How Many Meters of 34-Gauge Wire Can Be Made from 5.01 lb of Covellite?

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of how many meters of 34-gauge wire can be produced from 5.01 lb of covellite, which contains copper. Participants explore the necessary steps to solve the problem, including the application of volume formulas and the conversion of mass to volume based on density.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant outlines the initial problem, noting the mass of covellite and its copper content, and presents the formula for the volume of a cylinder.
  • Another participant suggests calculating the mass of copper from the covellite and then using the density of copper to find the volume available for wire production.
  • A participant expresses confusion regarding the variable used in the volume formula, mistaking it for height instead of length, and later clarifies their understanding with the help of another participant.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem, but there is a moment of confusion regarding the interpretation of variables in the volume formula. The discussion does not indicate any unresolved disagreements regarding the method itself.

Contextual Notes

Participants rely on specific assumptions about the density of copper and the percentage of copper in covellite, which may not be universally applicable without additional context. The interpretation of variables in the volume formula also highlights potential misunderstandings that could affect problem-solving.

Who May Find This Useful

This discussion may be useful for students working on similar homework problems involving material properties, volume calculations, and conversions between mass and volume in the context of physics or engineering.

metalmagik
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Homework Statement


Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10^-3 in) can be produced from the copper in 5.01 lb of covellite, an ore of copper that is 66% copper by mass? (Hint: Treat the wire as a cylinder. d of copper = 8.95 g/cm^3)


Homework Equations


V = (pi)(r^2)(h)


The Attempt at a Solution


So I used the formula for Volume of a cylinder, V = (pi)(r^2)(h) and plugged in half of the given diameter for r and then the whole diameter for h. And I got 1.968 x 10^-7 cubic inches. How should I go about doing the rest of the problem? It's really confusing me. Help please!
 
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Think of it this way. You are given a certain mass of covellite (5.01 lb = 2273 grams), of which 66% is copper (1500 grams). You know the density of copper, so you can calculate the volume of copper available, since

d = \frac{m}{V}

Once you calculate the volume, you want to form the copper into the shape of a cylinder (wire), and the volume of the cylinder = volume of copper available. For a cylinder, the volume can be written

V = pi r^{2} L

where L is the length we wish to find. Take V from the first part and solve this equation for L.
 
Ah man I totally get it now. I wasn't thinking of the variable in the Volume formula as length, my book labels it as h, which I interpreted for height.

Thanks so much edziura! That really helped me to clarify my thinking, all this stuff with mass is so difficult, its all over my homework which is due tomorrow, and my professor hasn't even gotten past dimensional analysis in lecture! Thanks again.
 
Your welcome.
 

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