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Homework Help: Question on Thermal Physics. [it feels impossibly difficult to me]

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    This question has got me REALLY confused. I'll be thankful for any help. I don't even know where to start, though I'll mention what my thoughts were.

    Q. A jeweller wishes to harden a sample of pure gold by mixing it with some silver so that the mixture contains 5.0% silver by weight the jeweller melts some pure gold and then adds the correct weight of silver. the initial temperature of the silver is 27 °C. Use the data given to calculate the initial temperature of the pure gold so that the final mixture is at the melting point of pure gold.


    Melting point for gold, 1340 K.

    Melting point for silver, 1240 K.

    Specific heat capacity for gold (solid or liquid), 129 J kg-1 K-1.

    Specific heat capacity for silver (solid or liquid), 235 J kg-1 K-1.

    Specific latent heat of fusion for gold, 628 kJ kg-1.

    Specific latent heat of fusion for silver, 105 kJ kg-1.

    2. Relevant equations

    The only two equations I suppose I know I can use here are:

    E=mcΔθ and E=mL

    3. The attempt at a solution

    The first and last thought that came to my mind and that I wrote down while I was trying to answer the question was, 'Assuming the gold weights 9.5kg and the silver weighs 0.5kg.'

    And after that I was stuck.

    Where do I start from? I'd love to be able to solve this myself but unfortunately I don't know what to do with the data or how to use use it.

    Could someone please help?

    Edit: I have the answer if anyone needs it to confirm their solution.
  2. jcsd
  3. Apr 20, 2012 #2
    OK, so you want to find some [itex]T_{0g}[/itex] (g stands for gold) so that [itex]T_f = 1340[/itex] K. In this problem you have the end state and you have to work backwards to get to the beginning. See if you can work out what the initial gold temperature had to have been, to mix with the silver in the right way to get to the right end state.

    Since you don't have the actual mass in kg, you have to pretend you do for a little while, and usually when it's not given to you in a problem it ends up not mattering. So just let [itex]M = m_g + m_s[/itex] such that [itex]m_g = 0.95 M[/itex] and [itex]m_s = 0.05 M[/itex], where M is the mass of the total mixture.
  4. Apr 20, 2012 #3
    Let me give it a spin and get back to you on that. Thanks!
  5. Apr 20, 2012 #4
    Umm, this is kind of embarrassing [then again a person should never be embarrassed to admit their own weakness], but, the ONLY thing I've been able to think of is that the initial temperature of the silver is 300 K, and that that has something to do with the melting point of gold is 1340 K and these two are connected somehow.

    And if I use E=mcΔθ, I don't know how to use it for each metal. I mean what temperatures do I put in? AND then how I'm supposed to use the two values for energy. My mind is forming absolutely no links.
  6. Apr 20, 2012 #5
    Hint: Figure out how much energy is needed to bring silver from 300K to 1340K. The energy requirement is based on mass but you know the mass is 5% that of the gold.
    The energy needed comes from gold being above its melting point.................so.......
  7. Apr 20, 2012 #6
    So the heat lost by the gold should be 0.95m x 129 x Δθ right? And the heat lost by the gold is the heat gained by the silver. [at least that's what I could make out from the hint]

    Once more, I'm stuck. So I've got a formula for the heat lost by the gold. What should I be doing with that or after that?

    And sure the gold is above it's melting point, but I don't see how that's helping me in finding out by how MUCH the gold is above it's melting point.
  8. Apr 20, 2012 #7
    You have not determined this yet.
  9. Apr 20, 2012 #8
    Isn't that supposed to be 0.05m x 235 x 1040? It can't be anything else can it?
  10. Apr 20, 2012 #9
    Yes it can and is. You have silver at 300K. It must go to 1240 K. That takes energy.
    Then it must melt. That takes more energy. Then it must go from 1240 to 1340. That takes energy.
  11. Apr 20, 2012 #10
    So the heat gained by the silver is supposed to be...

    0.05m x 235 x 900 + 0.05m x 105000 (keeping everything in Joules) + 0.05m x 235 x 100?

    [by the way I had absolutely NO idea about using the latent heat. I thought they'd put it in the question just for confusion's sake.]

    And all this energy comes from the gold? Am I right in assuming that?
  12. Apr 20, 2012 #11
    Temperature rise is 940 not 900 K.

    Yes, all the energy comes from the gold. So figure out how hot it needs to be to do this so it' final temperature is 1340 K.
    The latent heat of the gold is extraneous.
  13. Apr 20, 2012 #12
    Oh my. This is magic. [well actually it's logic but the way it all just worked out was magic!]

    Thank you!

    [I equalled the gain of the silver to the loss of the gold and got a value for temperature from there and I added that to the melting point of the gold. I hope that's correct.]
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