6.35
The entropy of the steam is m(s2 - s1). Find s1 from sf + xsfg @ 100 kPa and x = .25. The s1 = 2.8168 kJ/kg-K. To find s2, v1 = v2 and it's a saturated vapor. The v1 = vf + xvfg = 0.4243 m3/kg. So s2 = 6.8649 kJ/kg-K. And DS = 8.0962 kJ/K.

6.42
This is an isentropic process. The two properties that set your system for the initial state are P and T. The two properties that set your final state are P and s. Find mass by m = V/v = 0.5 m3/0.6339 m3/kg = 0.0789 kg. Then W = -m(u2 - u1). Find u1 @ 300 kPa and 150 °C = 2570.8 kJ/kg. Find u2 @ 1 MPa and s2 = s1 = 7.0778 kJ/kg. It should be around 2774.2 kJ/kg. This gives a work = -16.05 kJ.

6.48
Here you need to find the entropy of the iron, the copper and the lake. The DSiron = mCln(T2/T1) = 50kg * 0.45 kJ/kg-K ln (288/353) = -4.579 kJ/k. DScopper = mCln(T2/T1) = 20kg * 0.86 kJ/kg-K ln (288/353) = -1.571 kJ/K. The DS of the lake = Q/T0. The Q is found from the Q lost by the metal = (mC(T1 - T2)iron) + (mC(T1 - T2)copper) = 1964 kJ. Then the total entropy change = -4.579 kJ/K + -1.571 kJ/K + 1964 kJ/288 K = 0.670 kJ/K

6.57
a) Ds = CvlnT2/T1 + R lnV2/V1 = 0.690 kJ/kg-K ln (560/298) + (0.2598 kJ/kg-K)ln(.1/.8) = -0.105 kJ/kG-K
b) T1 = 298 K s10 = 205.33 kJ/kmol-K/32 kg/kmol = 6.4073 kJ/kg-K. T2 = 560 K s2o = 224.146 kJ/kmol-K/32 kg/kmol = 7.0046 kJ/kg-K. Ds = (7.0046 - 6.4073) - R ln(P2/P1). Use the ideal gas law to find P = RT/v for each pressure. Then Ds = -0.107 kJ/kg-K.

6.65
a) T2 = T1 (P2/P1)k-1/k = 290 K * (800/100)0.4/1.4 = 525.3 K. Then w = -CvDT = 0.727 kJ/kg-K* (525.3 - 290 K) = -171.1 kJ/kg. b) T1 = 90 K Pr1 = 1.311 u1 = 206.91 kJ/kg and Pr2 = P2/P1 * Pr1 = 800/100 * 1.2311 = 9.849. This sets the T at 522.4 K and u2 = 376.16 kJ/kg. Then w = u2 - u1 = 169.25 kJ/kg.