How does statistical mechanics explain temperature?

[sgstudent]

Since heat is transferred from a region of high temperature to low temperature if a few molecules have more energy, it would be transferred to the surrounding molecules. So if a molecule has 10J of heat it can only transfer some energy such that they have the same temperatures as each other.

So how can heat be transferred to the surface if the heat should be dissipated such that the temperatures are equal, how can the temperature of the surface be hotter than the rest of the molecules?

Thanks for the help :)

 

[russ_watters]

Heat transfer takes time and it isn't completely uniform.

 

[Philip Wood]

... how can the temperature of the surface be hotter than the rest of the molecules?

What makes you think that the surface is hotter than the bulk of the liquid?

 

[sgstudent]

What makes you think that the surface is hotter than the bulk of the liquid?

Hi Philip Wood thanks for the reply

Because at the surface of the liquid is where the a tiny amount of water molecules reaches 100 degrees? But that doesn't make sense to me because shouldn't heat be transferred away before that much heat can be accumulated?

 

[AbsoluteZer0]

Because at the surface of the liquid is where the a tiny amount of water molecules reaches 100 degrees

A substance, water for example, doesn't need to be at 100°C for evaporation to take place. 100°C, for water at least, is the boiling point.

So how can heat be transferred to the surface if the heat should be dissipated such that the temperatures are equal, how can the temperature of the surface be hotter than the rest of the molecules?

It doesn't mean that all the molecules in the substance have an equivalent level of energy. Temperature is the average kinetic energy in the molecules of a substance.

 

[sgstudent]

A substance, water for example, doesn't need to be at 100°C for evaporation to take place. 100°C, for water at least, is the boiling point.



It doesn't mean that all the molecules in the substance have an equivalent level of energy. Temperature is the average kinetic energy in the molecules of a substance.

But don't some of those particles have to have enough energy to boil off such that they have the mcΔT+ml to evaporate away? So even if the average kinetic energy is the same, those particles with slightly higher kinetic energies would only have a limited amount of energy they can transfer to the other particles?

 

[Philip Wood]

Boiling is a very special case of evaporation. It involves bubbles of vapour forming below the surface of the water. The bubbles usually form on minute concavities on the inside of the vessel below the liquid level, or on lime-scale or other solids below the water level. [The water is evaporating into minute air pockets in these concavities.] The bubbles are able to grow, and then break away and rise to the surface, when the saturated vapour pressure inside them is equal to the external pressure on them. This pressure is that of the atmosphere (about 101000 Pa) plus the extra pressures (usually negligible) of the liquid column above them and of surface tension at the bubble surface. So at what temperature is the svp of water equal to 101000 Pa? 100°C !

Ordinary evaporation from the exposed surface of a liquid is the unco-ordinated escape of individual molecules. They do not have to fight against atmospheric pressure, except inasmuch as they will collide with the molecules of air once they have escaped. So ordinary evaporation does not require the svp of the liquid to equal atmospheric pressure, as it does not involve bubbles of vapour having to grow and escape through the liquid against atmospheric pressure. Therefore ordinary evaporation can - and does - occur at much lower temperatures than 100°C. Even clothing put out to dry on a washing line in the open air in temperatures below 0°C will dry, though very slowly!

 

[sgstudent]

Boiling is a very special case of evaporation. It involves bubbles of vapour forming below the surface of the water. The bubbles usually form on minute concavities on the inside of the vessel below the liquid level, or on lime-scale or other solids below the water level. [The water is evaporating into minute air pockets in these concavities.] The bubbles are able to grow, and then break away and rise to the surface, when the saturated vapour pressure inside them is equal to the external pressure on them. This pressure is that of the atmosphere (about 101000 Pa) plus the extra pressures (usually negligible) of the liquid column above them and of surface tension at the bubble surface. So at what temperature is the svp of water equal to 101000 Pa? 100°C !

Ordinary evaporation from the exposed surface of a liquid is the unco-ordinated escape of individual molecules. They do not have to fight against atmospheric pressure, except inasmuch as they will collide with the molecules of air once they have escaped. So ordinary evaporation does not require the svp of the liquid to equal atmospheric pressure, as it does not involve the escape of bubbles of vapour having to grow and escape through the liquid against atmospheric pressure.

Oh but could you explain how the particles gain enough energy to evaporate? Since heat transfers from a region of high temperature to a region of low temperature how would it apply to the heat transfer to allow evaporation? I'm having trouble understanding how the heat energy can accumulate at the surface of the liquid..

