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Chapter 12: Problem 3

The specific gravity of cast iron is \(7.20\). Find its density and the mass of \(60.0 \mathrm{~cm}^{3}\) of it. Make use of $$ \text { sp } \mathrm{gr}=\frac{\text { Density of substance }}{\text { Density of water }} \quad \text { and } \quad \rho=\frac{m}{V} $$ From the first equation, Density of iron \(=(\) sp gr \()\) (Density of water) \(=(7.20)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=7200 \mathrm{~kg} / \mathrm{m}^{3}\) and so \(\quad\) Mass of \(60.0 \mathrm{~cm}^{3}=\rho V=\left(7200 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(60.0 \times 10^{-6} \mathrm{~m}_{3}\right)=0.432 \mathrm{~kg}\)

Short Answer

Expert verified
The mass of 60.0 cm鲁 of cast iron is 0.432 kg.

Step by step solution

01

Understanding Specific Gravity Definition

Specific gravity (sp gr) is a measure that compares the density of a substance to the density of water. Mathematically, it is given by the formula: \( \text{sp gr} = \frac{\text{Density of substance}}{\text{Density of water}} \). For cast iron, the specific gravity is given as 7.20.
02

Calculating Density of Cast Iron

To find the density of cast iron, multiply its specific gravity by the density of water (since water's density is approximately \(1000 \mathrm{~kg/m^3}\)). Thus, the density of cast iron is: \( \text{Density of cast iron} = 7.20 \times 1000 \mathrm{~kg/m^3} = 7200 \mathrm{~kg/m^3} \).
03

Using the Density Formula

The density \(\rho\) is related to mass \(m\) and volume \(V\) by the formula: \( \rho = \frac{m}{V} \). We want to find the mass of a specific volume of cast iron, so we rearrange this to find mass: \( m = \rho V \).
04

Converting Volume Units

The volume given is \(60.0 \mathrm{~cm}^{3}\). Convert this to cubic meters (m鲁) since the density is in kilograms per cubic meter: \( 60.0 \mathrm{~cm}^{3} = 60.0 \times 10^{-6} \mathrm{~m}^{3} \).
05

Calculating Mass of Cast Iron

Substitute the density of cast iron and the converted volume into the mass formula: \( m = 7200 \mathrm{~kg/m^3} \times 60.0 \times 10^{-6} \mathrm{~m}^{3} = 0.432 \mathrm{~kg} \). So, the mass of 60.0 cm鲁 of cast iron is 0.432 kg.

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Key Concepts

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

Specific Gravity
Specific gravity is a simple yet crucial concept in understanding the properties of substances. It's a dimensionless quantity that indicates how dense a substance is compared to water. The specific gravity is calculated by dividing the density of the substance by the density of water.
For example, if cast iron has a specific gravity of 7.20, it means it is 7.20 times denser than water.
The specific gravity helps in determining whether an object will float or sink when placed in water, and is widely used in material science and engineering.
Density of Substances
Density is defined as the mass per unit volume of a substance. It is a fundamental physical property denoted by the symbol \( \rho \).
The density equation is expressed as \( \rho = \frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume.
For example, to find the density of cast iron, we use its specific gravity and multiply it by the density of water: \( 7.20 \times 1000 \text{ kg/m}^3 = 7200 \text{ kg/m}^3 \).
This calculation gives us the ability to compare it to other materials and ascertain its applicability in different contexts.
Volume Conversion
Volume conversion is essential when you are working with density and mass calculations that require consistent units. Many times, you will need to convert volume from one unit to another, like from cubic centimeters (cm鲁) to cubic meters (m鲁), to align with other measurements like density in kg/m鲁.
For instance, converting 60.0 cm鲁 to m鲁 involves multiplying by \( 10^{-6} \), giving us \( 60.0 \times 10^{-6} \text{ m}^3 \).
It's vital to ensure all units are consistent to avoid errors in your calculations, enabling accurate results for real-world applications.
Mass Calculation
Mass calculation is often needed once you have determined the density and volume of a substance. Using the rearranged density formula \( m = \rho V \), you can find the mass from the known density and volume.
For example, using a density of \( 7200 \text{ kg/m}^3 \) for cast iron, and a volume of \( 60.0 \times 10^{-6} \text{ m}^3 \), the mass is calculated as follows: \( m = 7200 \times 60.0 \times 10^{-6} = 0.432 \text{ kg} \).
This method ensures you can compute the mass of any material accurately, provided you have its density and volume, serving various practical needs.

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Most popular questions from this chapter

What is the density of the material in the nucleus of the hydrogen atom? The nucleus can be considered to be a sphere of radius \(1.2 \times 10^{-15} \mathrm{~m}\), and its mass is \(1.67 \times 10^{-27} \mathrm{~kg}\). The volume of a sphere is \((4 / 3) \pi r^{3}\).

A \(60-\mathrm{kg}\) motor sits on four cylindrical rubber blocks. Each cylinder has a height of \(3.0 \mathrm{~cm}\) and a crosssectional area of \(15 \mathrm{~cm}^{2}\). The shear modulus for this rubber is \(2.0 \mathrm{MPa} .(a)\) If a sideways force of \(300 \mathrm{~N}\) is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

A thin, semitransparent film of gold \(\left(\rho=19300 \mathrm{~kg} / \mathrm{m}^{3}\right)\) has an area of \(14.5 \mathrm{~cm}^{2}\) and a mass of \(1.93 \mathrm{mg}\). (a) What is the volume of \(1.93 \mathrm{mg}\) of gold? ( \(b\) ) What is the thickness of the film in angstroms, where \(1 \AA=10^{-10} \mathrm{~m} ?(c)\) Gold atoms have a diameter of about \(5 \AA .\) How many atoms thick is the film?

A solid cube of aluminum is \(2.00 \mathrm{~cm}\) on each edge. The density of aluminum is \(2700 \mathrm{~kg} / \mathrm{m}^{3}\). Find the mass of the cube. $$ \text { Mass of cube }=\rho V=\left(2700 \mathrm{~kg} / \mathrm{m}^{3}\right)(0.0200 \mathrm{~m})^{3}=0.0216 \mathrm{~kg}=21.6 \mathrm{~g} $$

A box-shaped piece of gelatin dessert has a top area of \(15 \mathrm{~cm}^{2}\) and a height of \(3.0 \mathrm{~cm}\). When a shearing force of \(0.50 \mathrm{~N}\) is applied to the upper surface, the upper surface displaces \(4.0 \mathrm{~mm}\) relative to the bottom surface. What are the shearing stress, the shearing strain, and the shear modulus for the gelatin? $$ \begin{array}{l} \sigma_{s}=\frac{\text { Tangential force }}{\text { Area of face }}=\frac{0.50 \mathrm{~N}}{15 \times 10^{-4} \mathrm{~m}^{2}}=0.33 \mathrm{kPa} \\ \varepsilon_{s}-\frac{\text { Displacement }}{\text { Height }}-\frac{0.40 \mathrm{~cm}}{3.0 \mathrm{~cm}}-0.13 \\ S=\frac{0.33 \mathrm{kPa}}{0.13}=2.5 \mathrm{kPa} \end{array} $$
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