# Why does melting and boiling occur at a specific temperature?

by sgstudent
Tags: boiling, melting, occur, specific, temperature
 P: 440 Melting does not happen at a specific temperature. Except when special conditions are met: 1: Slow heating 2: Stuff that is heated must not be asphalt or butter or other similar stuff.
PF Gold
P: 1,113
 Quote by jartsa Melting does not happen at a specific temperature. Except when special conditions are met: 1: Slow heating
As I mentionned earlier, we are of course considering that the system is considered to always be in internal equilibrium. That said, even if not all parts of the system are at the same temperature, it is those parts where the melting temperature is reached that will melt.

 Quote by jartsa 2: Stuff that is heated must not be asphalt or butter or other similar stuff.
We are obviously talking only about pure substances, not mixtures.
 P: 836 At a microscopic level, molecules are constantly evaporating and condensing. These are reverse reactions. The concept of boiling only exists at a larger scale, since it is a thermodynamics concept, which deals with statistical averages. When the temperature is raised, the rate of evaporation increases since more molecules have the energy to escape the surface. The rate of condensation increases with the partial pressure of the vapor since more pressure = more molecules of the vapor hitting the surface and getting stuck. In a closed vessel, the liquid and vapor will reach some kind of equilibrium. But in an open vessel, the atmosphere is assumed to be an infinite reservoir. Depending on atmospheric humidity, which depends on the partial pressure of the vapor and the temperature, the liquid will eventually evaporate completely or the vapor will condense as dew. You don't have to reach the boiling point for all the liquid to evaporate away. But evaporation only happens at the surface, so it is a relatively slow process. Something special happens at the boiling point. This is the point where the pressure of the atmosphere is low enough that the vapor pressure matches the atmospheric pressure. Slightly above this point, bubbles can expand in the liquid because the gas generated by the reaction has enough pressure to push the liquid up and out of the way. Therefore, evaporation can occur throughout a large volume of the liquid, not just the surface. If the liquid is in a tall vessel, the bottom of the liquid is at a higher pressure than the top of the liquid and requires a higher temperature for the bubbles to grow. Someone correct me if I'm wrong: I think the dew point and the boiling point are essentially the same thing when the atmosphere consists completely of one type of vapor. The melting point conceptually similar to the boiling point, except that it is less sensitive to pressure since the density of the liquid phase and solid phase are similar. You can still have a concept analogous to partial pressure in the liquid if it's a mixture (like salt water). I suppose there should be two kinds of melting points for a mixed liquid: one analogous to the dew point and one analogous to the boiling point. But I might be totally wrong on this.
P: 440
 Quote by jartsa Melting does not happen at a specific temperature. Except when special conditions are met: 1: Slow heating 2: Stuff that is heated must not be asphalt or butter or other similar stuff.
 Quote by DrClaude We are obviously talking only about pure substances, not mixtures.

2: Stuff that is heated must not be asphalt or butter or other similar stuff.
Neither may it be glass (silicon dioxide, not a mixture).
But bronze, which is a mixture of copper and tin, it may be.
Also water has a sharp freezing point, although lot of energy is released when water molecules start sticking more tighly togerher, when water is cooled from 10 C to 5 C, for example.
P: 637
 Quote by DrClaude What does it mean to have the same kinetic energy? Are you thinking that all molecules in a gas at a given temperature have the same speed? Because they don't. Think of two classical atoms held together by a spring, just below the breaking point. The system has both PE and KE. Add an epsilon of energy, the bond breaks, at your are left with only KE. So the total energy, which included a PE component, is now converted to KE. You have an equilibrium between the two. Molecules are constantly going from one phase to the other. If you start with a solid and increase its temperature to exactly the melting point, you still have a solid. Any additional energy you add will be used to overcome the strong attraction in the solid and put molecules in the liquid phase. Stop adding energy, and the mixture of solid and liquid will keep the same proportions, but some molecules are constantly going from one phase to the other at the interface. Continue adding energy and the fraction of liquid will increase up to the point where it is all liquid, still at the melting temperature. Additional energy will then increase the temperature of the liquid.
Hi thanks again for the response :)

I was thinking the kinetic energy of the particles would be about the same not speed. So the 1/2mv2 would be roughly the same? And that value would be proportional to the temperature. But u guess that's a wrong way to look at it.

So for the liquid to gas case, the KE of the gas is greater than the liquid at the same temperature?

