Answer
742 kJ
Work Step by Step
The heat energy change during the conversion of $78^{\circ}C$ ethyl alcohol gas to its liquid form in the same temperature is
$Q_{1}=m(-L_{V})=0.510\,kg\times-879\,kJ/kg=-448\,kJ$ (heat of vaporization and heat of condensation are equal, but condensation is just the reverse process of vaporization and therefore we add a negative sign)
The energy change while reducing its temperature from $78^{\circ}C$ to $-114^{\circ}C$ is
$Q_{2}=mc\Delta T$
$=0.510\,kg\times2.43\,kJ/kg\cdot K\times(-114-78)K=-238\,kJ$
The energy change during the conversion of $-114^{\circ}C$ ethyl alcohol liquid to its solid form in the same temperature is
$Q_{3}=m(-L_{f})=0.510\,kg\times-109\,kJ/kg=-55.6\,kJ$ (negative sign is added with the latent heat of fusion because we are freezing ethyl alcohol and not fusing it)
Total energy Q=$Q_{1}+Q_{2}+Q_{3}$
$=-448\,kJ+(-238\,kJ)+(-55.6\,kJ)=-742\,kJ$
The negative sign indicates that the heat is to be removed.
742 kJ energy must be removed.
