Suppose i have a bucket full of water in a closed room,the water is

[nouveau_riche]

suppose i have a bucket full of water in a closed room,the water is kept undisturbed for say 5 days,after those 5 days i found that the temperature of water was lower than as it was 5 days before and so it does for the room
all this (for the system) isn't going in accordance with second law of thermodynamics
where i am wrong(if i am)?

 

[timthereaper]

So where did the heat go from the water? Wherever that heat went, the entropy increased there more than the loss of entropy in the water.

The SLOT is more than just a good idea; it's the law.

 

[Avijeet]

I believe the temperature of the room will increase. As you mentioned it a closed room and the heat lost by the bucket of water has to go in increasing in temperature of the room. This would be in accordance with the second law as now the energy is distributed among the enormously large (compared to that of water in the bucket) degrees of freedom of the gas molecules of the room.

 

[Drakkith]

Your system is not closed. The heat from the water was absorbed into the room which was then absorbed into the outside. Nothing here disagrees with thermodynamics.

 

[timthereaper]

Oh, and something else to think about:

1) Closing the door to the room doesn't close the system.
2) Walking away from the system and leaving it alone doesn't mean the system is isolated.

 

[Dale]

suppose i have a bucket full of water in a closed room,the water is kept undisturbed for say 5 days,after those 5 days i found that the temperature of water was lower than as it was 5 days before and so it does for the room
all this (for the system) isn't going in accordance with second law of thermodynamics
where i am wrong(if i am)?

If the room is truly isolated thermodynamically, and if the bucket and the dry air were initially at the same temperature then you would indeed find that the bucket of water and the air were at a lower temperature 5 days later.

So, since we know that the SLOT is correct, how did entropy increase in a situation where the temperature decreased and no heat was exchanged with the surroundings?

 

[nouveau_riche]

If the room is truly isolated thermodynamically, and if the bucket and the dry air were initially at the same temperature then you would indeed find that the bucket of water and the air were at a lower temperature 5 days later.

So, since we know that the SLOT is correct, how did entropy increase in a situation where the temperature decreased and no heat was exchanged with the surroundings?

that is what i am asking for

 

[Dale]

Yes, I know. Think about it a bit and try to figure out the answer. Make a guess under the assumption that physicists have encountered buckets of water and designed the SLOT accordingly. You might surprise yourself and find out that you know physics better than you suspect.

 

[Andrew Mason]

Try first applying FLOT - the first law of thermodynamics. That will allow you to determine Q: the heat flow from/to the room and the heat flow from/to the water

How is the change in entropy related to this heat flow and temperature? Once you figure out the heat flows you can work out the entropy change and perhaps then you will be able to see how FLOT fits into SLOT.

AM

 

[russ_watters]

C'mon, this is an easy one: what does water do in dry air?

 

[Andrew Mason]

C'mon, this is an easy one: what does water do in dry air?

And the issue is: what is the heat flow that is required in order for it to do that and where does it come from?

AM

 

[nouveau_riche]

C'mon, this is an easy one: what does water do in dry air?

if the water molecules go into the air,the flow of heat takes place,then this heat must increase room temperature?

 

[Drakkith]

Look up Phase Change on wikipedia.

 

[Dale]

if the water molecules go into the air

Excellent. Yes, the water molecules go into the air. And what about entropy, what has more entropy, a water molecule in the bucket or in the air?

 

[nouveau_riche]

Excellent. Yes, the water molecules go into the air. And what about entropy, what has more entropy, a water molecule in the bucket or in the air?

obviously the molecules in air,but the point is the downfall in temperature of room?

 

[Andrew Mason]

obviously the molecules in air,but the point is the downfall in temperature of room?

What direction is the heat flow relative to the water when it vaporizes? What does that tell you about the sign of the change in entropy of the water? What is the direction of heat flow relative to the air in the room? What does that tell you about the sign of the change in entropy of the air? What is the temperature of the water and room while this heat flow is occurring? What does that tell you about the relative magnitudes of the two entropy changes?

AM

 

[Dale]

obviously the molecules in air

OK, so you now understand how the second law of thermo and entropy work in this situation, correct?

but the point is the downfall in temperature of room?

The first thing to consider is the KE of a water molecule in the liquid phase at room temperature and the KE of a water molecule in the vapor phase at room temperature. Clearly, even though they are at the same temperature, the KE of the vapor is much higher than the KE of the liquid. This energy is called the heat of vaporization. Because energy is going from the liquid into the vapor phase, the temperature of the liquid phase must decrease.

As long as the air is dry there will be more water molecules going from the liquid to the vapor phase than vice versa. As long as that happens, the temperature of the liquid must decrease in order to supply the heat of vaporization.

 

[nouveau_riche]

OK, so you now understand how the second law of thermo and entropy work in this situation, correct?

