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A copper refinery produces a copper ingot weighing \(150 \mathrm{lb}\). If the copper is drawn into wire whose diameter is \(7.50 \mathrm{~mm}\), how many feet of copper can be obtained from the ingot? The density of copper is \(8.94 \mathrm{~g} / \mathrm{cm}^{3}\). (Assume that the wire is a cylinder whose volume \(V=\pi r^{2} h,\) where \(r\) is its radius and \(h\) is its height or length.)

Short Answer

Expert verified
Approximately 707.26 feet of copper wire can be obtained from the 150 lb copper ingot.

Step by step solution

01

Convert weight of the copper ingot to grams

First, we should convert the weight of the copper ingot from pounds (lb) to grams (g) as the density of copper is given in grams per cubic centimeter (g/cm³). We know that 1 lb is equal to 453.592 g, so we can use this conversion factor: \(150 \mathrm{lb} \times \frac{453.592 \mathrm{g}}{1 \mathrm{lb}} = 68038.8 \mathrm{g}\) The copper ingot weighs 68038.8 g.
02

Calculate the volume of the copper ingot

Now, we should use the density formula to calculate the volume of the copper ingot: \(\rho = \frac{m}{V}\) Where \(\rho\) is the density, \(m\) is the mass, and \(V\) is the volume. We can rearrange the formula to solve for \(V\): \(V = \frac{m}{\rho}\) Inserting the given information, we get: \(V = \frac{68038.8 \mathrm{g}}{8.94 \mathrm{\frac{g}{cm^3}}}\) \(V = 7610.49175 \mathrm{cm^3}\) The volume of the copper ingot is 7610.49175 cm³.
03

Determine the radius of the copper wire

Next step is to determine the radius of the copper wire whose diameter is given as 7.50 mm. Recall that radius is half of the diameter, so: \(r = \frac{d}{2}\) \(r = \frac{7.50 \mathrm{mm}}{2}\) \(r = 3.75 \mathrm{mm}\) Now, we should convert the radius from millimeters to centimeters: \(3.75 \mathrm{mm} \times \frac{1 \mathrm{cm}}{10 \mathrm{mm}} = 0.375 \mathrm{cm}\) The radius of the copper wire is 0.375 cm.
04

Find the length (height) of the copper wire

The volume formula for a cylinder is given as \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height (length in our case). We should use this formula to find the length of the copper wire that can be obtained from the ingot. We have the values of the volume (\(V\)) and radius (\(r\)), so we can rearrange the formula to solve for the height (\(h\)): \(h = \frac{V}{\pi r^2}\) Substitute the values in the formula: \(h = \frac{7610.49175 \mathrm{cm^3}}{\pi (0.375 \mathrm{cm})^2}\) \(h = 21572.6679 \mathrm{cm}\) Now, we should convert the height (length) from centimeters to feet. Since 1 cm is equal to 0.0328084 ft, we can use this conversion factor: \(21572.6679 \mathrm{cm} \times \frac{1 \mathrm{ft}}{30.48 \mathrm{cm}} = 707.25834 \mathrm{ft}\) So, approximately 707.26 feet of copper wire can be obtained from the 150 lb copper ingot.

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Key Concepts

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

Density of Copper
Understanding the density of copper is crucial when working with this versatile metal. Density is a measure of how much mass a material has in a given volume. The density of copper is commonly listed as 8.94 g/cm³. This value helps to determine how heavy or light an object made of copper will be based on its size.

When solving problems involving copper's density, remember that the unit g/cm³ indicates that density is a ratio of grams to cubic centimeters. This is an essential step for calculating the volume of copper objects when their mass is known. For instance, to find the volume of a copper ingot, you would divide its mass by the density of copper to determine how much space that ingot occupies. Accurate knowledge of copper's density is also essential in industries such as construction and electrical wiring, where weight and volume constraints are often critical.
Volume of a Cylinder
The volume of a cylinder can be found using the formula \(V = \rho r^{2} h\), where \(V\) is volume, \(r\) is the radius of the circular base, and \(h\) is the height of the cylinder. In practical terms, for a copper wire, which is essentially a long, thin cylinder, this formula lets us determine how much copper is present or how long a wire we can create from a certain quantity of the metal.

