Cylinder in a heat bath @ 0C contains water and ice & is compressed

[zemaitistrys]

Homework Statement

A cylinder is fitted with a piston and is in thermal contact with a heat bath at
273 K. Initially the volume in the cylinder is filled with 10 kg of pure H2O and about
half of this is liquid and the other half is solid. The piston is lowered so as to reduce
the volume by 2 × 10−5 m3. What is the sign and magnitude of the heat transfer to the
bath?
[Density of liquid water 1.0 × 103 kg m−3; density of ice 0.92 × 103 kg m−3; latent heat
of fusion of ice 333.0 kJ kg−1.]

Relevant Equations

First law of thermodynamics.

You can use the change in volume to find how much ice turned into water, and then find the energy required to melt that ice - that I have no problems with.

But then work done by piston = change in internal energy of ice/water combination (which we found) + heat released to the bath.

And we don't know the work done...

 

[Chestermiller]

I think you are supposed to assume that the pressure doesn't change much from 1 atm.

 

[zemaitistrys]

I think you are supposed to assume that the pressure doesn't change much from 1 atm.

In which case work would just be change in volume * 1 atm, right?

I guess it just appeared counter-intuitive that you may assume p to be 1 atm if the cyllinder compresses in the first place.

 

[Chestermiller]

In which case work would just be change in volume * 1 atm, right?

I guess it just appeared counter-intuitive that you may assume p to be 1 atm if the cyllinder compresses in the first place.

Well, you’re assuming that the specific volumes are constant, right?

 

[Steve4Physics]

What is the sign and magnitude of the heat transfer to the bath?

Hi. Can you answer these questions?

1. If there are both water and ice always present in the mix, does the temperature change if some of the ice melts (or if some of the water freezes)?

2. For the given figures, are both ice and water still present in the final state?

3. Do you require a temperature difference for net heat-transfer between the mix and the heat-bath?

 

[Chestermiller]

Hi. Can you answer these questions?

1. If there are both water and ice always present in the mix, does the temperature change if some of the ice melts (or if some of the water freezes)?

No. The ice and water are in contact with a constant temperature bath.

2. For the given figures, are both ice and water still present in the final state?

Yes

3. Do you require a temperature difference for net heat-transfer between the mix and the heat-bath?

Any small non-uniform changes in temperature of the system will induce small heat exchanges with the bath (to cancel these changes), and if the volume change is carried out very slowly, we need not be concerned with the rate of heat transfer (and still obtain significant net heat transfer).

 

[Steve4Physics]

If there are both water and ice always present in the mix, does the temperature change if some of the ice melts (or if some of the water freezes)?

No. The ice and water are in contact with a constant temperature bath.

Hi. I agree the answer is 'no, as you say. But this has nothing to do with the constant temperature bath. An ice-water mix intrinsically has a constant temperature (0ºC). Temperature remains constant during a phase change. (For example. that's why melting ice is used in many experiments requiring a constant temperature of 0ºC to be maintained over time.)

A simple calculation shows not all the ice get melted in this problem.

Knowing that the ice-water mix is always at the same temperature as the constant temperature bath, we can answer the original question with no further calculation.

 

[Chestermiller]

Hi. I agree the answer is 'no, as you say. But this has nothing to do with the constant temperature bath. An ice-water mix intrinsically has a constant temperature (0ºC). Temperature remains constant during a phase change. (For example. that's why melting ice is used in many experiments requiring a constant temperature of 0ºC to be maintained over time.)

A simple calculation shows not all the ice get melted in this problem.

Knowing that the ice-water mix is always at the same temperature as the constant temperature bath, we can answer the original question with no further calculation.

I don't disagree with any of this. Isn't that what I implied by my answers?

 

[Steve4Physics]

I don't disagree with any of this. Isn't that what I implied by my answers?

I thought you implied that the constant temperature was due to the constant temperature bath (CTB). I disagreed with this: I believe that the CTB has nothing whatsoever to do with the constant temperature of the ice-water mix.

If I am correct, the final answer is simply 'zero' because the mix and the CTB happen to be at the same temperature.

