:\[\text{Volume of gold} = \frac{38.25 \text{ g}}{19.3 \text{ g/cm}^3} = 1.982 \text{ cm}^3 \]
This calculation helps to determine how much space the gold would occupy if it were solid, without any hollow parts. It's important to ensure the units for mass and density are compatible when performing such calculations to avoid errors.
Apparent Weight Loss
Apparent weight loss in water is a fundamental concept tied to buoyancy. When an object is submerged in a fluid, it appears to weigh less due to the buoyant force, which is equal to the weight of the fluid displaced by the object.

In our exercise, the gold piece appears to lose 2.03 g when submerged, calculated by subtracting the weight in water from the weight in air (38.25 g - 36.22 g = 2.03 g). This difference represents the weight of the displaced water, demonstrating the principle of buoyancy.
  • The apparent weight loss equals the buoyant force.
  • This force corresponds to the weight of the fluid displaced.
  • For objects submerged in water: \( \text{Volume displaced (cm}^3 \text{)} = \text{Weight loss (g)} \)
Understanding this concept allows one to infer information about the submerged object's volume and structure, such as detecting hollow sections within otherwise solid materials.

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Most popular questions from this chapter

A covered cubic tank \(5.00 \mathrm{~m}\) by \(5.00 \mathrm{~m}\) by \(5.00 \mathrm{~m}\) is completely filled with water through a threaded hole in its lid. A hollow vertical pipe \(5.00 \mathrm{~m}\) tall is screwed into the hole. The pipe has a cross-sectional opening area of \(8.00 \mathrm{~cm}^{2}\). If the pipe is then filled to a height of \(4.00 \mathrm{~m}\) with an additional amount of water, what change in pressure, if any, will be read by a gauge in the side of the tank?

A vertical test tube has \(2.0 \mathrm{~cm}\) of oil \(\left(\rho=0.80 \mathrm{~g} / \mathrm{cm}^{3}\right)\) floating on \(8.0 \mathrm{~cm}\) of water. What is the pressure at the bottom of the tube due to the liquid in it? $$ \begin{aligned} P &=\rho_{1} h h_{1}+\rho_{2} h_{2}=\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2} \times 0.020 \mathrm{~m}\right)+\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.080 \mathrm{~m}) \\ &=0.94 \mathrm{kPa} \end{aligned} $$

On a day when the pressure of the atmosphere is \(1.000 \times 10^{5} \mathrm{~Pa}\), a chemist distills a liquid under slightly reduced pressure. The pressure within the distillation chamber is read by an oil-filled manometer (density of oil \(\left.=0.78 \mathrm{~g} / \mathrm{cm}^{3}\right)\). The difference in heights on the two sides of the manometer is \(27 \mathrm{~cm}\). What is the pressure in the distillation chamber?

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) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~m})=29 \mathrm{kPa}\)

A 60 -kg rectangular box, open at the top, has base dimensions of \(1.0 \mathrm{~m}\) by \(0.80 \mathrm{~m}\) and a depth of \(0.50 \mathrm{~m}\). (a) How deep will it sink in fresh water? (b) What weight \(F_{W b}\) of ballast will cause it to sink to a depth of \(30 \mathrm{~cm}\) ? (a) Assuming that the box floats, \(F_{B}=\) Weight of displaced water \(=\) Weight of box \(\left.1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times y)=(60 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) where \(y\) is the depth the box sinks. Solving yields \(y=0.075 \mathrm{~m}\). Because this is smaller than \(0.50 \mathrm{~m}\), our assumption is shown to be correct. (b) \(F_{B}=\) weight of box \(+\) weight of ballast But the \(F_{B}\) is equal to the weight of the displaced water. Therefore, the above equation becomes \(\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times 0.30 \mathrm{~m})=(60)(9.81) \mathrm{N}+F_{W b}\) from which \(F_{W b}=1760 \mathrm{~N}=1.8 \mathrm{kN}\). The ballast must have a mass of \((1760 / 9.81) \mathrm{kg}=180 \mathrm{~kg}\).
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