Hello again!
The problem:
The heat of ethanol condensation at 15°С equals to -27,62 kJ/mol. Average specific heat capacities of liquid ethanol and its vapor between 0 and 78°С equal to 2,418 and
1,597 J/(g*K), respectively. Find the amount of heat needed to evaporate 500 g of ethanol at 60°С.
The answer in the book: 281,7 kJ.
My solution:
C2H5OH (l)
C2H5OH (v)
500 g / 46 (g/mol) = 10.87 mol of ethanol
-27620 * 10.87 = -300229.4 J, but since we need to evaporate the ethanol rather than condensate it's 300229.4 J.
Δ H for liquid ethanol = 2.418*500*(333-288) = 54405 J
Δ H for ethanol vapor = 1.597*500*(333-288) = 35932.5 J
300229.4 + 35932.5 - 54405. = 281756.9 J = 281.8 kJ
Could you check please, if everything's correct?