The specific enthalpy of liquid water at typical ambient conditions, like T=25°C and P=1 bar, is not given in the steam tables. However, the specific enthalpy of saturated liquid at P=1 bar is given.
A) Using the approximation that C_p for liquid water is constant at 4.19 J/g·K, estimate the specific enthalpy of liquid water at T=25°C and P=1 bar.
B) Compare the answer you obtained in part A to the specific enthalpy of saturated liquid water at T=25°C.
A) Saturated liquid at 1 bar (99.6°C)\rightarrow 417.50 \frac{\mathrm{kJ}}{\mathrm{kg}}
\begin{gathered}\widehat{\mathrm{H}}_{25^{\circ} \mathrm{C}}=\widehat{\mathrm{H}}_{99.6^{\circ} \mathrm{C}}+\int_{99.6^{\circ} \mathrm{C}}^{25^{\circ} \mathrm{C}} \mathrm{C}_{\mathrm{p}} \mathrm{dT} \\\widehat{\mathrm{H}}_{25^{\circ} \mathrm{C}}=417.5 \frac{\mathrm{kJ}}{\mathrm{kg}}+\left(4.19 \frac{\mathrm{J}}{\mathrm{g} \mathrm{K}}\right)\left(25^{\circ} \mathrm{C}-99.6^{\circ} \mathrm{C}\right)=\mathbf{1 0 4 . 4 3}\left(\frac{\mathrm{kJ}}{\mathrm{kg}}\right)\end{gathered}
Note: A change in temperature in degrees Celsius is identical to a change in temperature in degrees Kelvin. The equation above involves only the change in temperature (ΔT =–74.6°C). Conversion to degrees Kelvin would be necessary if the temperature appeared in the equation as an absolute quantity, rather than a change in T.
Note: 1 J/g = 1 kJ/kg
B) Saturated liquid at 25°C (0.0317 bar) \rightarrow 104.8\left(\frac{\mathrm{kJ}}{\mathrm{kg}}\right)
This value and our answer from part a) are very close. This illustrates that the pressure of a liquid does not have much influence on its enthalpy, at least over a small P interval like 1 bar vs. 0.0317 bar.