Thermodynamics: Dead state assumptions

[jdawg]

Homework Statement

Im having a little trouble with knowing what to assume for the dead state! In the question that I posted, I don't understand how they knew to assume that the water in the system is a super heated vapor? Or how they knew that the dead state is a compressed liquid?

Homework Equations

The Attempt at a Solution

 

[SteamKing]

Homework Statement

Im having a little trouble with knowing what to assume for the dead state! In the question that I posted, I don't understand how they knew to assume that the water in the system is a super heated vapor? Or how they knew that the dead state is a compressed liquid?

Homework Equations

The Attempt at a Solution

You have to match the conditions given in the problem statement with the particular portion of the steam tables you are using.

You can determine the state of the water at P = 1 bar and T = 100 °C by checking the saturation line. Remember, 1 bar is not quite the same pressure as 1 atmosphere (1 atm. = 1.01325 bar), so these conditions suggest that the water is in vapor form. Checking the saturation line for 1 bar will give the corresponding saturation temperature, which will be slightly below 100 °C. Therefore, you can conclude that the vapor is technically superheated, but only by the barest of margins.

As for the conditions of water in the surroundings, P = 1 bar and T = 20 °C, it's safe to assume that these conditions reasonably approximate standard conditions, and water is a liquid. That this is called "compressed liquid" in the steam tables is, I think, a way to distinguish the liquid phase from the saturated condition, where vapor and liquid coexist in equilibrium. The temperature of 20 °C is certainly well below the saturation temperature of water at a pressure of 1 bar.

 

[jdawg]

Thanks! That makes so much sense!