Question

How much heat (kJ) is required to convert 3.12 moles of liquid benzene at 75.1°C to gaseous benzene at 115.1°C? The following information may be useful.
b.p = 80.1C
C_m (liquid benzene) = 136.0 J/ (mol * °C)
ΔH_vap = 30.72 kJ/mol
C_m (gaseous benzene) = 36.6 J/ (mol * °C)

Answer 1

liquid goes from 75.1°C to 80.1°C; then boils at 80.1°C; and then vapor goes from 80.1°C to 115.1°C

This requires 3 types of heat

Q1 = sensible heat of liquid

Q2 = latent heat of avporization

Q3 = sensible heat of gas

Then

Q1 = n*Cpliquid * (Tb-Tinitial)

Q2 = n*Latent Heat

Q3 = n*Cpvapor * (Tfinal - Tb)

substitute

Q1 = n*Cpliquid * (Tb-Tinitial) = 3.12 * 136 * (80.1-75.1) = 2121.6 J

Q2 = n*Latent Heat = 3.12*30.72*10^3= 95846.4 J

Q3 = n*Cpvapor * (Tfinal - Tb) = 3.12*36.6*(115.1-80.1) =3996.72

Total

Q = 2121.6 +95846.4 +3996.72

Q = 101964.72 J

Q = 101.96 KJ