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Latent Heat + Specific Heat Problem

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose molten (liquid) lead, mass ml = 9.83 kg, is at its melting point. The lead is poured into water of mass mw = 585 g and initial temperature Tw = 16.2 C. Find mwb, the amount of the water that boils.
    NOTE: Reaching the boiling point is not enough; the question asks for the amount of water that vaporizes as well.
    ASSUME: no water evaporates; it only vaporizes at the boiling point.

    2. Relevant equations
    Constants: M.P Lead = 327 C, MP Water = 100 C
    Latent Heat Lead = 33 kJ/kg
    Latent Heat Q = mL
    Specific Heat Eq'n: Q=mcΔT

    3. The attempt at a solution
    mL + mcΔT = mcΔT + mL
    9.83 kg (33 kJ/kg) + 9.83 kg(0.130 kJ/kg)(327C - 100C) = 0.585(4.186)(100 -16.2) + m(2260 kJ/kg)
    Solving for m, we get m = 0.215 kg.

    Did I go wrong in my reasoning here? Many thanks in advance for your help!
     
  2. jcsd
  3. Dec 16, 2012 #2

    TSny

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    Your equation looks good to me. But I don't get the same numerical answer.
     
  4. Dec 16, 2012 #3
    What do you get when you plug and chug in the values? I worked on the similar type of problem, and I got stuck on it from using that equation.
     
  5. Dec 16, 2012 #4

    TSny

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    I get about 0.181 kg. What do you get for the numerical value of the left side of the equation? What do you get for the simplified form of the right side?
     
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