The first thing to consider is the KE of a water molecule in the liquid phase at room temperature and the KE of a water molecule in the vapor phase at room temperature. Clearly, even though they are at the same temperature, the KE of the vapor is much higher than the KE of the liquid. This energy is called the heat of vaporization. Because energy is going from the liquid into the vapor phase, the temperature of the liquid phase must decrease.

As long as the air is dry there will be more water molecules going from the liquid to the vapor phase than vice versa. As long as that happens, the temperature of the liquid must decrease in order to supply the heat of vaporization.

okay,suppose at moment say 't' the temperature of room is "T",this is the state of initial equilibrium,the water molecules can vaporize at this temperature,they do so by increasing the K.E of the dry air,this will ,however, can still allow the flow of heat from air to water which cannot happen because the temperature available cannot condense it
it's like the system is disturbed from the initial equilibrium to throw away heat into air?

 

[nouveau_riche]

What direction is the heat flow relative to the water when it vaporizes? What does that tell you about the sign of the change in entropy of the water? What is the direction of heat flow relative to the air in the room? What does that tell you about the sign of the change in entropy of the air? What is the temperature of the water and room while this heat flow is occurring? What does that tell you about the relative magnitudes of the two entropy changes?

AM

well i guess you get your answer in my last post to dalespam

 

[Drakkith]

The water molecules "colliding" with the surface of the water can lose their energy to the liquid and become liquid again themselves. In equilibrium the amount of water evaporating and condensing back into the water is equal over time. It is only when you factor in the average amount of molecules evaporating vs the ones condensing that you see this. A snapshot could show that either the air or the water has more or less energy that it does a second later.

 

[Dale]

okay,suppose at moment say 't' the temperature of room is "T",this is the state of initial equilibrium,the water molecules can vaporize at this temperature,they do so by increasing the K.E of the dry air,this will ,however, can still allow the flow of heat from air to water which cannot happen because the temperature available cannot condense it

OK, this is basically just expanding on Drakkith's response, with a little bit of handwaving stastical mechanics for detail.

At any moment consider the molecules of water in the bucket. They are all jiggling around and bouncing into and off of each other. They have an average speed which determines the temperature, but they do not all have exactly that speed, there is a statistical distribution of speeds. A very small fraction of the molecules have enough speed and are at the surface going in the correct direction to leave the liquid phase and turn to vapor. Once a high-speed molecule leaves, the remaining liquid has a lower average speed and therefore a lower temperature. The liquid phase has transferred thermal energy to the vapor phase.

The reverse also happens. At any given time some small fraction of the vapor molecules in the air are in the right place traveling in the right direction and may be "captured" by the water. These fast-moving molecules, once captured, increase the average speed and therefore raise the temperature of the liquid. The vapor phase has transferred thermal energy to the liquid phase.

At equilibrium there is enough vapor in the air (partial pressure) that the same number of molecules go each direction and there is no net transfer of either molecules or energy. If the air is wet then there will be more condensation than evaporation and the liquid will have a net gain in both molecules and energy. If the air is dry then there will be more evaporation than condensation and the liquid will have a net loss in both molecules and energy.

it's like the system is disturbed from the initial equilibrium to throw away heat into air?

The point is that the system is NOT initially in equilibrium because the air is dry. Therefore there are more water molecules vaporizing than condensing.

 

[nouveau_riche]

OK, this is basically just expanding on Drakkith's response, with a little bit of handwaving stastical mechanics for detail.

At any moment consider the molecules of water in the bucket. They are all jiggling around and bouncing into and off of each other. They have an average speed which determines the temperature, but they do not all have exactly that speed, there is a statistical distribution of speeds. A very small fraction of the molecules have enough speed and are at the surface going in the correct direction to leave the liquid phase and turn to vapor. Once a high-speed molecule leaves, the remaining liquid has a lower average speed and therefore a lower temperature. The liquid phase has transferred thermal energy to the vapor phase.

The reverse also happens. At any given time some small fraction of the vapor molecules in the air are in the right place traveling in the right direction and may be "captured" by the water. These fast-moving molecules, once captured, increase the average speed and therefore raise the temperature of the liquid. The vapor phase has transferred thermal energy to the liquid phase.

At equilibrium there is enough vapor in the air (partial pressure) that the same number of molecules go each direction and there is no net transfer of either molecules or energy. If the air is wet then there will be more condensation than evaporation and the liquid will have a net gain in both molecules and energy. If the air is dry then there will be more evaporation than condensation and the liquid will have a net loss in both molecules and energy.

i get you,but the problem is that you are allowing the flow of heat from air to water by moving the molecules in water ,refer to your second paragraph,can that happen?

 

[Dale]

i get you,but the problem is that you are allowing the flow of heat from air to water by moving the molecules in water ,refer to your second paragraph,can that happen?

Yes, it can and it does.

