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Homework Help: Specific Heat molten lead Question

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    If you pour 0.50 kg of molten lead at 328°C into 2.5 liters of water at 20°C, what will be the final temperatures of the water and the lead? The specific heat of (solid) lead has an average value of 3.4 X 10-2 kcal/kg • °C over the relevant temperature range.


    2. Relevant equations
    Q=mass*c* [tex]\Delta[/tex] T
    Qfus=m*cfus
    masspb=0.50 Kg @328oC
    cpb=34 cal/KgoC
    Tfinal=?
    cfusion=6800 cal
    3. The attempt at a solution

    Qfusion=6800 cal/Kg * 0.50 Kg= 3400 cal
    Qpb = 3400 cal + .50Kg * 34 cal/KgoC* (328-x)
    Qpb=3400+5576-17x

    For Water:
    QH2O=2.5 Kg * 1000 cal/KgoC * (x-20oC)
    QH2O=2500x-50000

    Setting these equal to each other I get:

    2500x-50000=3400-17x+5576
    2517x=58976 cal
    x=23.4311oC

    The back of the book has the answer of 22oC.

    I can get 22oC as an answer only if I subtract the latent heat of fusion:

    2500x-50000=17x+5576
    2517x=55576
    x=22.0803oC

    The math seems pretty straight forward, so it makes me wonder whether the book made an error, or I did.

    Any help would be appreciated! Thanks.
     
  2. jcsd
  3. Jan 24, 2010 #2

    ideasrule

    User Avatar
    Homework Helper

    I think your answer is right. There's no reason to subtract the heat of fusion.
     
  4. Jan 24, 2010 #3
    Thanks for the reply. This is the second time I've seen errors in the book, that's why I needed to make sure I hadn't made an error.
     
  5. Jan 25, 2010 #4

    Borek

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    Staff: Mentor

    I would not say "subtract", I would say "ignore". But it has to be taken into account.
     
  6. Dec 12, 2011 #5
    I realize this is a year old. But I just did the same problem but my book says the specific heat of lead is 140 J/kg*°C. Just in case somebody else reads this.
     
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