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Calorimetry: Finding mass of ice, water, steam left

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  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Given a copper calorimeter that has a mass of 446 grams containing 95 grams of ice at 0.0°C. If 35 grams of steam at 100.0°C and 1.00 atm pressure is added to the can, what is the final temperature of the can and its contents? At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

    2. Relevant equations
    Q=mcΔT

    3. The attempt at a solution
    Finding the final temperature (I think I got this part right already but here it is):
    Qloss = Qgain
    Where Steam = Calorimetry + Ice
    msLv + mscs(100°C) - mscsTf = mcccTf - mccc(0°C) + miLf + miciTf - mici(0°C)
    Arrange the equation to find Tf
    Tf = miLf - msLv - mscs(100°C) / (- mscs - mccc - mici)
    Tf = (95g)(80cal/g)-(35g)(540cal/g)-(35g)(1cal/g⋅C°)(100.0°C) / (-(35g)(1cal/g⋅C°)-(446g)(0.0923cal/g⋅C°)-(95g)(1cal/g⋅C°))
    Tf = 86.47°C

    How do I find the mass of ice, water, and steam left?
    Since the temperature is around 86.47°C, doesn't it mean that the ice are all melted?
    Qice = (95g)(80cal/g) = 7600 cal
    Qsteam = (35g)(540cal/g)-(35g)(1cal/g⋅C°)(86.47°C)+(35g)(1cal/g⋅C°)(100.0°C) = 19373.55 cal
    Since Qsteam > Qice, all ice is melted.

    For mass of steam left:
    7600 cal = ms(540cal/g)-(35g)(1cal/g⋅C°)(86.47°C)+(35g)(1cal/g⋅C°)(100.0°C)
    ms = 13.20 grams (amount of steam condensed)
    35g - 13.20g = 21.8 grams (amount of steam left)

    For the mass of liquid left:
    13.20 grams (amount of steam condensed) + 95grams (melted ice) = 108.2 grams
     
  2. jcsd
  3. Mar 11, 2017 #2

    haruspex

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    Yes.
    Didn't your calculation of Tf assume all steam condenses?
     
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