Problem # p040

State based on entropy and pressure

One kilogram of Refrigerant 12 undergoes a process starting from p1=4 bars, T1=100oC to a state 2 where the pressure is p2=1 bar. During the process the specific entropy of the refrigerant changes drops due to heat loss : s2-s1 = -0.05 kJ/(kg K).
Use the table below and linear interpolation to determine for the state 2 :

  1. The temperature T2.

  2. The internal energy u2.

T [oC] p [bar] v[m3/kg] u[kJ/kg]h[kJ/kg]s[kJ/(kg K)]
101.00.190 1791980.807
201.00.197 1842040.828
301.00.205 1902100.849
401.00.212 1952160.869
 
1004.00.0617 2282530.883
1204.00.0655 2412670.919