Q:

internal energy calculation of superheated vapor

I know that steam behaves like an ideal gas. Thus, the internal energy is a function only of temperature. I also know that $U=C_{V}(T_2-T_1)$, where 
$C_V$ is the specific heat at constant volume. My question is:
Since $C_V$ is a function of temperature, which value of $C_V$ should I take, or should I just use the values of internal energies on the steam tables? Thanks in advance!

A:

When the specific heat varies with temperature, you need an integral to compute the change in internal energy:
$$
\Delta U = \int_{T_1}^{T_2} c_V(T) \, dT
$$
If you have a table of the specific heats at various temperatures, you can make a numerical estimate of this integral using the trapezoidal rule.  For example, suppose that we are trying to find the change in internal energy between 400 K and 600 K and we know that the specific heat is $1.901 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}$ at 400 K; $1.954 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}$ at 500 K; and $2.015 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}$ at 600 K.  Then $\Delta U$ is approximately
\begin{align}
&\frac{1.901 \frac{\text{kJ}}{\text{kg} \cdot \text{K}} + 1.954 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}}{2} \cdot (500 \,\text{K} - 400 \,\text{K}) \\
+ \,&\frac{1.954 \frac{\text{kJ}}{\text{kg} \cdot \text{K}} + 2.015 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}}{2} \cdot (600 \,\text{K} - 500 \,\text{K}) \\
= \,&391.2 \,\frac{\text{kJ}}{\text{kg}}
\end{align}

