# Thermodynamics question. In Dewar flask liquid nitrogen.

## 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

Last edited:

I like Serena
Homework Helper
Welcome to PF, AudriusR! You're reasoning looks fine.

Only one note.
Your question does not seem to say how much mass of nitrogen is evaporating.