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Topic: Another gold ingot problem  (Read 6639 times)

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afchick7689

  • Guest
Another gold ingot problem
« on: September 08, 2005, 01:04:46 PM »
The problem reads:

A copper refinery produces a copper ingot weighing 150 lb.  If the copper is drawn into wire whose diameter is 8.25 mm, how many feet of copper can be obtained from the ingot? The density of copper is 8.94 g/cm3.

So i think i have the process correct, but my answer is not matching up with the answer given in the back of the book. The book says the answer should be 467 feet.  Here's my work and answer-- if someone could please tell me where my error is...


150 lb | 453.59g = 68038.5 g
          | 1 lb
68038.5 g / 8.94 g/mL = 7610.57 mL of copper
pi*r*r*h= volume of a wire
8.25 mm| .1 cm = .825 cm
             | 1 mm
7610.57 mL = pi * .825 cm * .825 cm * h
h =3559.26 cm
3559.26 cm | 1 in       | 1 ft   = 116.79 feet
                 | 2.54 cm  | 12 in

afchick7689

  • Guest
Re:Another gold ingot problem
« Reply #1 on: September 08, 2005, 01:12:53 PM »
never mind  :), i found my own error

( i was doing diameter squared instead of radius squared :P)

thanks anyway!

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