Heat rejected to the cooling water, Q = – 50 kJ/kg
(–ve sign since heat is rejected)
Work input, W = – 100 kJ/kg
(–ve sign since work is supplied to the system)
Using the relation, Q = (u2 – u1) + W
– 50 = (u2 – u1) – 100 or
u2 – u1 = – 50 + 100 = 50 kJ/kg
Hence, gain in internal energy = 50 kJ/kg.