How many kilojoules of heat are required to raise the temperature of 48.7 grams of ethanol (C2H5OH, 46.1 g/mol) from 0 °C to °25 C and then vaporize it completely?

Question

How many kilojoules of heat are required to raise the temperature of 48.7 grams of ethanol (C₂H₅OH, 46.1 g/mol) from 0 °C to °25 C and then vaporize it completely? 

The heat capacity of C₂H₅OH at 25°C = 112.4 J/(mol.K)

The molar heat of vaporization of C₂H₅OH = +38.56 kJ/mol

Answer is 2015 kJ

Answer 1

NOTE: The heat capacity is given at 25ºC and in order to proceed, it will be assumed that it is constant over the temperature range of 0º-25ºC. Also, you cannot vaporize the ethanol completely at 25ºC since the boiling point of ethanol under standard conditions is 78.37ºC

Since the heat capacity and ∆H_vap are both given in units containing moles, and the mass of ethanol is given in grams, we should change one or the other to have the same units. In this case, I'll convert 48.7 g to moles and then proceed.

moles ethanol = 48.7 g x 1 mole/46.07 g = 1.057 moles

Step 1: heat needed to raise 1.057 moles EtOH(ethanol) from 0ºC to 25ºC = q = mC∆T

q = (1.057 moles)(112.4 J/mol/deg)(25 deg) = 2970 J = 2.97 kJ

Now, IF YOU COULD vaporize the ethanol at 25º, then you would have...

Step 2: heat needed to vaporize 1.057 moles EtOH at 25ºC = q = m x ∆H_vap

q = (1.057 moles)(38.56 kJ/mol) = 40.76 kJ

But, since you CANNOT vaporize ethanol at 25º and normal pressure, we will add another step, and that is to get the ethanol up to the boiling point before vaporizing it.

Step 3: heat needed to raise 1.057 moles EthOH from 25º to 78.37º C = q = mC∆T

q = (1.057 moles)(112.4 J/mol/deg)(53.37 deg) = 6341 J = 6.34 kJ

Step 4: add them together = 2.97 kJ + 40.76 kJ + 6.34 kJ = 50.1 kJ (to 3 significant figures)