Thanks for the help :)

 

[Nugatory]

Oh but could you explain how the particles gain enough energy to evaporate? Since heat transfers from a region of high temperature to a region of low temperature how would it apply to the heat transfer to allow evaporation? I'm having trouble understanding how the heat energy can accumulate at the surface of the liquid.

Heat energy doesn't "accumulate" at the surface of the liquid.

Temperature measures the average speed of the molecules, so even when the temperature is uniform throughout the liquid, some of the molecules are moving a bit faster than average and others a bit slower. At the surface of liquid, if one of the "bit faster" ones happens to moving in an upwards direction, it can escape.

The effect is to slightly cool the surface; the faster-moving molecules are more likely to escape so the average energy of the ones left behind goes down. But as you say... heat moves from areas of high temperature to lower, so as the surface cools, heat flows from the warmer interior of the liquid and eliminates the temperature differential again.

 

[AbsoluteZer0]

I'm having trouble understanding how the heat energy can accumulate at the surface of the liquid..

Thanks for the help :)

If you put a glass of water out into the sunlight for example, the radiation from the sun can increase the energy in the water. Areas of the water that possesses greater average kinetic energy rise due to a process known as convection. Convection is heat transfer due to the movement of the substance itself.

When you heat water, for example, it expands and the density therefore decreases. As a consequence of this, it is able to rise to the surface.

 

[Philip Wood]

First: please don't talk about heat "accumulating". Heat is energy in transit from a region of higher temperature to a region of lower temperature. If you are considering a defined region, then the energy residing in that region is classed as internal energy.

Evaporation is, as I said in my last post, the unco-ordinated escape of molecules from the exposed surface of a liquid. At any instant, molecules in the liquid will have different kinetic energies. Their mean kinetic energy is usually roughly proportional to the Kelvin temperature. Most of the molecules in the liquid won't have enough energy to escape from the liquid surface, against the (shortish-range) attractive forces from other molecules in the liquid surface. But some molecules (in a short time interval) will have experienced a succession of 'lucky' hits from other molecules, and so will have acquired energies well above the mean. Even enough energy for them to escape (if they are in the surface)! This is evaporation!

 

[Chestermiller]

The driving force for evaporation is equal to the equilibrium vapor pressure of the liquid at the interface minus the partial pressure of the same substance in the gas phase. Where does the heat come from to supply the heat of evaporation? It comes from the liquid below and the gas above. The temperature at the interface drops a little below that of the bulk liquid. This causes heat to flow from the bulk liquid to the interface. The interface might also be cooler than the gas above. This too would result in heat flow toward the interface. The rate of heat flow from the liquid below and the gas above will equal the rate of evaporation times the heat of evaporation.

 

[sgstudent]

First: please don't talk about heat "accumulating". Heat is energy in transit from a region of higher temperature to a region of lower temperature. If you are considering a defined region, then the energy residing in that region is classed as internal energy.

Evaporation is, as I said in my last post, the unco-ordinated escape of molecules from the exposed surface of a liquid. At any instant, molecules in the liquid will have different kinetic energies. Their mean kinetic energy is usually roughly proportional to the Kelvin temperature. Most of the molecules in the liquid won't have enough energy to escape from the liquid surface, against the (shortish-range) attractive forces from other molecules in the liquid surface. But some molecules (in a short time interval) will have experienced a succession of 'lucky' hits from other molecules, and so will have acquired energies well above the mean. Even enough energy for them to escape (if they are in the surface)! This is evaporation!

Oh but I'm having trouble understanding this 'lucky' bump concept. Usually if I a large metal block at 100 degrees and a piece of wood at 20 degrees, the heat will transfer from the metal to the piece of wood until both their temperature is the same and there would not be any heat transfer anymore.

So when i apply this to the evaporation, it wouldn't make sense the surface molecules could gain more heat than the rest of the other molecules around it such that they are able to break free from the bulk of the liquid. i think i have a misconception here, but i can't seem to understand it.

Thanks for the help :)

 

[AbsoluteZer0]

Oh but I'm having trouble understanding this 'lucky' bump concept. Usually if I a large metal block at 100 degrees and a piece of wood at 20 degrees, the heat will transfer from the metal to the piece of wood until both their temperature is the same and there would not be any heat transfer anymore.

So when i apply this to the evaporation, it wouldn't make sense the surface molecules could gain more heat than the rest of the other molecules around it such that they are able to break free from the bulk of the liquid. i think i have a misconception here, but i can't seem to understand it.