Apologies if I misinterpreted what you wrote.

 

[Chestermiller]

I thought you implied that the constant temperature was due to the constant temperature bath (CTB). I disagreed with this: I believe that the CTB has nothing whatsoever to do with the constant temperature of the ice-water mix.

If I am correct, the final answer is simply 'zero' because the mix and the CTB happen to be at the same temperature.

Apologies if I misinterpreted what you wrote.

No. That's exactly what I did mean. What do you think happens to the latent heat released or absorbed when the proportions of ice and liquid water change within the cylinder? Locally, there must be hotter or colder regions formed, and heat would have to be exchanged with the constant temperature bath in order for the temperature within the cylinder to be re-homogenized at 0 C. Otherwise, the first law could not be satisfied when applied to the cylinder contents.

Obviously, unless an ice-water mixture is at thermodynamic equilibrium, it is easy to create local hotter regions within the liquid water (say, for example., away from an ice-water interface).

 

[Steve4Physics]

No. That's exactly what I did mean. What do you think happens to the latent heat released or absorbed when the proportions of ice and liquid water change within the cylinder? Locally, there must be hotter or colder regions formed, and heat would have to be exchanged with the constant temperature bath in order for the temperature within the cylinder to be re-homogenized at 0 C. Otherwise, the first law could not be satisfied when applied to the cylinder contents.

Obviously, unless an ice-water mixture is at thermodynamic equilibrium, it is easy to create local hotter regions within the liquid water (say, for example., away from an ice-water interface).

Agreed on all of that! Do we also agree that the final answer (net heat transfer to the constant temperature bath) is exactly zero?

 

[zemaitistrys]

Agreed on all of that! Do we also agree that the final answer (net heat transfer to the constant temperature bath) is exactly zero?

I sure don't, haha. From the first law of thermodynamics the final answer depends on the work done by the cyllinder's piston. The difference between the work and the energy needed to melt the ice will be turned into heat and transferred to the heat bath. And I don't see why that amount should be zero

 

[Steve4Physics]

I sure don't, haha. From the first law of thermodynamics the final answer depends on the work done by the cyllinder's piston. The difference between the work and the energy needed to melt the ice will be turned into heat and transferred to the heat bath. And I don't see why that amount should be zero

I look at it this way. If the process is peformed (infinitely) slowly, thewhole of the ice-water mix stays at 0ºC. This would be true even if the mix was in an insulated container.

There is no temperature difference between the mix and the bath - hence no heat transfer.

I guess we'll have to agree to disagree.

EDIT. You said "The difference between the work and the energy needed to melt the ice will be turned to heat..."

But I believe the work and energy needed to melt the ice are exactly equal. The work done equals the increase in internal (potential) energy during the constant-temperature phase change.

 

[Chestermiller]

Agreed on all of that! Do we also agree that the final answer (net heat transfer to the constant temperature bath) is exactly zero?

No. The problem statement indicates that the volume of ice and liquid water in the container decreases, and the implication is that this is accommodated by some of the ice (of lower density) melting to form liquid water (of higher density). Even though the final temperature of the system is 0 C, since the internal energy of liquid water is higher than that of ice (by the heat of fusion), the final internal energy of the contents of the container is higher than its initial internal energy. From the 1st law of thermodynamics, neglecting the work done to compress the contents, in order of the internal energy to have increased, heat would have had to have flowed from the surroundings to the system.

I'll get the calculation start by determining the amount of ice that melts. The initial volume of the 5 kg of ice and the 5 kg of water in the container is $$V_i=\frac{5}{\rho_{ice}}+\frac{5}{\rho_{liquid}}$$The final volume of the ice and water is
$$V_f=\frac{5-\delta m}{\rho_{ice}}+\frac{5+\delta m}{\rho_{liquid}}$$where $\delta m$ is the mass of ice that melts.
So the change in volume is
$$\delta V=\left(\frac{1}{\rho_{liquid}}-\frac{1}{\rho_{ice}}\right)\delta m=-\frac{(\rho_{liquid}-\rho_{ice})}{\rho_{liquid} \rho_{ice}}\delta m$$

 

[zemaitistrys]

I look at it this way. If the process is peformed (infinitely) slowly, thewhole of the ice-water mix stays at 0ºC. This would be true even if the mix was in an insulated container.

