There is no entry in the steam tables at 350 kPa (3.5 bar). Interpolation between the data at 3 and 4 bar, and 200 ˚C, can be used to find the specific enthalpy at 3.5 bar and 200 ˚C. But what we actually seek is the change in specific enthalpy, so we will instead look at the temperatures above and below 200 ˚C.
Steam at 350 kPa and 150°C →Interpolation Needed Interpolate
\begin{array}{|c|c|c|c|} \hline & 1 & & 2\\\hline \text{Enthalpy (y) }&2761.2 & ??? & 2752.8\\\hline \text{Pressure (x) }& 3 & 3.5 & 4\\\hline\end{array}
\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\\mathrm{y}=\frac{(2752.8-2761.2)(3.5-3)}{(4-3)}+2761.2=2757.0 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{gathered}
Steam at 350 kPa and 250°C → Interpolation Needed Interpolate
\begin{array}{|c|c|c|c|}\hline & 1 & & 2\\\hline \text{Enthalpy (y) }&2967.9 & ??? & 2964.5\\\hline \text{Pressure (x) }& 3 & 3.5 & 4\\\hline\end{array}
\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\\mathrm{y}=\frac{(2964.5-2967.9)(3.5-3)}{(4-3)}+2967.9=2966.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{gathered}
Constant pressure heat capacity \left(\mathrm{C}_{\mathrm{P}}\right) can be estimated as
\mathrm{C}_{\mathrm{P}} \approx \frac{\Delta \mathrm{H}}{\Delta \mathrm{T}}=\frac{2966.2 \frac{\mathrm{kJ}}{\mathrm{kg}}-2757.0 \frac{\mathrm{kJ}}{\mathrm{kg}}}{250^{\circ} \mathrm{C}-150^{\circ} \mathrm{C}}=2.092 \frac{\mathrm{kJ}}{\mathrm{kg}^{\circ} \mathrm{C}}
If we attempt to apply a directly analogous procedure at 700 kPa (7 bar) there is a complication:
Water at 700 kPa and 150°C →Liquid
Since these two intensive properties do not yield a vapor, we must use the enthalpy of saturated vapor at 700 kPa. This temperature is 164.9°C
Steam at 700 kPa and 164.9°C → 2762.8 \frac{\mathrm{kJ}}{\mathrm{kg}}
Steam at 700 kPa and 250°C → Interpolation Needed
Interpolate
\begin{array}{|c|c|c|c|} \hline & 1 & & 2\\\hline \text{Enthalpy (y) }&2957.6 & ??? & 2950.4\\\hline \text{Pressure (x) }& 6 & 7 & 8\\\hline\end{array}
\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\\mathrm{y}=\frac{(2950.4-2957.6)(7-6)}{(8-6)}+2957.6=2954.0 \frac{\mathrm{kJ}}{\mathrm{kg}}\\\mathrm{C}_{\mathrm{P}} \approx \frac{\Delta \mathrm{H}}{\Delta \mathrm{T}}=\frac{2954.0 \frac{\mathrm{kJ}}{\mathrm{kg}}-2762.8 \frac{\mathrm{kJ}}{\mathrm{kg}}}{250^{\circ} \mathrm{C}-164.9^{\circ} \mathrm{C}}=2.247 \frac{\mathrm{kJ}}{\mathrm{kg}^{\circ} \mathrm{C}} \end{gathered}
These answers are fairly similar, but different enough to demonstrate that \mathrm{C}_{\mathrm{P}} can in reality be a function of pressure as well as temperature. This is why the distinction between “ideal gas heat capacity” and simply “heat capacity” is important; ideal gas heat capacity is only a function of temperature.
Conceptually, we could get more accurate estimates of C_{P} by using values of \widehat{H} that were closer to the actual temperature of interest; e.g., 190 and 210°C instead of 150 and 250°C. But that data is not included in the available version of the steam table.