Thanks for the help :)

I believe your primary misconception regards you're assumption that once the molecules reach thermal equilibrium, heat transfer comes to a stand still and all molecules in the water possesses the same quantity of energy. If a sample of water has 10j of energy and has 5 molecules it may be divided as:

-2j, 2j, 2j, 2j, 2j
-2j, 2j, 2j, 3j, 1j

and so forth.

What Phillip Wood meant by 'lucky bump' was that some molecules may have undergone a collision with other molecules that provided them with sufficient energy to vaporize. Heat transfer of this form is known as Conduction. Conductionis heat transfer due to collision with other particles.

Water molecules are attracted to each other by intermolecular forces known as the Van Der Waals forces. More specifically, in this case it is the hydrogen bond between the oxygen in one molecule and the hydrogen in the other.

H₂O---H₂O (the --- represents the hydrogen bond.)

If you provide sufficient energy you will increase molecular vibrations to an extent where the Van Der Waals forces can be overcome. Vaporization will take place as a result.

Info on Van Der Waals Forces:
http://www.physlink.com/education/askexperts/ae206.cfm

 

[sgstudent]

I think your primary misconception is that your assuming that once the molecules reach thermal equilibrium, heat transfer comes to a stand still and all molecules in the water possesses the same quantity of energy. If a sample of water has 10j of energy and has 5 molecules it may be divided as:

-2j, 2j, 2j, 2j, 2j
-2j, 2j, 2j, 3j, 1j

and so forth.

What Phillip Wood meant by 'lucky bump' was that some molecules may have undergone a collision that provided them with sufficient energy to vaporize.

Water molecules are attracted to each other by intermolecular forces known as the Van Der Waals forces. More specifically, in this case it is the hydrogen bond between the oxygen in one molecule and the hydrogen in the other.

H₂O---H₂O (the --- represents the hydrogen bond.)

If you provide sufficient energy you will increase molecular vibrations to an extent where the Van Der Waals forces can be overcome. Vaporization will take place as a result.

Oh, so it's possible for a particle to gain a lot more energy than the rest allowing it to evaporate?

So if i have 5 molecules and for one molecule to evaporate it must have 5J of heat while the total energy of 10J

So they can split this way: 5J 1.25J 1.25J 1.25J 1.25J? If so why won't it work like this, shouldn't it be similar in the manner that once temperature is equalized than there should not be any transfer of energy anymore?

 

[AbsoluteZer0]

Oh, so it's possible for a particle to gain a lot more energy than the rest allowing it to evaporate?

Yes.

E_k = \frac{1}{2} mv^2

What this equation states is that Kinetic energy is equivalent to half of the mass times the square of the velocity. When molecules collide with each other, they can either increase in velocity or decrease, changing their kinetic energy.

why won't it work like this, shouldn't it be similar in the manner that once temperature is equalized than there should not be any transfer of energy anymore?

Molecules are always in motion and are always colliding with each other. There will always be a difference in kinetic energy between most molecules in a substance. One molecule in a substance may have the same temperature as another, but this commonality would be very short lived.

 

[sgstudent]

Yes.

E_k = \frac{1}{2} mv^2

What this equation states is that Kinetic energy is equivalent to half of the mass times the square of the velocity. When molecules collide with each other, they can either increase in velocity or decrease, changing their kinetic energy.



Molecules are always in motion and are always colliding with each other. There will always be a difference in kinetic energy between most molecules in a substance. One molecule in a substance may have the same temperature as another, but this commonality would be very short lived.

Oh I think i understand it now :) to confirm my understanding the overall temperature will be the same even when some particles have more KE than others. Since the average KE remains the same its possible for certain molecules to gain a lot more KE than the rest allowing them to escape.

However, usually we would say that the temperature of the liquid would drop after evaporation. Would the explanation be something like this: after they transfer some KE to the surface molecules they have a smaller KE, hence once those surface molecules escape to form a gas the average KE of the liquid decreases and hence the temperature drops?

 

[AbsoluteZer0]

to confirm my understanding the overall temperature will be the same even when some particles have more KE than others.

So long as their kinetic energies average to get the specified temperature, yes.

However, usually we would say that the temperature of the liquid would drop after evaporation. Would the explanation be something like this: after they transfer some KE to the surface molecules they have a smaller KE, hence once those surface molecules escape to form a gas the average KE of the liquid decreases and hence the temperature drops?

Exactly. The molecules left behind have a lesser kinetic energy than those that vaporized.

 

[sgstudent]

So long as their kinetic energies average to get the specified temperature, yes.



Exactly.

