C.V. Bottle. Flow in, no work, no heat transfer.
Continuity Eq m2 - m1 = min ;
Energy Eq. m2u2 - m1u1 = minhin
Now use the energy equation from the beginning to the final state
1Q3 = m2u3 - m1u1 - minhin = (12.022 - 0.29) 214.36 - 11.732 x 260.32 = -539.2 kJ
Entropy equation from state 1 to state 3 with change in s from Eq
Sgen = m2s3 - m1s1 - minsin - 1Q3/T = m2(s3 -sin) - m1(s1 - sin) - 1Q3/T
= 12.022[6.8693 - 6.7256 - R ln(4140/6000)]
- 0.29[6.8693 - 6.7256 - R ln(100/6000)] + 539.2/300 = 4.423 kJ/K
Problem could have been solved with constant specific heats in which case we would get the energy explicit in T2 (no iterations).