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Thermodynamics question. In Dewar flask liquid nitrogen.

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem is that we need to calculate, how much time it takes for nitrogen to evaporate.
    Given data: dewar flask is in a room, where is T= 293K temperature; In such flask m=0,2 kg of Hydrogen would evaporate in t= 3600s. Also there is given boiling point temperatures for Hydrogen and Nitrogen TH2= 20,4K ; TN2= 77,3K
    Heat of evaporations for both are : LH2= 4,5 x 105 J/kg
    LN2=2 x 105 J/kg


    2. Relevant equations

    I don't really know how to do it. Dewar flask don't allow to lose the heat, so no external loss Q=0 J
    I found only one way to find the answer, but I think it's not possible. I used this formula -


    k(T-T') t = Lm


    3. The attempt at a solution

    k(T-TH2) t = LH2m
    k(T-TN2) t' = LN2m
    So t' would be t'= LN2 ( T-TH2) t / ( LH2 (T-TN2))

    And I would get the answer of 2022 seconds or about 33,7 minutes

    I'm waiting for your opinion and attempt for the answer. :)
     
    Last edited: Nov 18, 2011
  2. jcsd
  3. Nov 20, 2011 #2

    I like Serena

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    Homework Helper

    Welcome to PF, AudriusR! :smile:

    You're reasoning looks fine.

    Only one note.
    Your question does not seem to say how much mass of nitrogen is evaporating.
    Your answer assumes it is also 0.2 kg.
    But perhaps you should give an answer in seconds per kilogram.
     
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