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A pure substance at 8 MPa and 400 °C is having a specific internal energy of 2864 kJ/kg and a specific volume of 0.03432 m³/kg. Its specific enthalpy (in kJ/kg) is ______

1. 3148.56
2. 3138.56
3. 3128.56
4. 3118.56

Correct answer: 3138.56

Solution

Specific enthalpy h = u + p*v. With u=2864 kJ/kg, p=8000 kPa, v=0.03432 m3/kg: p*v = 274.56 kJ/kg, so h = 2864 + 274.56 = 3138.56 kJ/kg.

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