/*! 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 26 The density of aluminum is \(2.7... [FREE SOLUTION] | 91Ó°ÊÓ

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The density of aluminum is \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What volume does \(2.00\) kg occupy?

Short Answer

Expert verified
The volume is approximately 740.74 cm³.

Step by step solution

01

Convert mass into grams

We start by converting the given mass of aluminum from kilograms to grams, since the density is given in units of grams per cubic centimeter. We know that there are 1000 grams in 1 kilogram. Therefore, for 2.00 kg, the conversion is:\[ 2.00 \text{ kg} \times 1000 \text{ g/kg} = 2000 \text{ g} \]
02

Apply the density formula

The formula to find the volume given the mass and the density is:\[ V = \frac{m}{\rho} \]where \( V \) is the volume,\( m \) is the mass, and\( \rho \) is the density.
03

Substitute values into the formula

Now substitute the known values into the formula:\[ m = 2000 \text{ g}, \quad \rho = 2.70 \text{ g/cm}^3 \]Then the volume is: \[ V = \frac{2000 \text{ g}}{2.70 \text{ g/cm}^3} \]
04

Calculate the volume

Perform the division to find the volume:\[ V = \frac{2000}{2.70} \approx 740.74 \text{ cm}^3 \]Therefore, the volume of aluminum that has a mass of 2.00 kg is approximately 740.74 cm³.

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Key Concepts

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

Density Calculation
Understanding density is a key concept in physics. Density essentially measures how much mass is contained in a given volume. It tells us how tightly matter is packed together in a substance. The formula for density (\( \rho \)) is given by:\[\rho = \frac{m}{V}\]where \( m \) is the mass and \( V \) is the volume. To rearrange this formula for finding volume, you can use:\[V = \frac{m}{\rho}\]This reorganization shows how volume relates to mass when the density is known. The calculation helps in finding out how much space an object will occupy given its mass and density. In our example with aluminum, knowing the density (2.70 g/cm³) allowed us to find the space that 2.00 kg of aluminum will occupy. This principle is useful across many real-world applications, from shipping materials to manufacturing processes.
Unit Conversion
Unit conversion is necessary to ensure that quantities are expressed in consistent units ready for calculations. Here, converting mass from kilograms to grams was crucial because the measurement for density was given in g/cm³. The general rule for converting kilograms to grams is straightforward as: - 1 kilogram is equivalent to 1000 grams.
- Therefore, multiply the mass in kilograms by 1000. For example, 2.00 kg of aluminum became 2000 g after conversion. Unit conversion is not only important in solving physics problems but also in everyday situations such as cooking or traveling. If ignored, different units can result in calculation errors. Adjusting units ensures proper understanding and eliminates potential mistakes during problem-solving steps.
Volume Determination
Determining volume means finding out how much space an object or material occupies. In physics, after converting units and knowing the density of a material, you can easily compute volume with the rearranged density formula:\[V = \frac{m}{\rho}\]In cases where you have mass and density, simply divide the mass by the density. This will give you the object's volume, considering the units are compatible.In our sample problem, we calculated the volume for 2.00 kg (or 2000 g) of aluminum. With a density of 2.70 g/cm³, substituting these values:\[V = \frac{2000 \, \text{g}}{2.70 \, \text{g/cm}^3} \approx 740.74 \, \text{cm}^3\]Volume determination plays a crucial role in various applications such as fluid dynamics, material science, and construction, where knowing the exact space an object occupies is vital.

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Most popular questions from this chapter

A calibrated flask has a mass of \(30.0\) g when empty, \(81.0\) g when filled with water, and \(68.0 \mathrm{~g}\) when filled with an oil. Find the density of the oil. First find the volume of the flask from \(\rho=m / v\) using the water data: $$ V=\frac{m}{\rho}=\frac{(81.0-30.0) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=51.0 \times 10^{-6} \mathrm{~m}^{3} $$ Then, for the oil. $$ \rho_{\mathrm{oil}}=\frac{m_{\mathrm{cil}}}{V}=\frac{(68.0-30.0) \times 10^{-3} \mathrm{~kg}}{51.0 \times 10^{-6} \mathrm{~m}^{3}}=745 \mathrm{~kg} / \mathrm{m}^{3} $$

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?

Atmospheric pressure is about \(1.01 \times 10^{5} \mathrm{~Pa} .\) How large a force does the atmosphere exert on a \(2.0-\mathrm{cm}^{2}\) area on the top of your head? Because \(P=F / A\), where \(F\) is perpendicular to \(A\), we have \(F=P A\). Assuming that \(2.0 \mathrm{~cm}^{2}\) of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), $$ F=P A=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.0 \times 10^{-4} \mathrm{~m}^{2}\right)=20 \mathrm{~N} $$

The compressibility of water is \(5.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}\). Find the decrease in volume of \(100 \mathrm{~mL}\) of water when subjected to a pressure of 15 MPa.

What volume does 300 g of mercury occupy? The density of mercury is 13600 . From \(\rho=m / v\) $$ V=\frac{m}{\rho}=\frac{0.300 \mathrm{~kg}}{13600 \mathrm{~kg} / \mathrm{m}^{3}}=2.21 \times 10^{-5} \mathrm{~m}^{3}=22.1 \mathrm{~cm}^{3} $$

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