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

I answered this earlier, and now after reflecting on it, and looking at it more carefully, I believe there is an error in the question, and/or in the answer. The question states that there are 48.7 GRAMS of ethanol, and if that is true, then my previous answer is correct. Here it is again:

heat = q = mass x specific heat x change in temperature + mass x ∆Hvap

q = (1.056 moles)(0.1124 kJ/mole/deg)(25 deg) + (1.056 moles)(38.56 kJ/mole) = 2.97 + 40.7 = 43.7 kJ

Now, if as I suspect, the question was supposed to state that there are 48.7 MOLES of ethanol, then this is what you have...

(48.7 moles)(0.1124 kJ/mole/deg)(25 deg) + (48.7 moles)(38.56 kJ/mole) = 137 + 1878 = 2015 kJ

Either YOU mistyped the question, or whoever gave you the problem made the mistake.