I have a question about the relation between enthalpy and the specific heat of water.

If we look in the property tables (Page 890) at let's say 20 degC, the properties of water are:

  • $P_{sat} = 2.34 kPa$,
  • $h_{f} = 83.915 kJ/kg$
  • $h_{g} = 2537.4 kJ/kg$
  • $Cp = 4.18 kJ/kg$

And if we say that the saturated liquid behaves like a incompressible liquid, the following expression applies:

$h_f = Cp_L T = 4.18 \cdot 20 = 83.6 kJ/kg$ which is similar to the one in the table.

But if we assume that the saturated vapor behaves as an ideal gas, this should apply:

$h_g = Cp_G T = 4.18 \cdot (20 + 273) = 1224 kJ/kg$, which is far from the value of $h_g$ from the table.

And okay, steam has a different specific heat ($C_{p,steam} = 1.87$ - Table A-2). Then

$h_g = Cp_G T = 1.87 \cdot (20 + 273) = 547 kJ/kg$

It still doesn't match! More than that, the value is very far.

I know that ideal gas is a pretty rough approximation, but the pressure and temperature are far from the critical point, so the error should not be that high. What am I doing wrong?


  • $\begingroup$ You are not taking into account the heat of vaporizaton. $h_g=h_f+\Delta h_v$ $\endgroup$ Commented Oct 15, 2016 at 14:17
  • $\begingroup$ $Delta h_v$ is the energy necessary to break to bonds, right? By definition it looks like that latent energy will stay in the gas (steam)... Is that right? $\endgroup$
    – Physther
    Commented Oct 15, 2016 at 14:28
  • 1
    $\begingroup$ Sure, since, when it's vaporized, there's no liquid left. For purposes of this table, there is a datum state at which u is taken to be equal to zero: liquid water at 0 C at either 1 atm or the equilibrium vapor pressure, depending on the table (probably the latter). Everything else is referenced to this state. $\endgroup$ Commented Oct 15, 2016 at 14:44
  • $\begingroup$ Okay. Then it makes sense... Thank you, Chester. You are always the first to answer and help me learn :) $\endgroup$
    – Physther
    Commented Oct 15, 2016 at 14:56

1 Answer 1


Okay. So thanks to Chester Miller, I have managed to answer the question.

So to get $h_g$ for steam, one can use:

$h_g = Cp_L T + h_{fg} = 4.18 \cdot 20 + 2453 = 2536.6 kJ/kg $

Where $h_{fg}$ is taken from that link.


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