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The weight of gold in a crown

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A jeweler has made a crown of an alloy of gold and copper. The crown weighs 0.358 N when measured in air and 0.338 N when submerged completely in water. The density of water is 1.00x10^3 kg/m^3, the density of copper is 8.92X10^3 kg/m^3, and the density of gold is 1.93x10^4 kg/m^3. Calculate the weight of the gold in the crown.


    2. Relevant equations
    Unknown. None are given and can't find any in my textbook.


    3. The attempt at a solution
    None.

    I can't seem to find any information about how exactly to do this problem in my textbook or online. Help please?
     
  2. jcsd
  3. May 15, 2012 #2

    gneill

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    Staff: Mentor

    If you can't find a specific example or a recipe for how to do the problem, you may have to resort to devising your own method :smile:

    Why not start by determining the volume of the crown and its density?
     
  4. May 16, 2012 #3
    OK, I attempted to find the volume and density. I got:
    mass(in air) as 36.5g
    appearant mass(in water) as 34.5g
    volume = 2.1x10^6 L(?)
    density = 1.83 x 10^4 kg/l

    I am assuming now that I need to find a percentage or something of the sort for the gold in the crown.
     
  5. May 16, 2012 #4

    gneill

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    Staff: Mentor

    Your mass looks fine, although "apparent mass in water" doesn't make physical sense; stick with the apparent weight.

    Your volume looks a bit odd; that's a LOT of liters! How did you arrive at it? Can you demonstrate your calculation?
     
  6. May 16, 2012 #5
    Whoops! I didn't catch that before I posted. The volume I calculated is supposed to be 2.0x10^-6 m^3 (got confused with the unit), not 10^6. and I wrote the apparent mass title based off the equation I used:

    m(in air) - m(apparent/in water) = d(water)*v , solving for v

    also density unit should be kg/m^3
     
    Last edited: May 16, 2012
  7. May 17, 2012 #6

    gneill

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    Okay. That clears up a few things :smile:

    To how many significant figures should your results be reported?
     
  8. May 17, 2012 #7
    I believe I need to use 2 significant digits. So that would make my density 1.8x10^4, right?

    Also, I did some calculations and managed to get what looks like a possibly correct answer. Here's what I did:

    Formula: D(copper)*(c-1)+D(gold)*c=D(crown), where c is the ratio of gold to crown
    Solved for c - c=[D(crown)+D(copper)]/[D(copper)+D(gold)]
    Substituted values - c=[1.8*10^4+8.92*10^3]/[8.92*10^3+1.93*10^4]
    And solved - c=.965

    then i took weight of crown times c and got 3.52*10^-2 kg or .345 N

    I am unsure is this is right or not though, since the amount of gold is so high, but it is my best bet at the moment.
     
  9. May 17, 2012 #8

    gneill

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    Staff: Mentor

    How many significant figures are in the numbers that are given in the problem statement?
    Good idea, and almost correct :smile:

    The fractions of the whole should sum to the whole, so your terms c and c-1 should add up to 1. That is, they should be c and 1-c.
     
  10. May 17, 2012 #9
    Significant Digits = 3

    And the formula should be D(copper)*(1-c)+D(gold)*c=D(crown), right?

    If so, I got 3.52*10^-2 kg or .345 N

    How am I doing?
     
  11. May 17, 2012 #10

    gneill

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    Staff: Mentor

    Your approach looks fine now. I'm a bit concerned with the final value though, it's in the right ballpark but seems a tad high to me. Could it be rounding or truncation error creeping in from intermediate values? Did you keep sufficient extra digits of precision in your intermediate steps?
     
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