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Calorimetry homework: mixing steam and liquid water

  1. Jun 5, 2007 #1
    1. The problem statement, all variables and given/known data
    What mass of steam at 100 degree celsius must be passed into 5.4 kg of water at 30 degree celsius to raise temperature of water to 80 degree celsius?


    2. Relevant equations

    C(water)=4.2 J/(g C)
    L(steam)=2268 J/g

    3. The attempt at a solution
    Assuming no heat loss,
    Heat gained by water=Heat given by steam by changing to water+ heat released by water so formed to fall from 100 degree to 80 degree

    5.4*50*4200=m*2268000+m*4200*20
    1134000=m(2352000)
    m=0.482142857 Kg

    My book gives me the answer as 0.5 Kg. It neglects the heat given by water formed from steam, it seems. why?????
     
    Last edited: Jun 5, 2007
  2. jcsd
  3. Jun 5, 2007 #2
    What does this mean?
     
  4. Jun 5, 2007 #3
    I mean that the water formed from condensation of steam will also provide heat to the water at 30 degree, isnt it?
     
  5. Jun 5, 2007 #4
    Yes, youre right... but does all of the steam need to be condensed? If it doesnt, then only steam will give the heat first because it has more energy....
     
  6. Jun 5, 2007 #5
    What do you think that the question wants me to do?
     
  7. Jun 6, 2007 #6
    That is latent heat which has already been included in your solution.
     
  8. Jun 6, 2007 #7
    No that isnt laten heat. Somebody help!!:surprised
     
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