In principle you could obtain the latent heat of vaporization of the mixture by Differential Scanning Calorimetry (DSC).  But it is needed a specific experimental skill to obtain good results.

Hope to have helped you

Regards

Federico

MEA?

MEA = Monoethanol amine

It depends how precise and accurate a result you need.  First of all realise that what will vaporize will not be the same as the liquid.  At 90 deg C what will come off will be about 3% MEA, so what will be vaporizing is water. Thus to a first approximation, the enthalpy of vaporization is that of water.

This is a  well-known system, so you can easily find papers on its vapor-liquid equilibrium.

You can find the result in terms of the Clausisus-Clapeyron (CC) equation in the excel file based on the data of "J. Chem. Eng. Data 2009, 54, 2312–2316" and on the assumprtons that :

a) 7:3 means 0.3 MEA molar fraction;

b) 0.274 is close enough to 0.3, the desired mole fraction

c) DeltavapH is constant in the temp interval 283-363 K

The CC eq is ln pv= a0 -DvapH/R * 1/T therefore it is easy to get the value you want from the slope of this equation. Since the regression parameters are quite good I don't think you need to consider the variation of the vap. enthalpy with T.

If you mean 7:3 in terms of wheight then you need to get more data to transform  the data I used into w/w fractions. You can find that in "Journal of Chemical and Engineering Data, Vol. 48, No. 3, 2003".

If you need to calculation of DeltavapH exactly fot a 0.3 mixture than you nee to first interpolate for 0.3 from the data in the spreadsheet.

What do you mean by "and then the evaporation..."?

I Hope this is what you need regarding the first part of your question.

Thank you for your question...

Next time don't expect to get my answer!