Ive tried this question 100 times and cannot get the right answer anyone want to help me understand the steps?

Question

What quantity of heat, in kJ, is required to convert 36.0 g of ethanol (C₂H₅OH) at 23.0°C to a vapor at 78.3°C (its boiling point)? (specific heat capacity of ethanol = 2.46 J/g • C; ∆Hvap = 39.3 kJ/mol)

Answer 1

Do this in a step-wise fashion.

Step 1: heat needed to raise the temp of 36.0 g ethanol from 23º to 78.3º

Step 2: heat needed to convert 36.0 g ethanol from liquid @ 78.3º to gas@ 78.3º

1). q = mC∆T

q = heat = ?

m = mass = 36.0 g

C = specific heat = 2.46 J/gº

∆T = change in temperature = 78.3 - 23.0 = 55.3º

q = (36.0 g)(2.46 J/gº)(55.3º) = 4897 J = 4.897 kJ

2). q = m∆Hvap (this is a phase change and there is NO CHANGE in temperature)

since ∆Hvap is given in kJ/mole, we'll change it to kJ/g so we can use 36.0 g (we could also change g to mols)

39.3 kJ/mol x 1 mol ethanol / 46.1 g = 0.852 kJ/g

q = (36.0 g)(0.852 kJ/g) = 30.7 kJ

Add the heat from steps 1 and 2 to get total heat in kJ...

4.897 kJ + 30.7 kJ = 35.6 kJ (3 sig. figs.)

NOTE: be sure to check my math as I was in a hurry when answering this.