Lastly, I think I'm understanding it already. But I was wondering when the ice has partially melted can the water part rise is temperature or are we assuming that thermal equilibrium is always maintained so if the water gains any excess heat it would immediately transfer it to the ice such that the temperatures are always fixed at Tm?

Thanks so much for the help :)
PF Gold
P: 1,113
 Quote by sgstudent I was thinking the kinetic energy of the particles would be about the same not speed. So the 1/2mv2 would be roughly the same? And that value would be proportional to the temperature. But u guess that's a wrong way to look at it.
Look up the Maxwell-Boltzmann distribution.

 Quote by sgstudent So for the liquid to gas case, the KE of the gas is greater than the liquid at the same temperature?
Yes, average KE for a gas is higher because it has less PE.

 Quote by sgstudent Lastly, I think I'm understanding it already. But I was wondering when the ice has partially melted can the water part rise is temperature or are we assuming that thermal equilibrium is always maintained so if the water gains any excess heat it would immediately transfer it to the ice such that the temperatures are always fixed at Tm?
Yes, such a theoretical description assumes internal equilibrium: exchange of heat within the system is much faster than exchange of heat with the environment.

Of course, reality is hardly ever ideal. If you have an open container that is heated on all sides, then that container can heat up directly the gas once boiling has started, and you can get the gas hotter than the liquid. Likewise, since it is open some gas molecules will escape the container, and they tend to be the ones with the most energy (those going the fastest). In a previous post, Khashishi also pointed out some other things to consider for a real system.
P: 637
 Quote by DrClaude Look up the Maxwell-Boltzmann distribution. Yes, average KE for a gas is higher because it has less PE. Yes, such a theoretical description assumes internal equilibrium: exchange of heat within the system is much faster than exchange of heat with the environment. Of course, reality is hardly ever ideal. If you have an open container that is heated on all sides, then that container can heat up directly the gas once boiling has started, and you can get the gas hotter than the liquid. Likewise, since it is open some gas molecules will escape the container, and they tend to be the ones with the most energy (those going the fastest). In a previous post, Khashishi also pointed out some other things to consider for a real system.
Hi thanks for the reply :)

I read about this before and it shows that some molecules have more KE than others but most have a moderate amount while few has either a lot or very little KE?

So the gas at the same temperature of boiling point has a greater KE than the liquid but does the gas have a greater total internal energ? If so does it mean now the KE the gas has is like the KE of the liquid +PE of liquid + extra energy from the heating?

Oh I think I understand the Gibbs free energy explanation better now but as I was reading my book, it shows that dG=0 so dH=TdS and for dH and dS they used the dH naught and dS naught to solve for T. But I thought when dG=0 the composition of solid and liquid is varied? Like it can be 100% solid and 0% liquid so how can we use the dH naught and dS naught which assumes 100% change?

or is it because even at 100% liquid and 0% solid the dG can still be zero as long as we do not increase the temperature and let it stay at melting point then the dG would still be zero? But in this case when we use the dGo=-RTlnKeq to solve for our Keq our Keq would be a constant but at 100% solid isnt the Keq 0?

thanks so much for the help :)
PF Gold
P: 1,113
 Quote by sgstudent So the gas at the same temperature of boiling point has a greater KE than the liquid but does the gas have a greater total internal energ? If so does it mean now the KE the gas has is like the KE of the liquid +PE of liquid + extra energy from the heating?
The gas has to have slightly higher total energy, coming from the breaking of the bonds between the molecules. But the difference will be small.

 Quote by sgstudent Oh I think I understand the Gibbs free energy explanation better now but as I was reading my book, it shows that dG=0 so dH=TdS and for dH and dS they used the dH naught and dS naught to solve for T. But I thought when dG=0 the composition of solid and liquid is varied? Like it can be 100% solid and 0% liquid so how can we use the dH naught and dS naught which assumes 100% change? or is it because even at 100% liquid and 0% solid the dG can still be zero as long as we do not increase the temperature and let it stay at melting point then the dG would still be zero? But in this case when we use the dGo=-RTlnKeq to solve for our Keq our Keq would be a constant but at 100% solid isnt the Keq 0?
I'm sorry, but I don't undertsand what you mean here.
P: 637
 Quote by DrClaude The gas has to have slightly higher total energy, coming from the breaking of the bonds between the molecules. But the difference will be small. I'm sorry, but I don't undertsand what you mean here.