Answer
18260 kJ
Work Step by Step
Heating the liquid:
$\Delta H=12\ kg\times 4.7\ kJ/kg.K\times (-33.3--50)\ K=941.88\ kJ$
Boiling:
$\Delta H=12\ kg\times23.33\ kJ/mol\div 17.03\ g/mol\times1000\ g/kg=16439\ kJ$
Heating the gas:
$\Delta H=12\ kg\times 2.2\ kJ/kg.K\times (0--33.3)\ K=879.12\ kJ$
Total heat required: 18260 kJ