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Determine how long it takes to cover both sides of the plate

  1. May 9, 2015 #1
    A square plate has a side of 100mm. It is connected to the cathode in a copper electroplating bath and a current of 4A is passed through the bath. Neglecting the thickness of the plate, determine how long it takes to cover both sides of the plate with copper to thickness of 0.1mm

    Density of copper to be 9000kg/m^3
    The electrochemical equivalent of copper to be 0.330mg/C

    Answer given = 227 min 16 sec

    Volume = area x height
    = 20 000x10^-7 X 0.23x10^-3
    = 4x10^-7

    Mass = volume x density
    = 4x10^-7 X 9000kg/m^3
    = 3.6x10^-3kg = 3600mg

    m = zIt
    t = m/zI
    = 3600/(0.330)(4)
    = 43636.36364s = 727 min and 16 sec

    I am quite confident in my answer but it is way off from the supplied answer - unless that was a typo.
    Any input would be appreciated.

    Thanks,
    Devon
     
  2. jcsd
  3. May 9, 2015 #2

    ehild

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    Where are the red numbers from?
     
  4. May 9, 2015 #3

    mfb

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    Where do those numbers come from and what are their units?

    Edit: ehild beat me.
     
  5. May 9, 2015 #4
    square plate has 100mm side, therefore area = 100mmx100mm=10000mm^2 X 2 sides = 20000mm^2 = 20000x10^-7
    copper thickness = 0.1mm each side = 0.2mm = 0.2x10^-3

    Sorry should have been 0.2x10^-3 not 0.23x10^-3
     
  6. May 9, 2015 #5

    mfb

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    1mm^2 = 10-7? An area is a number?
    If that is supposed to be square meters, the number is wrong.

    I can confirm the given answer.
     
    Last edited: May 9, 2015
  7. May 9, 2015 #6

    ehild

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    You counted the two sides twice. And 1 mm2=10-6 m2.
     
  8. May 9, 2015 #7
    What is 20000mm^2 in m^2?
     
  9. May 9, 2015 #8
    How did I count them twice? 1 side = 100mmx100mm = 10 000mm2 so 2 sides = 20 000mm2
     
  10. May 9, 2015 #9

    mfb

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    1mm = 10-3m, square both sides, or see post 6.

    To get closer to the given answer, you might also want to use a more precise density estimate for copper, but that just changes the result by about one minute.
     
  11. May 9, 2015 #10

    ehild

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    After that, you said that 0.1 mm layer on both sides, so it is 0.2 mm on both sides.
     
  12. May 9, 2015 #11
    Alright so 20 000mm2 = 20 000 x10-6?
    If I use that number I get an answer of 7272 min
    The density was given as 9000kg/m3
     
  13. May 9, 2015 #12
    I see what you mean, thank you.

    volume = 20 000x10-7 X 0.1x10-3
    = 2x10-7

    mass = 2x10-7 X 9000kg/m3
    = 1.8x103kg = 1800mg

    t = m/zI
    = 1800/(0.33)(4)
    = 1363.636364s = 22 min 43 seconds
     
  14. May 9, 2015 #13

    mfb

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    Now you used the wrong 10-7 again. Apart from that it is right.
     
  15. May 10, 2015 #14
    volume = 20 000x10-6 X 0.1x10-3
    = 2x10-6

    mass = 2x10-6 X 9000kg/m3
    = 0.018kg = 18000mg

    t = m/zI
    = 18000/(0.33)(4)
    = 13636.36364s = 227 min 16 seconds

    My goodness finally. Thank you very much mfb and ehild for your help!
     
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