Change of Temperature from Work in Hydrostatic System

[BobaJ]

Hi,
I have a little conceptual question. I have an idea about the topic, but I would need to formalize it a little bit.
The problem is the following:

Imagine the following scenario: You put a little bit of water in a bottle, you assure yourself that everything as at a constant temperature and measure the temperature of the water with a thermometer. You close the bottle and shake it as hard as you can during some minutes. When you are tired and feel that you can't go on, you continue to shake the bottle for a few minutes more. After this you measure again the temperature. Make an estimation for the change of temperature.

Well, if I'm not mistaken, the shaking should count as work. The system should be a classic hydrostatic system and so the thermodynamic coordinates are P, V and T. So I'm doing some work over the system, the volume stays constant. What happens with the pressure, does it stay constant or does it change?
And with respect to the temperature, I would say that it increases. But I do not know how to answer the question the correct way and so that it sounds good.

Maybe you can help me.

 

[ZapperZ]

Hi,
I have a little conceptual question. I have an idea about the topic, but I would need to formalize it a little bit.
The problem is the following:

Imagine the following scenario: You put a little bit of water in a bottle, you assure yourself that everything as at a constant temperature and measure the temperature of the water with a thermometer. You close the bottle and shake it as hard as you can during some minutes. When you are tired and feel that you can't go on, you continue to shake the bottle for a few minutes more. After this you measure again the temperature. Make an estimation for the change of temperature.

Well, if I'm not mistaken, the shaking should count as work. The system should be a classic hydrostatic system and so the thermodynamic coordinates are P, V and T. So I'm doing some work over the system, the volume stays constant. What happens with the pressure, does it stay constant or does it change?
And with respect to the temperature, I would say that it increases. But I do not know how to answer the question the correct way and so that it sounds good.

Maybe you can help me.

This "thought experiment" is a bit unfocused.

Remember that, assuming that the container isolates the temperature of the content from the outside once you started shaking (i.e. you're not transferring heat via other means), there are two entities in the bottle, the water and the AIR trapped inside the bottle. So when you asked about the "temperature", it is assumed that you are asking about the temperature of the water. But when you are asking about the "pressure", which entity are you referring to? The pressure of the water would not change that noticeably, but the pressure of the trapped air might.

So why can't you test this out yourself. Use one of the plastic water bottle, screw the top tightly, and shake away? When you're done shaking, pay attention to what you hear when you slowly unscrew the top. Did you hear a slight hissing sound at some point? What do you think is the cause of that?

Zz.

 

[BobaJ]

So, the pressure of the water would stay basically constant. Does that mean, that by shaking the bottle, the pressure of the air increases?

 

[ZapperZ]

So, the pressure of the water would stay basically constant. Does that mean, that by shaking the bottle, the pressure of the air increases?

When you shake the bottle, you are increasing the temperature of the water slightly. This increase in temperature will also affect the temperature of the air since it is in equilibrium with the water. If the water warms up just a bit and expands, will also push against the air. The pressure of the water may change, but this change, as I said above, may be too small to be noticeable, at least using the simple apparatus describe above. How accurate do you intend to do all this?

This is why I said that, depending on how this is all done, you may detect the increase in pressure of the air simply by slowly unscrewing the top. This is by no means an accurate measurement.

Zz.

 

[BobaJ]

And why exactly does the temperature of the water increase? I'm sorry, I'm quite new to the topic.

 

[Chestermiller]

And why exactly does the temperature of the water increase? I'm sorry, I'm quite new to the topic.

When a viscous fluid like water is deformed, the mechanical energy used to deform the water is converted to internal energy of the fluid. This is because, on the smaller scale, forces are acting within the fluid and corresponding displacements are occurring. Joule showed this in his famous experiment where he had a stirrer within the water and accurately measured the amount of mechanical energy expended. He also measured the temperature rise, and established the conversion between work and internal energy. A viscous fluid (particularly an incompressible fluid like water) always irreversibly converts viscous stresses to internal energy.

 

[BobaJ]

Oh ok. So, to sum up. I assume that the bottle is an adiabatic container, so there is no exchange of temperature between the interior of the bottle (system) and its surroundings. This way I eliminate all other factors that could increase de temperature of the water. The shaking of the bottle is positive mechanical work that is done on the system and (taking the internal-energy function), this way the internal energy of the system increases.

As an "side effect", I can say, that by increasing the temperature of the water, it slightly expands so can affect the air in the bottle (slight increase in pressure).

So, to answer the original question. The estimation of the change of temperature would be that it increasing. The amount of increase depends on the amount of work (time and strength of shaking) done on the bottle. More work = more increase in temperature.

Correct?

 

[Chestermiller]

Oh ok. So, to sum up. I assume that the bottle is an adiabatic container, so there is no exchange of temperature between the interior of the bottle (system) and its surroundings. This way I eliminate all other factors that could increase de temperature of the water. The shaking of the bottle is positive mechanical work that is done on the system and (taking the internal-energy function), this way the internal energy of the system increases.

As an "side effect", I can say, that by increasing the temperature of the water, it slightly expands so can affect the air in the bottle (slight increase in pressure).

So, to answer the original question. The estimation of the change of temperature would be that it increasing. The amount of increase depends on the amount of work (time and strength of shaking) done on the bottle. More work = more increase in temperature.

Correct?

Yes, completely, except for the water expanding to pressurize the air. The increased temperature of the water results in heat transfer to the air, which also then rises to the same temperature as the water at equilibrium. The increased temperature of the air results in an increased pressure, due to the ideal gas law. The thermal expansion effect of the water is very very small compared to this.

 

[BobaJ]

ok. Thank you very much. I think now I understood the topic.

 

[russ_watters]

But when you are asking about the "pressure", which entity are you referring to? The pressure of the water would not change that noticeably, but the pressure of the trapped air might.

This is confusingly put. Except for a gravitational component, the pressure of the air and water must be identical. As Chet said, the water doesn't expand much as its temperature or pressure changes, so the pressure change in the water is driven by the pressure change in the air, not vice versa.

Conversely, the temperature change in the air is from conduction with the water; and the water increases its temperature via viscous friction.

 

[Chestermiller]

This is confusingly put. Except for a gravitational component, the pressure of the air and water must be identical. As Chet said, the water doesn't expand much as its temperature or pressure changes, so the pressure change in the water is driven by the pressure change in the air, not vice versa.

Conversely, the temperature change in the air is from conduction with the water; and the water increases its temperature via viscous friction.

In addition to this, for a mixing situation like the present one, the pressure within the liquid phase is not uniform spatially. This spatial pressure variation is driven by viscous forces and by inertial effects. The only place where the pressure of the liquid is fixed is at the interface with the air.