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You purchase a rectangular piece of metal that has dimensions 5.0 \(\times\) 15.0 \(\times\) 30.0 mm and mass 0.0158 kg. The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated?

Short Answer

Expert verified
The density is 7.02 g/cm³, which is not gold. You were cheated.

Step by step solution

01

Calculate the volume of the rectangular piece

To find the volume of a rectangular solid, use the formula: \( V = l \times w \times h \). Here, \( l = 5.0 \) mm, \( w = 15.0 \) mm, and \( h = 30.0 \) mm. So, \( V = 5.0 \times 15.0 \times 30.0 = 2250 \) cubic millimeters.
02

Convert the volume from cubic millimeters to cubic centimeters

Since density is typically expressed in \(\text{g/cm}^3\), convert the volume from \(\text{mm}^3\) to \(\text{cm}^3\). There are 1000 cubic millimeters in one cubic centimeter. Thus, \(2250 \text{ mm}^3\) is equal to \(2250 / 1000 = 2.25 \text{ cm}^3\).
03

Calculate the density of the metal piece

Density is calculated as mass divided by volume. The mass of the metal is 0.0158 kg, which is 15.8 grams. Using the volume from Step 2, the density is \(\frac{15.8 \text{ g}}{2.25 \text{ cm}^3} \approx 7.02 \text{ g/cm}^3\).
04

Compare calculated density with the density of gold

The density of gold is \(19.32 \text{ g/cm}^3\). Compare this with the calculated density of \(7.02 \text{ g/cm}^3\). Clearly, they do not match.

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Key Concepts

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

Volume Conversion
When dealing with different measurement units, converting the volume from one unit to another is often necessary. In this problem, we originally calculated the volume of a rectangular solid in cubic millimeters (\(\text{mm}^3\)). However, to properly evaluate density, we need our volume in cubic centimeters (\(\text{cm}^3\)).

To convert between these units, remember the conversion factor: 1 cubic centimeter equals 1000 cubic millimeters. Thus, if you have a volume in \(\text{mm}^3\), simply divide by 1000 to get \(\text{cm}^3\). This is because converting from a smaller unit to a larger one typically involves division.

In our example, the conversion from 2250 \(\text{mm}^3\) to cubic centimeters is done by calculating \(2250 / 1000 = 2.25 \text{ cm}^3\). Conversion allows you to easily compare or use other standard measurements applicable to your problem.
Density Comparison
Density provides a way to identify or verify a material by comparing its theoretical density against known values. It's calculated as \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Here, we convert the mass from kilograms to grams first because density is often expressed in \(\text{g/cm}^3\). Since 1 kg is 1000 grams, the mass of 0.0158 kg converts to 15.8 grams.

With our volume previously converted to \(2.25 \text{ cm}^3\), the density will be \(\frac{15.8 \text{ g}}{2.25 \text{ cm}^3} \approx 7.02 \text{ g/cm}^3\). You can use this with known densities to verify what a material might be.

For example, since gold has a density of \(19.32 \text{ g/cm}^3\), a density much higher than \(7.02 \text{ g/cm}^3\) indicates that the metal in the exercise is not gold. This significant difference confirms you were not sold pure gold.
Rectangular Solid Volume Calculation
Calculating the volume of a rectangular solid is straightforward by using the formula given: \(V = l \times w \times h\). This means you multiply the length (\(l\)), width (\(w\)), and height (\(h\)) of the rectangular solid to obtain the volume in cubic units.

In our situation, the dimensions provided are in millimeters: length = \(5.0 \text{ mm}\), width = \(15.0 \text{ mm}\), and height = \(30.0 \text{ mm}\). Thus, the calculation would be \(5.0 \times 15.0 \times 30.0 = 2250 \text{ mm}^3\).

This volume calculation is crucial for further computations, such as determining density, which inherently relies on accurate measurement of the solid's dimensions. Always ensure measurements are in the correct units and appropriately converted if necessary.

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