If we consider a pure, single-phase substance (which we will do for all of our calculations), we can show (later) that 2 intensive variables fix the state of the system.
Since we know that internal energy is NOT a strong function of P, we might be tempted to say that $\hat U = F(T,\hat V)$.
If we keep $\hat V$ constant, we can then find the dependence of $\hat U$ on T only.
We have no reason to suspect that this is a straight line (and in fact it isn't), but for a small enough $\Delta T$, we can safely say that $\Delta \hat U$ = (slope) $\Delta T$.
This slope is called the specific heat at constant volume (or heat capacity at constant volume) and is denoted as cv
As you all recall from calculus, integration simply means adding up lots of little pieces. Since we know that $\Delta \hat U$ = (slope) $\Delta T$ as long as $\Delta T$ is small enough, in the limit that $\Delta T \rightarrow dT$ (infintely small change), we can write an exact answer mathematically as:
$\Delta \hat U = \int c_v dT \approx c_v \Delta T$
This expression is equally valid for (ideal) gases, solids, and liquids!
The difference between energy and enthalpy, of course, is the "flow work" term. If we take $\hat H=F(T,P)$ (since we can write it in terms of any two variables) and we think about getting to our final state first by keeping P constant (rather than $\hat V$, as we did last time), and then letting P change, the flow work term can be changed from $\Delta (P \hat V)$ to $\hat V \Delta P$ (since the changes in $\hat V$ would have occured in the constant P part!).
By analogy to the section above, keeping P constant and letting T change, we can get the expression for the constant P part as:
$\Delta H = \int c_P dT \approx c_P \Delta T$ (at constant P)
cp in the expression above is the specific heat at constant pressure (or heat capacity at constant pressure) and, as before, it represents the slope of the $\hat H$ vs. T curve.
We, finally, then need to consider the flow work part.
For an ideal gas, there is no "friction" to the flow (since they are essentially infinitely dilute), so there is no flow work involved! This gives us that:
$\Delta \hat H = \int c_P dT \approx c_P \Delta T$
For a "real" gas, we either need to use the charts (see next lecture) or, for small values of $\Delta P$ we can use the following (where we will take the average $\hat V$ to be the correct (constant) value to use):
$\Delta \hat H = \int c_P dT + \hat V \Delta P \approx c_P \Delta T + \hat V \Delta P$
Solids and liquids are incompressible. Therefore, the flow work term ignores changes in pressure (even if there is not change in temperature) so we get:
$\Delta \hat H = \int c_P dT + \hat V\Delta P \approx c_P \Delta T + \hat V\Delta P$
So, we have now given ourselves expressions for each of the types of changes proposed for our fictitious paths: T changes at constant P (for gases, liquids, or solids), P changes at constant T (for gases, liquids, or solids), and we already know that we can simply "look up" the heat of phase changes!
Use heat capacities and phase data to calculate H and U from fictitious paths
Actually perform the calculations from your previous test yourself: Get the enthalpy (and internal energy) of steam at 200C and 1.5atm relative to a reference state of ice at -10C and 1 atm.