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Chapter 1: Problem 82

A copper wire (density \(=8.96 \mathrm{g} / \mathrm{cm}^{3} )\) has a diameter of 0.25 \(\mathrm{mm}\) . If a sample of this copper wire has a mass of 22 \(\mathrm{g}\) , how long is the wire?

Short Answer

Expert verified
The length of the copper wire is approximately \(2077.44 \textrm{ cm}\).

Step by step solution

01

Calculate the volume of the copper wire

To find the volume of the copper wire, we can use the formula: Volume \(= \frac{\text{mass}}{\text{density}}\) We are given the mass of the wire as 22 g and the density of copper as \(8.96 \textrm{ g/cm}^3\). Plugging in the values, we get: Volume \(= \frac{22 \textrm{ g}}{8.96 \, \textrm{g/cm}^3}\)
02

Calculate the cross-sectional area of the wire

The copper wire has a circular cross-section, and we are given its diameter as 0.25 mm. To calculate the cross-sectional area, we first need to find the wire's radius. The radius of the wire is half of the diameter: Radius \(= \frac{\text{diameter}}{2} = \frac{0.25 \textrm{mm}}{2} = 0.125 \textrm{mm}\) Now, to find the cross-sectional area of the wire, we use the formula for the area of a circle: Area \(= \pi (\text{radius})^2\) Plugging in the value of the radius, we get: Area \(= \pi (0.125 \textrm{mm})^2\) It is important to convert the radius to cm before proceeding: Radius \(= 0.0125 \textrm{ cm}\) Area \(= \pi (0.0125 \textrm{cm})^2\)
03

Calculate the length of the wire

Now that we have the volume and cross-sectional area of the wire, we can find its length using the formula: Length \(= \frac{\text{volume}}{\text{area}}\) Plugging in the values, we get: Length \(= \frac{\frac{22}{8.96}}{\pi (0.0125)^2}\) Now, solve for the length: Length \(= \frac{22}{8.96 \times \pi (0.0125)^2}\textrm{ cm}\) Length \(= 2077.44 \textrm{ cm}\) So, the length of the copper wire is approximately 2077.44 cm.

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Key Concepts

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

Volume Calculation
To solve for the volume of a copper wire, you need to use a very simple formula involving mass and density. The formula is:
  • Volume = \( \frac{\text{mass}}{\text{density}} \)
In this particular problem, the mass of the copper wire is given as 22 grams, and its density is provided as \(8.96 \, \text{g/cm}^3\). By substituting the given values into the formula:
  • Volume = \( \frac{22 \, \text{g}}{8.96 \, \text{g/cm}^3} \)
When you calculate, it results in:
  • Volume 鈮� 2.455 \( \text{cm}^3 \)
The volume signifies the amount of space the copper wire occupies. Calculating the volume is crucial because it allows us to find out other dimensions like the length when the cross-sectional area is known.
Cross-Sectional Area
Understanding the cross-sectional area of a copper wire involves some basic geometry. Since a wire is generally cylindrical, you can imagine its cross-section as a circle. The formula to calculate the circular area is:
  • Area = \( \pi (\text{radius})^2 \)
First, obtain the radius of the wire. Given that the diameter is 0.25 mm, calculate the radius:
  • Radius = \( \frac{0.25 \, \text{mm}}{2} = 0.125 \, \text{mm} \)
It's important to convert millimeters to centimeters for consistency in units:
  • Radius = 0.0125 cm
Now, plug this value into the area formula:
  • Area = \( \pi (0.0125 \, \text{cm})^2 \)
  • Area 鈮� 0.0004909 \( \text{cm}^2 \)
The cross-sectional area helps determine how much space the copper wire uses in a plane perpendicular to its length, which is necessary for calculating its length.
Wire Length Calculation
Now, with the volume and cross-sectional area known, the length of the copper wire can be found using another simple formula:
  • Length = \( \frac{\text{volume}}{\text{area}} \)
Using the calculations from earlier:
  • Volume = 2.455 \( \text{cm}^3 \)
  • Area 鈮� 0.0004909 \( \text{cm}^2 \)
You substitute these values into the formula to find length:
  • Length = \( \frac{2.455}{0.0004909} \)
This results in:
  • Length 鈮� 5000 cm
The length calculation tells you how far the copper wire would stretch out if it were laid in a straight line. This step demonstrates the application of geometry and physics in solving practical problems related to physical dimensions.

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Most popular questions from this chapter

To determine the volume of a cube, a student measured one of the dimensions of the cube several times. If the true dimension of the cube is 10.62 cm, give an example of four sets of measurements that would illustrate the following. a. imprecise and inaccurate data b. precise but inaccurate data c. precise and accurate data Give a possible explanation as to why data can be imprecise or inaccurate. What is wrong with saying a set of measurements is imprecise but accurate?

A parsec is an astronomical unit of distance where 1 parsec \(=\) 3.26 light years \((1 \text { light year equals the distance traveled by }\) light in one year). If the speed of light is \(186,000 \mathrm{mi} / \mathrm{s}\) , calcu- late the distance in meters of an object that travels 9.6 parsecs.

Diamonds are measured in carats, and 1 carat \(=0.200 \mathrm{g} .\) The density of diamond is 3.51 \(\mathrm{g} / \mathrm{cm}^{3}\) . a. What is the volume of a 5.0 -carat diamond? b. What is the mass in carats of a diamond measuring 2.8 \(\mathrm{mL} ?\)

Precious metals and gems are measured in troy weights in the English system: $$\begin{aligned} 24 \text { grains } &=1 \text { pennyweight (exact) } \\ 20 \text { pennyweight } &=1 \text { troy ounce (exact) } \\ 12 \text { troy ounces } &=1 \text { troy pound (exact) } \\ 1 \operatorname{grain} &=0.0648 \mathrm{g} \\ 1 \text { carat } &=0.200 \mathrm{g} \end{aligned}$$ a. The most common English unit of mass is the pound avoirdupois. What is 1 troy pound in kilograms and in pounds? b. What is the mass of a troy ounce of gold in grams and in carats? c. The density of gold is 19.3 \(\mathrm{g} / \mathrm{cm}^{3} .\) What is the volume of a troy pound of gold?

Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound d. element e. physical change f. chemical change
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