To find the volume of copper wire, the diameter is often provided, and it must be halved to obtain the radius, a necessary component of the volume formula. Calculating the volume of a cylinder is a fundamental task in many areas of study and professions, such as engineering, architecture, and manufacturing, and is essential in understanding how much space cylindrical objects occupy.
Unit Conversion
Unit conversion is a process that allows us to express a quantity in different units without changing its value. It's vital when dealing with measurements, as frequently, different systems or scales are in use. For example, converting from pounds to grams or from centimeters to feet is necessary to solve the problem involving the length of a copper wire drawn from an ingot.

Important conversion factors include knowing that there are 453.592 grams in a pound and 2.54 centimeters in an inch (or 30.48 centimeters in a foot). When we need to calculate something as specific as the length of a wire, these conversions ensure that all measurements are in consistent units, enabling accurate calculations. This skill is not only important in academic settings but also in everyday life, especially for those working in fields that involve international trade, science, engineering, and construction.

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Most popular questions from this chapter

(a) A sample of tetrachloroethylene, a liquid used in dry cleaning that is being phased out because of its potential to cause cancer, has a mass of \(40.55 \mathrm{~g}\) and a volume of \(25.0 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\). What is its density at this temperature? Will tetrachloroethylene float on water? (Materials that are less dense than water will float.) (b) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "supercritical fluid" that is a much safer drycleaning agent than tetrachloroethylene. At a certain pressure, the density of supercritical \(\mathrm{CO}_{2}\) is \(0.469 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the mass of a \(25.0-\mathrm{mL}\) sample of supercritical \(\mathrm{CO}_{2}\) at this pressure?

(a) A cube of osmium metal \(1.500 \mathrm{~cm}\) on a side has a mass of \(76.31 \mathrm{~g}\) at \(25^{\circ} \mathrm{C}\). What is its density in \(\mathrm{g} / \mathrm{cm}^{3}\) at this temperature? (b) The density of titanium metal is \(4.51 \mathrm{~g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\). What mass of titanium displaces \(125.0 \mathrm{~mL}\) of water at \(25^{\circ} \mathrm{C} ?\) (c) The density of benzene at \(15^{\circ} \mathrm{C}\) is \(0.8787 \mathrm{~g} / \mathrm{mL} .\) Calculate the mass of \(0.1500 \mathrm{~L}\) of benzene at this temperature.

In 2009 , a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is \(0.20 \mathrm{~g} / \mathrm{cm}^{3}\), and its surface area is \(1242 \mathrm{~m}^{2}\) per gram of material. (a) Calculate the volume of a 10.0 -mg sample of this material. (b) Calculate the surface area for a 10.0 \(\mathrm{mg}\) sample of this material. (c) A \(10.0-\mathrm{mL}\) sample of contaminated water had \(7.748 \mathrm{mg}\) of mercury in it. After treatment with \(10.0 \mathrm{mg}\) of the new spongy material, \(0.001 \mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?

Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry. (a) Consider a piece of gold jewelry that weighs \(9.85 \mathrm{~g}\) and has a volume of \(0.675 \mathrm{~cm}^{3} .\) The jewelry contains only gold and silver, which have densities of \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\) and \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. If the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains, calculate the percentage of gold (by mass) in the jewelry. (b) The relative amount of gold in an alloy is commonly expressed in units of carats. Pure gold is 24 carat, and the percentage of gold in an alloy is given as a percentage of this value. For example, an alloy that is \(50 \%\) gold is 12 carat. State the purity of the gold jewelry in carats.

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains \(1.50 \mathrm{~g}\) of carbon and \(2.00 \mathrm{~g}\) of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains \(6.35 \mathrm{~g}\) of carbon. How many grams of oxygen does it contain? Which law are you assuming in answering this question?

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