Hi thanks for the help

What about melting could it be possible for those lucky bumps to occur as well allowing it to turn into a liquid?

 

[nasu]

Hi thanks for the help

What about melting could it be possible for those lucky bumps to occur as well allowing it to turn into a liquid?

A solid can evaporate as well, at temperatures well below the melting point.
See "sublimation".

 

[sgstudent]

A solid can evaporate as well, at temperatures well below the melting point.
See "sublimation".

But what about turning into a liquid before its melting point? Like how the liquid turns into a gas before it reaches its boiling point.

Thanks for the help :)

 

[Lsos]

I'm no expert in this subject, but wanted to put my 2 cents into it. When a solid or liquid turns into a gas, the molecules escape (along with their energy) and are lost forever. I'm not sure about them turning into liquid (can one molecule even ever be considered a liquid?) but even if it does happen, it will, by definition, not "evaporate" away. It will instead remain on the surface, where it will turn back into solid almsot immediately.

 

[jbriggs444]

But what about turning into a liquid before its melting point? Like how the liquid turns into a gas before it reaches its boiling point.

Consider the surface of water in a covered pot simmering on the stove. At boiiling temperature the rate of evaporation of water into the saturated vapor is equal to the rate of condensation of the saturated vapor back into the water. There is an equilibrium.

In the case of your clothing drying on the line in sub-zero temperatures the partial pressure of water vapor in the air is very low. So low that the boiling temperature of water at that pressure would be even further below zero. There is no equililibrium. The clothes dry out.

The point here is that the "boiling point" of a liquid is a function of pressure. In some sense, a liquid does not turn into a gas before it reaches its boiling point.

In the case of the transition between water and ice there is some pressure dependency. It is possible to melt ice by applying pressure rather than by increasing temperature. It is also possible to melt ice by applying salt. Also, some solids soften as their melting temperature is approached.

 

[sgstudent]

I'm no expert in this subject, but wanted to put my 2 cents into it. When a solid or liquid turns into a gas, the molecules escape (along with their energy) and are lost forever. I'm not sure about them turning into liquid (can one molecule even ever be considered a liquid?) but even if it does happen, it will, by definition, not "evaporate" away. It will instead remain on the surface, where it will turn back into solid almsot immediately.

Oh would the reason it turns into a solid immediately be that since it is at a liquid state where it does not have enough energy to escape like a gas. So it's own kinetic energy would be transferred to the other particles causing it to solidify again?

 

[sgstudent]

Yes.

E_k = \frac{1}{2} mv^2

What this equation states is that Kinetic energy is equivalent to half of the mass times the square of the velocity. When molecules collide with each other, they can either increase in velocity or decrease, changing their kinetic energy.



Molecules are always in motion and are always colliding with each other. There will always be a difference in kinetic energy between most molecules in a substance. One molecule in a substance may have the same temperature as another, but this commonality would be very short lived.

Hi AbsoluteZer0, I was thinking about the evaporation and when those few molecules gain enough kinetic energy and escape as a gas, what would their temperatures be? When they were still in a liquid state despite having more KE, the temperature still remained the same. Once they leave the liquid to becomes a gas, how will their temperatures considered to be?

Thanks for the help :)

 

[Lsos]

Oh would the reason it turns into a solid immediately be that since it is at a liquid state where it does not have enough energy to escape like a gas. So it's own kinetic energy would be transferred to the other particles causing it to solidify again?

That's what I'm thinking. I didn't read this anywhere though, just arriving to conclusons...

 

[AbsoluteZer0]

When they were still in a liquid state despite having more KE, the temperature still remained the same

Are you talking about the temperature of the whole system or the temperature of the individual molecules?

 

[sgstudent]

Are you talking about the temperature of the whole system or the temperature of the individual molecules?

I guess it would be those individual atoms? Since they have a greater kinetic energy so will they be considered to be at 100 degrees?

 

[AbsoluteZer0]

I guess it would be those individual atoms? Since they have a greater kinetic energy so will they be considered to be at 100 degrees?

They wouldn't necessarily be at 100°C. Evaporation can take place at temperatures lower than the boiling point.

http://en.wikipedia.org/wiki/Temperature#Statistical_mechanics_approach_to_temperature

Classical thermodynamics concerns mostly the macroscopic (systems as a whole.)
As a consequence, it's inefficient to discuss molecules in terms of their thermal temperature. It's easier to discuss their statistical temperature. Statistical Thermodynamics deals with large populations of particles.

http://en.wikipedia.org/wiki/Statistical_mechanics