Thermodynamics question. In Dewar flask liquid nitrogen.

  • Thread starter AudriusR
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Homework Statement


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


Homework 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


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:

Answers and Replies

  • #2
I like Serena
Homework Helper
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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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