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Homework Help: Latent heat and melting lead

  1. Aug 2, 2008 #1
    1. The problem statement, all variables and given/known data
    A sculptor has come to you with a problem, because she knows that you are such a good physics student. She wishes to make a lead casting of an object which will require 2 kg of melted lead. If the lead was stored in her basement at approximately 55˚F, and it is to be melted in a 1/4" thick steel ladle/pot of approximate cross sectional area 12in^2, how long will it take to melt the lead if it is placed in a kiln at 617.5˚F, the melting point of lead?

    Lf = 0.25x10^5 J/kg
    Lv = 8.7x10^5 J/kg
    c = 130 J/kg*C˚
    k = 40 J/s*m*C˚

    2. Relevant equations
    P = kA(ΔT/d)
    Q = mcΔT
    Q = mL
    P = Q/Δt

    3. The attempt at a solution
    12in^2 = 0.0077 m^2
    55˚F = 12.7˚C
    617.5˚F = 325.3˚C
    0.25 in = 0.00635 m

    P = kA(ΔT/d)
    P = (40 J/s*m*C˚)(0.0077m^2)((325.3˚C-12.7˚C) / 0.00635 m)
    P = 15162.3

    Q1 = mcΔT
    Q1 = (2 kg)(130 J/kg*˚C)(325.3˚C-12.7˚C)
    Q1 = 81276 J

    Q2 = mL
    Q2 = (2 kg)(0.25x10^5 J/kg)
    Q2 = 50000 J

    Q1+Q2 = 81276 J + 50000 J = 131,276 J

    P = Q/Δt
    15162.3 = (131,276 J)t
    t = 0.1155 s

    So... I REALLY don't think this is correct, but I can't seem to think of another way around this. Any help would be appreciated!
  2. jcsd
  3. Aug 3, 2008 #2


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    EDIT: I'm taking back part of what I had here. I guess you need the thermal conductivity because you have nothing else that sets a time scale for the heating. I believe you also need to consider that the steel pot itself must heat up to 325.3º C. It looks like you can ignore the walls of the pot and just find the mass of steel for the section described.

    But look at your algebra again here:

    [tex]P = Q/\Delta t[/tex]
    15162.3 = (131,276 J)t
    t = 0.1155 s

    How did t end up in a product? Wouldn't that be 131,276 / t ?
    Last edited: Aug 3, 2008
  4. Aug 3, 2008 #3
    I'm not sure I entirely understand. Are you saying I need to find heat (Q) for the lead pot as well? And then add that value to the other two heat values (Q1+Q2)? How can I find the heat when I don't have the mass value of lead? Plus... how do I find the mass if you're suggesting I DO need to find heat?

    And as for the equation.... perhaps I've been working on math too long, but I assumed that, by using the equation P=Q/delta-t to find t I need to divide both sides of the equation by Q. So then the equation is t = P/Q... which is what I did.
  5. Aug 3, 2008 #4


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    It's a steel pot, no? You're given the dimensions involved, so you can get a volume; you'll then need to look up a density and heat capacity for steel (or iron will probably do).

    Q is being divided by delta-t, so if you're going to solve P = Q / delta-t for the time delta-t , won't it be Q/P ? (Look at the units: Q is in Joules and P is in Joules/sec .)
  6. Aug 3, 2008 #5
    Ahhh, I see now. Thank you. :)
  7. Aug 3, 2008 #6


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    I think something I need to ask is what level of physics course this problem is from. It seems they want you to use thermal conductivity, but there is a complication. Since the rate of conductive heat transfer depends on the temperature difference between the two sides of the "barrier", the rate of transfer will decrease as the pot heats up to the temperature of the kiln. If you assume the rate of transfer is constant at the value you found, the time required to melt the lead will be an underestimate. If this is an introductory course, though, you may be permitted to simply ignore this.
  8. Aug 3, 2008 #7
    This is a college level introductory physics course; Physics 101
  9. Aug 4, 2008 #8


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    I take it that you had something about heat transfer in your course, since you brought in the expression for the rate of heat conduction. I'm guessing, though, that they aren't looking for an elaborate solution. Is this a problem from a book or from your instructor? I'm asking to see how much they might be expecting you to consider.

    I would say that you probably need to include the heat required to raise the temperature of the steel pot to the temperature of the kiln, along with the lead to be melted, but you can use a constant rate of heat conduction, while stating that such an assumption will lead to an underestimate for the time to melt the lead (since the rate of heat flow will decline as the pot and lead approach the temperature of the kiln).
  10. Aug 5, 2008 #9
    The question was made up by the instructor. It is actually part of the take-home portion of our final, which is exactly why I attempted doing it before posting the question.

    The rest of the final is just as complicated, if not more so, and I can't say I'm entirely certain as to how elaborate a solution he is expecting. This is because the final was not made by MY instructor. My instructor is more of an adjunct professor and teaches Ecology versus Physics except in the summer sometimes.
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