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Melting and Latent Heat

  1. Mar 30, 2006 #1
    Am I on the right track here?


    How much energy is required to melt 400 grams of lead, if the initial temperature is 25 oC?

    m=0.4kg deltaT 328-25= 303 SpecHeat = 128


    Q=mCdeltaT

    Q=.4*128*303

    15513.6J
     
  2. jcsd
  3. Mar 30, 2006 #2

    Doc Al

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    Staff: Mentor

    To melt that lead mass you'll need to do two things:
    (1) Increase the temperature to the melting point of lead. (Hint: specific heat)
    (2) Melt the lead. (Hint: latent heat of fusion)​
    Find the energy for each step separately and add them.
     
  4. Mar 30, 2006 #3

    OK so I have the step 1 part OK????

    How do I set up part two? Latent heat (fusion) requires a seperate equation presumably, I mean different to the Specific Heat in part 1?

    So how does this look?

    Heat Energy = m X L + m X c X q
     
    Last edited: Mar 30, 2006
  5. Mar 30, 2006 #4

    Doc Al

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    Staff: Mentor

    Looks OK.

    Heat required to melt a substance = mass X latent heat of fusion

    I assume that is meant to represent the energy of both steps? If so, the first term is correct for step 2, but the second term has a q where it should have a delta t. I'll rewrite it:
    [tex]Q = Q_1 + Q_2 = m c \Delta t + m L [/tex]
     
  6. Mar 30, 2006 #5
    m=0.4kg deltaT 328-25= 303 SpecHeat = 128

    (1) Increase the temperature to the melting point of lead.


    Q=mCdeltaT

    Q=.4*128*303

    Energy Required to heat: 15513.6J





    (2) Melt the lead. (latent heat of fusion)

    L=0.25x10^5 Taken for a list of properties where lead is listed as 0.25x10^5 for Melt ie:
    Lt(J/kg)


    Heat Energy = m * L = 10,000J

    Add the energy for each step 10,000+15514=25514J

    Answer: 25514J
     
  7. Mar 30, 2006 #6

    Doc Al

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    Staff: Mentor

    Looks reasonable to me. (Be sure to round off your answer to a sensible number of significant figures.)
     
  8. Mar 30, 2006 #7

    OK thanks for your help.
     
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