Hi, I asked this question before, but wanted to check my solution is correct:
Question: A cylinder fitted with a piston has a volume of 0.1 m3 and contains 0.5 kg of steam at 0.4 MPa. Heat is transferred to the steam until the temperature is 300 °C, as a result the volume increases to three times and the change in internal energy is 680.1 kJ. Determine the work and heat transfer for this process. Neglect the change in kinetic and potential energies.
Solution: I don't think we're interested in tables, what do you think?
Work = P.dV = P(V2 – V1)
Also, V2 = V1 x 3 - So V2 = 0.3m3
Work = 400 kPa x (0.3 – 0.1)
W = 80 kJ
dQ = dU + dW
dQ = 680.1 + 80
dQ = 760.1 kJ