How much heat (kj) is required to convert 3.12 moles of liquid benzene at 75.1â°c to gaseous benzene at 115.1â°c?

The total energy includes sensible heat to raise the temperature from 75.1°C to the boiling point. It also includes the latent heat to convert the liquid to gas. Then, it also includes sensible heat from he boiling point to 115.1°C. The equation is:

Energy = nCp,liquid(T,bp - T₁) + nΔH + nCp,gas(T₂ - T,bp)
where
n is the number of moles
T,bp is the boiling point of benzene at 80.1°C
Cp,liquid = 134.8 J/mol·°C
Cp,gas = 82.44 J/mol·°C
ΔH = 87.1 J/mol

Energy = (3.12 moles)(134.8 J/mol·°C)(80.1°C - 75.1°C) + (3.12 moles)(87.1 J/mol) + (3.12 moles)(82.44 J/mol·°C)(115.1°C - 80.1°C)
Energy = 11,377.08 J

It would be (at) + cu (s)

Because a model would make the investigation easier to understand