This example of working out of water, 208 Celsius, and 2.15 megabascals. So if you go to the saturated water table, we are in this region here. So our saturation pressure is somewhere between 15 and 19 bar. Our pressure is 2.15 megabascals, 21.5 bar. So we are certainly above saturation pressure. And we will be in the compressive equation phase. Now we could proceed by going to the compressive liquid table. But I would like to show you a more general and more flexible way of doing it, that relies on an approximation and only requires the saturated water table. So this is the way it works. If we think about a TV diagram for water. And let's draw some constant pressure curves on that. And the thing about the constant pressure curves on the TV diagram is they all collapse down to a single curve when they get into the compressive liquid region. In fact, they pretty much collapse onto the saturated liquid side of the saturation dome. Even the supercritical pressures do the same sort of thing. And if this was accurate TV diagram, it would be even more obvious that all these curves would just disappear into one vertical line on the left-hand side. So what this means is that there is no difference in volume between the highest pressures and the lowest pressures in the compressive liquid region. And that's just another way of saying something that everybody knows. Liquid water is incompressible, almost completely incompressible. If you increase the pressure on it, its volume does not change. So the volume only depends on the temperature. It does not depend on pressure. So mathematically, we could say that the is a function of temperature only. It's not a function of the second thermogenic function. Like most things are for compressive liquid. And a bit more specifically, V is equal to VF, the volume of saturated liquid at the same temperature. And it turns out that the same is true for internal energy. And the same is true for entropy for compressed liquid. And I should have approximately equal sign up there as well. Okay, so let's put that to work in this example. So we're asked to find the enthalpy. The enthalpy, H is equal to U plus PV. So we're going to find U and V. This little approximation doesn't work for enthalpy. So we have to do it by finding U and finding V. We already know P, so we'd calculate enthalpy. So at 2A to receive, RU is approximately UF at 2A to receive. We're going to need to interpolate, find that. So where are we? We are interpolating between these two values here. So we're going to take V as being approximately VF at 200N to raise. So we're now interpolating between these two. And that, of course, is 10 to the minus 3. Let me just keep it short for a few to the ground. So our enthalpy is 878.55. And we need to take care with the units here. Plus pressure times volume. 8181.06 kilojoules per kilo-gram.