# Latent heat and unstable system

1. May 23, 2004

### MiGUi

Latent heat is the variation of entalpy between two phases (or the heat that you must provide to the system to change the phase).

But, if we are in a unstable system (such as water at 1 atm of pressure and 270 K of temperature), we need no heat to change the phase, simply we have to variate infinitesimally the conditions to make the system change.

So my question is: That reason is correct? Latent heat in unstable conditions is equal to zero?

MiGUi

2. May 24, 2004

### Staff: Mentor

No. You provided the definition in your first sentence and then discarded it: heat added/subtracted to change the temperature is sensible heat and a different issue. Latent heat content can change at constant temperature.

Think about what happens to a glass of water in the freezer. First, the temperature decreases by 1 degree C for each calorie of energy removed per gram. When the water gets to 0 C, energy continues to be pulled out while it freezes. The heat of fusion for water is 80 cal/g-C, so for each 80cal taken out, 1g freezes. Once frozen, the temperature of the ice will begin to drop again.

Last edited: May 24, 2004
3. May 25, 2004

### MiGUi

But if you have liquid water (not ice) at 270 K and 1 atm (unstable), latent heat exists?

4. May 25, 2004

### turin

Perhaps you should think of latent heat as more of a process than a property.

I'm assuming that, by "270 K" and "1 atm," you intend to specify the conditions that indicate the boundary between solid and liquid. In this case, it is crucial to determine/specify whether or not the system is isolated. If it is truly isolated (all exchanges and interactions are completely contained), then the ratio of solid to liquid H2O will persist (statistically). But, just the slightest perturbation will change this ratio. If you have 100% solid H2O, and there is an addition of heat from the environment, then this will liquify some of that H2O. Then, you have to consider what happens to the pressure to determine where the state goes.

Last edited: May 25, 2004
5. May 27, 2004

### MiGUi

We can cool water and keep it 100% liquid at 1 atm and 270 K. But the Clausius-Clapeyron equation says that this is not an stable system.

If we change a bit the properties of the system, then water freezes spontaneously and pressure may be constant. We do not have to give aditional heat to the system to make it change.

6. May 27, 2004

### turin

Yeah, that sounds correct (except that I don't know what the Clausius-Clapeyron equation is). Hmm. I probably should not have been so hasty with my previous post. Sorry about that. I'll leave it to the experts.

7. May 27, 2004

### Staff: Mentor

edit: missed the key. Got it now.

You're talking about super-cooling. If you are careful not to provide a means for chrystals to form, it is possible to cool water below its freezing point (273K). But purturb it (literally, bump the vessel) and some will freeze. The temperature of the remaining water will go back up to 273K and the system will remain in thermal equilibrium.

Last edited: May 27, 2004
8. May 29, 2004

### MiGUi

Well, I'm not talking about chrystals... only water and the possibility of cool it below its freezing point. This is an unstable system.

The point is that: The latent heat is the heat you need to make a system, change the phase. For example, at 373K and 1 atm if you give 2260 J/g of heat, you make water boil and the liquid phase dissapears.

But, if we are at 375 K and 1 atm ... we can cool very slow the system and we can have water vapour at 370 K and 1 atm. To convert that vapour into liquid water, we only have to change slightly the conditions.

The system at stable conditions (vapour at 375K, 1atm) is like a "lego castle", and we have to use a lot of energy to break it. That energy is the latent heat.

The system at unstable conditions (vapour at 370K, 1 atm) is like a "card castle" and we touch a card very slightly, the castle collapses, because is a unstable system. No energy gave, no latent heat.