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Specific heat capacity, latent heat => ice water steam mixture

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A quantity of 100g of ice at 0°C and 50g steam at 100°C are added to a container that has 150g water at 30°C. What is the final temp of the container? Ignore the container itself in the calc.


    2. Relevant equations
    Q=mc (change in T)
    Q=mL


    3. The attempt at a solution
    I know how to figure out this kind of problem with only 2 inputs, but with three i struggle with a good approach.
    Can I just calculate the final temp for the water and the ice and then do the same for the resulting water temp and mass with the steam?
    MOST IMPORTANTLY: what happens to the energy of the steam, when it becomes water? does it add to the system ?


    regards
     
  2. jcsd
  3. Aug 18, 2009 #2

    kuruman

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    With three inputs it is the same thing. Some items see their temperature drop some see their temperature rise, but (and this is very important) no heat leaves the system. Now Q is positive if the temperature of an item rises and negative if it decreases. Then you can say that the sum of all Q changes must be zero.

    0 = m1c1ΔT1 + m2c2ΔT2+m3c3ΔT3

    You can put together as many items as you want this way in one equation.

    The energy of the steam stays in and contributes to the rise in the temperature of the other items. Of course you need to remember to add appropriate terms for the heat of fusion and heat of vaporization. You can write one equation taking everything into account.
     
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