If 500 calorie of heat energy is added to the system and the system does 350 calorie of work on the surroundings, the internal energy change of the system is

The correct option is B 150 cal
Heat absosrbed by the system (q) = 500 cal
Work done(w) = -350 cal
According to the first law of thermodynamics
dU = q + w
= 500 + (-350)
= 150 cal.