Question
How much energy (in kilojoules) is released when 26.5g of ethanol vapor at 86.0C is cooled to -12.0C? Ethanol has mp = -114.5C , bp = 78.4C, delta H vap = 38.56 (KJ/mol) and delta H fus = 4.60 (KJ/mol) . The molar heat capacity is 113J/(K.mol) for the liquid and 65.7 J/(K.mol) for the vapor.
Answers
DrBob222
You must calculate the heat evolved in stages, then add all of the heats together to obtain the sum. Here is the way you go about it.
q1 = heat lost in moving the vapor from the starting point to the boiling point.
q1 = mass vapor x specific heat vapor x (Tfinal-Tinitial) where Tfinal will be the boiling point and Tinitial is where you start.
q2 = heat lost in condensing the vapor at the boiling point.
q2 = mass vapor at the boiling point x delta Hvaporization.
q3 heat lost in cooling the liquid vrom the boiling point to the melting (freezing) point.
q3 = mass liquid x specific heat liquid x (Tfinal-Tinitial) where Tfinal is the melting point and Tinitial is the boiling point.
q4 = heat lost is freezing the liquid.
q4 = mass liquid x heat fusion.
q5 = heat lost in moving the temperature from the freezing point to the final T.
q5 = mass solid x specific heat solid x (Tfinal - Tinitial) where Tfinal is the final T and Tinitial is the melting point.
qtotal = q1 + .....q5.
Watch the units. I note most of the specific heats are in J or kJ/mol but the mass is in grams.
q1 = heat lost in moving the vapor from the starting point to the boiling point.
q1 = mass vapor x specific heat vapor x (Tfinal-Tinitial) where Tfinal will be the boiling point and Tinitial is where you start.
q2 = heat lost in condensing the vapor at the boiling point.
q2 = mass vapor at the boiling point x delta Hvaporization.
q3 heat lost in cooling the liquid vrom the boiling point to the melting (freezing) point.
q3 = mass liquid x specific heat liquid x (Tfinal-Tinitial) where Tfinal is the melting point and Tinitial is the boiling point.
q4 = heat lost is freezing the liquid.
q4 = mass liquid x heat fusion.
q5 = heat lost in moving the temperature from the freezing point to the final T.
q5 = mass solid x specific heat solid x (Tfinal - Tinitial) where Tfinal is the final T and Tinitial is the melting point.
qtotal = q1 + .....q5.
Watch the units. I note most of the specific heats are in J or kJ/mol but the mass is in grams.