Answer:
-900 kJ
Step-by-step explanation:
Q = ∆U + W
The specific internal energy decreases by 150 kJ/kg, therefore, change in internal energy (∆U = -150 kJ/kg × 3 kg = -450 kJ)
W = (P2V2 - P1V1)/1-n
P2 = P1(V1/V2)^n
P1 = 200 kPa
V1 = 1.5 m^3/kg × 3 kg = 4.5 m^3
V2 = 1 m^3/kg × 3 kg = 3 m^3
n = 2
P2 = 200(4.5/3)^2 = 450 kPa
W = (450×3 - 200×4.5)/1-2 = -450 kJ
Q = -450 + (-450) = -450 - 450 = -900 kJ