We can solve the problem by using the first law of thermodynamics, which states that:
[tex]\Delta U = Q-W[/tex]
where
[tex]\Delta U[/tex] is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system
In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
[tex]\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ[/tex]