 

[nouveau_riche]

Yes, it can and it does.

okay,even if it does there isn't a chance that could make room temperature to go down together with water,suppose initially water posses high temperature than air,the heat will flow from water to air,there will then be a rise in temperature of air by that very amount,also (as u say)some of water molecules will go back into water,this will increase the entropy of water molecules again,they will then gain some heat,the process follows till the flow of heat is equal(a stage of thermal equilibrium)

however the entropy of water molecules in air is more than in water,so the amount of heat delivered to the air will be more than that of water,so the average heat that might have flown until an equilibrium is established should have raised the temperature of air

but what i found was that both "the room and water was cooler than the air outside the room"

 

[Dale]

suppose initially water posses high temperature than air

No, let's stick with one scenario and not add additional variables unless you want to do a detailed mathematical computation.

but what i found was that both "the room and water was cooler than the air outside the room"

Exactly. Therefore we know that there is at least one error in your reasoning. Can you go back and spot it?

PS Here is a hint: If A is an array of the integers from 0 to 100 and B is an array of the numbers from 50 to 150 then how does the average of A change and how does the average of B change if we move the 100 from A to B?

 

[Dale]

Hi nouveau_riche, sorry about this, but my hint is completely misleading and should be ignored:

PS Here is a hint: If A is an array of the integers from 0 to 100 and B is an array of the numbers from 50 to 150 then how does the average of A change and how does the average of B change if we move the 100 from A to B?

The gas and vapor phases have the same temperature and therefore the same average KE
The important part in the phase change is not a difference in average KE between the liquid and vapor phases, but a difference in the intermolecular forces. Molecules in the liquid phase have strong intermolecular forces (that is what keeps them so densely packed) but molecules in the vapor phase are much further apart so the intermolecular forces are much weaker. This intermolecular force is a potential, meaning that the gas molecules have higher potential energy than the liquid molecules.

When a molecule leaves the liquid phase it must have enough KE to overcome the intermolecular forces, exchanging KE for PE. So during evaporation the temperature of the liquid goes down because it is preferentially losing high KE molecules, and the temperature of the gas does not go up because the molecules exchange KE for PE and thus don't increase the average gas KE.

 

[nouveau_riche]

Hi nouveau_riche, sorry about this, but my hint is completely misleading and should be ignored:
The gas and vapor phases have the same temperature and therefore the same average KE
The important part in the phase change is not a difference in average KE between the liquid and vapor phases, but a difference in the intermolecular forces. Molecules in the liquid phase have strong intermolecular forces (that is what keeps them so densely packed) but molecules in the vapor phase are much further apart so the intermolecular forces are much weaker. This intermolecular force is a potential, meaning that the gas molecules have higher potential energy than the liquid molecules.

When a molecule leaves the liquid phase it must have enough KE to overcome the intermolecular forces, exchanging KE for PE. So during evaporation the temperature of the liquid goes down because it is preferentially losing high KE molecules, and the temperature of the gas does not go up because the molecules exchange KE for PE and thus don't increase the average gas KE.

that demands constant average entropy,the basic says that the energy of molecule going into air is the seed of entropy being added

or how is entropy of molecule in air more than in water?

 

[arka.12]

You have mentioned that the room was closed. though it was closed it was not a absolutely closed system..... The tem. of the room has to increased a little bit as the tem of the bucket has falllen down......

 

[Dale]

that demands constant average entropy,the basic says that the energy of molecule going into air is the seed of entropy being added

The total entropy is not constant while the water is evaporating, it is increasing. The entropy only becomes constant once equilibrium is reached.

or how is entropy of molecule in air more than in water?

Here is a page I like for the entropy of a molecule of an ideal gas:
http://hyperphysics.phy-astr.gsu.edu/hbase/therm/entropgas.html

Unfortunately, I don't have a similar page for liquids, but you can look at tables and find that the standard molar entropy for liquid water is 70 J/mol/K and for water vapor is 189 J/mol/K.
http://en.wikipedia.org/wiki/Water_(data_page )

 

[nouveau_riche]

The total entropy is not constant while the water is evaporating, it is increasing. The entropy only becomes constant once equilibrium is reached.

Here is a page I like for the entropy of a molecule of an ideal gas:
http://hyperphysics.phy-astr.gsu.edu/hbase/therm/entropgas.html

Unfortunately, I don't have a similar page for liquids, but you can look at tables and find that the standard molar entropy for liquid water is 70 J/mol/K and for water vapor is 189 J/mol/K.
http://en.wikipedia.org/wiki/Water_(data_page )

well i am not much into mathematical physics,the explanation i had is that the inter molecular forces in water are more tighter,so the energy imparted to the molecules in air after the collision will create more disorder ,but my point is the energy spend in increasing that entropy,however ,will be conserved
so why does the lowering in temperature of room and water both after five days ?