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Energy lost in conversion of steam to ice

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data

    How many J of energy must be removed when 145.0 g of steam, at a temperature of 188.0°C, is cooled and frozen into 145.0 g of ice at 0°C? Take the specific heat of steam to be 2.1 kJ/(kg·K).


    2. Relevant equations

    Q=m*L

    3. The attempt at a solution

    Energy = Mgas:liquid*Lvaporization + Mliquid:solid * Lfusion

    .145*22.6E5 + .145*33.5E4 = 3.76E5 J
     
  2. jcsd
  3. Nov 16, 2007 #2

    mgb_phys

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    You also need to account for the energy to cool the steam from 188C to and then cool the water from 100C to 0C.
    Use Q = m*c*T
     
  4. Nov 16, 2007 #3
    hm... it's still not working.

    I added .145(2.1)(88) + .145*4186*100 = 60723.769 J

    The total I found was 4.37E5 J.
     
  5. Nov 16, 2007 #4

    mgb_phys

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    cool steam = 0.145 * (188-100) * 2100
    condense steam = 0.145 * 2272E3
    cool water = 0.145 * (100-0) * 4181
    freeze water = 0.145 * 334E3

    Just a small typo, the heat capcity of steam is 2.1KJ/kg = 2100 J/kg
     
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