J.R. S. answered • 05/27/21
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To answer this question, one needs to know the specific heat of glacial acetic acid. Then use the equation,
q = mC∆T
q = heat = 725 J
m = mass = 27.5 g
C = specific heat = 2.04 J/gº (I looked this up on Google)
∆T = change in temperature = ?
Solving for ∆T ...
∆T = q / mC = 725 J / (27.5 g)(2.04 J/gº)
∆T = 12.9º
J.R. S.
tutor
123.2 J/molxK is the same as 2.04 J/gº when you use 60 g/mol as the MW of acetic acid. Also, unless we know the initial temperature of the acetic acid, as well as the boiling point, the ∆Hvap does us no good and we still cannot determine the final temperature, or the temperature increase, as I see it. (∆Hf doesn't come into play at all). Do you have any other additional data that you left out of the question?
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05/29/21
Claire T.
Sorry! I forgot to add the thermal data. C= 123.1 J/(mol x K), Hf= 11.74kj/mol, and Hv = 24.39 kJ/mol05/28/21