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ΔHrxn Liquid -> Solid

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the heat change in J associated with 80.7 g of liquid water at 5.00 ° C changing to solid water at -5.00 °C?


    2. Relevant equations
    c(H2 O) (liq) = 4.184 J/(g.K)

    c(H2O) (s) = 2.09 J/ (g.K)

    DHfus(H2 O) = 6.02 kJ/mol


    3. The attempt at a solution
    Δt = 5°

    While liquid
    (80.7g)(4.184J/g.k)(5°) = 1688.2 J
    (-6.02kJ/mol)(80.7/18 = 4.48 mol) = -27.0 kJ
    (80.7g)(2.09J/g.k)(5°)= 843.3 J

    2531.5 - 27000
    -2.44E4
    (correct answer is -2.95E-4)

    I mean I was pretty sure that is how you went about doing this type of problem :/.
    The Δt = 5° came from (5 - 0) (and 0-(-5))
     
  2. jcsd
  3. Oct 2, 2009 #2

    chemisttree

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    I believe you are messing up in your sign convention. You are removing heat throughout this process and thus all of your signs must be negative. Delta T is -5, not 5.
     
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