There is no temperature difference between the mix and the bath - hence no heat transfer.

I guess we'll have to agree to disagree.

EDIT. You said "The difference between the work and the energy needed to melt the ice will be turned to heat..."

But I believe the work and energy needed to melt the ice are exactly equal. The work done equals the increase in internal (potential) energy during the constant-temperature phase change.

I think the same motive applies in a problem where you throw a rock off a certain height, and need to calculate the change in entropy of the universe as a result. Even if you say that the rock and the surface that it falls down into are at the same temperature, the kinetic energy of the rock is turned into heat that flows into the surroundings.

The reason why you can have this heat flow "without temperature difference" is because there is one microscopically, it's just that as it flows into the heat bath, it causes an infinitesimal increase in temperature because of the relatively infinite heat capacity of the heat bath (by definition)

 

[Steve4Physics]

I think the same motive applies in a problem where you throw a rock off a certain height, and need to calculate the change in entropy of the universe as a result. Even if you say that the rock and the surface that it falls down into are at the same temperature, the kinetic energy of the rock is turned into heat that flows into the surroundings.

The reason why you can have this heat flow "without temperature difference" is because there is one microscopically, it's just that as it flows into the heat bath, it causes an infinitesimal increase in temperature because of the relatively infinite heat capacity of the heat bath (by definition)

I understand what you are saying. My problem is that the question specifically refers to a piston being pushed down and a volume change. I don't see why we should ignore (or treat as negligible) the work done. In which case, the original question is unanswerable without knowing the pressure due to the piston. (As noted by the OP.)

I need to go away and think about this more. Many thanks for your insights - I can now see the hole in my reasoning!

 

[zemaitistrys]

I understand what you are saying. My problem is that the question specifically refers to a piston being pushed down and a volume change. I don't see why we should ignore (or treat as negligible) the work done. In which case, the original question is unanswerable without knowing the pressure due to the piston. (As noted by the OP.)

I need to go away and think about this more. Many thanks for your insights - I can now see the hole in my reasoning!

I am the OP : ' D (insert Palpatine meme)

P.S. Yes totally agreed, hence my best guess is Chestermiller's idea of assuming the external pressure is roughly constant at 1 atm.

 

[Steve4Physics]

I am the OP : ' D (insert Palpatine meme)

P.S. Yes totally agreed, hence my best guess is Chestermiller's idea of assuming the external pressure is roughly constant at 1 atm.

Hi. My previous answers were way-off (many thanks to Chestermiller for the insights). But if I can redeem myself a bit, here’s my new version of what I believe happens.

1. The effect of increased pressure is to *reduce the freezing point* so the ice (at 0ºC) is now slightly above its freezing point and starts to melt, taking latent heat from it’s surroundings.

2. The melting process (if my maths is correct) requires 7590J to be supplied to the mixture.
EDIT: That should be 75900J

3. For a pressure of 1 atm, the work done is only about 2J. So for reasonable pressures, we can neglect the contribution from the work. The constant temperature bath effectively supplies all the energy (as heat) to the ice-water mix.

 

[Chestermiller]

Hi. My previous answers were way-off (many thanks to Chestermiller for the insights). But if I can redeem myself a bit, here’s my new version of what I believe happens.

1. The effect of increased pressure is to *reduce the freezing point* so the ice (at 0ºC) is now slightly above its freezing point and starts to melt, taking latent heat from it’s surroundings.

2. The melting process (if my maths is correct) requires 7590J to be supplied to the mixture.

3. For a pressure of 1 atm, the work done is only about 2J. So for reasonable pressures, we can neglect the contribution from the work. The constant temperature bath effectively supplies all the energy (as heat) to the ice-water mix.

Please check you arithmetic. I get 10X that amount of heat.

 

[Steve4Physics]

Please check you arithmetic. I get 10X that amount of heat.

Agreed. Not my day. (Updated with an edit.)