Carnot Refrigerator at Phase Boundry

[Smacal1072]

Hi All,

My textbook mentions that the coefficient of refrigeration for a Carnot refrigerator acting between a hot reservoir [tex]T_{H}[/tex] and a cold reservoir [tex]T_{C}[/tex] is

[tex]K_{Carnot} = \frac{T_C}{T_H - T_C}[/tex]

This got me wondering - Suppose you had 1 kg of ice at it's melting point, and 1 kg of water at it's freezing point. So suppose the ice is [tex]T_{C}=0[/tex] Celsius and the water is [tex]T_H=0^{o}[/tex] Celsius.

Now suppose I have a Carnot fridge (I know they don't exist...), and I slowly heat the ice until all of it melts and slowly cool the water until all of it freezes. Since the temperature difference is zero during this process, the coefficient of refrigeration is infinity. Am I allowed to do this, expending almost no work?

 

[Andrew Mason]

Hi All,

My textbook mentions that the coefficient of refrigeration for a Carnot refrigerator acting between a hot reservoir [tex]T_{H}[/tex] and a cold reservoir [tex]T_{C}[/tex] is

[tex]K_{Carnot} = \frac{T_C}{T_H - T_C}[/tex]

This got me wondering - Suppose you had 1 kg of ice at it's melting point, and 1 kg of water at it's freezing point. So suppose the ice is [tex]T_{C}=0[/tex] Celsius and the water is [tex]T_H=0^{o}[/tex] Celsius.

Now suppose I have a Carnot fridge (I know they don't exist...), and I slowly heat the ice until all of it melts and slowly cool the water until all of it freezes. Since the temperature difference is zero during this process, the coefficient of refrigeration is infinity. Am I allowed to do this, expending almost no work?

You are not describing a thermodynamic cycle. What you are describing is a free flow of heat. I am not sure how you get the heat flowing from the water to the ice if there is no temperature difference. If I put the ice and water together in a thermos, does the ice melt? Does the water turn to ice? Why not?

AM

 

[Smacal1072]

Sorry I misspoke: In my scenario the ice should be the hot source, and the water should be the cold source.

I guess what I'm saying is that the coefficient of performance for an ideal refrigerator is infinity for two objects at the same temperature. If I'm understanding this right, it means I can hypothetically transfer an infinite amount of heat between two objects at the same temperature per unit of work I put in.

If my ice and water remain at the same temperature during a phase transition, then I can put in an arbitrarily small amount of work to completely freeze the water and melt the ice?

 

[Andrew Mason]

Sorry I misspoke: In my scenario the ice should be the hot source, and the water should be the cold source.

I guess what I'm saying is that the coefficient of performance for an ideal refrigerator is infinity for two objects at the same temperature. If I'm understanding this right, it means I can hypothetically transfer an infinite amount of heat between two objects at the same temperature per unit of work I put in.

If my ice and water remain at the same temperature during a phase transition, then I can put in an arbitrarily small amount of work to completely freeze the water and melt the ice?

So you want to run a Carnot refrigerator to extract thermal energy from the liquid water and put it into melting the ice. Since both reservoirs are at the same temperature, you say you could do this with no work being done.

First of all, the Carnot refrigerator is really just a limit. It can be approached arbitrarily closely but can never be reached - even in theory.

You have a situation in which you wish to have heat flowing from the liquid to the ice. In order to do that you have to have an infinitessimal temperature difference between the two. In other words, the ice has to be at a temperature that is slightly less than 0C for the heat to flow into it from the water at 0C. Or the water has to be slightly higher than 0C and the ice at 0C AT ALL TIMES. If that is the case, you cannot achieve the changes in phase that you are seeking.

Also, you could not create more ice with such a refrigerator. If you have a finite amount of water at 0C, as soon as you extact some thermal energy from it, part of it will turn to ice. And as soon as you add some thermal energy to the ice at 0C, some part of it will turn to water. So it does nothing more than preserve the same proportion of ice and liquid water. There is a net decrease in entropy if you are going to turn water into ice and end up with more ice. This means that some work must be done.

AM

 

[Smacal1072]

Thanks for your replies - I know I can't create more ice with such a refrigerator, it would basically just switch the state of the two objects.

I just find it interesting that theory says we can freeze all the water while simultaneously melting all the ice with so little work being done.

 

[Andrew Mason]

Thanks for your replies - I know I can't create more ice with such a refrigerator, it would basically just switch the state of the two objects.

I just find it interesting that theory says we can freeze all the water while simultaneously melting all the ice with so little work being done.

That is not exactly what the theory says can happen. You do not need a Carnot refrigerator here since there is not a finite temperature difference between the two reservoirs. You just have an isothermal heat flow from the water to the ice - that is to say a heat flow caused by an infinitessimal difference in temperature between the ice and the water.

The bottom line is that you have to have some temperature difference in order to have heat flow. If the ice and water are in thermal equilibrium, no heat will flow and no change will occur. So we have to assume a temperature difference between water and ice of dT.

Suppose the ice is set at temperature 0C-dT. In that case, the ice will never melt. Or, suppose that the water is set at temperature 0C + dT. In that case, the water will never freeze.

So the issue here is: how do you get the heat to flow from water to ice if there is not even an infinitessimal temperature difference for an isothermal heat transfer?

AM

 

[Smacal1072]

So the issue here is: how do you get the heat to flow from water to ice if there is not even an infinitessimal temperature difference for an isothermal heat transfer?

I'm totally with ya, heat only flows spontaneously when there is a temperature difference. No heat will spontaneously flow from the water to the ice if they are at the same temperature. Nothing is going to happen when I put these two bodies together and watch.

So, if I want to move heat from the 0 degree water to the 0 degree ice, I'm going to have to put a heat pump between the water and ice, and also put in some work to pump heat from the water to the ice.

What I found so surprising is that with an ideal heat pump, the number of Joules of work I have to put into this heat pump to move 1 Joule of heat from the water to the ice is:

[tex]\frac{1}{K_{Carnot}} = \frac{T_{H} - T_{C}}{T_{C}} = \frac{273.15 K - 273.15 K}{273.15 K} = 0 \mbox{ Joules}[/tex]

Of course ideal heat pumps are impossible, I just was interested at this result.

 

[Andrew Mason]

I'm totally with ya, heat only flows spontaneously when there is a temperature difference. No heat will spontaneously flow from the water to the ice if they are at the same temperature. Nothing is going to happen when I put these two bodies together and watch.

So, if I want to move heat from the 0 degree water to the 0 degree ice, I'm going to have to put a heat pump between the water and ice, and also put in some work to pump heat from the water to the ice.

And my point is that you cannot have a heat pump, even an ideal Carnot heat pump, operating without a temperature difference.

You can only pump heat between two reservoirs with a finite temperature difference. You can pump heat from the water to melt the ice, but the output reservoir will not be at 0C. It has to be something higher than 0C. Some work will always be required. Even in theory. Even with an ideal Carnot heat pump.

AM