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$begingroup$

One student of mine has found one question in a book of thermodynamics (1st year university level).

Work or Heat? You grab a bottle of juice and shake it thoroughly.
Is this an example of a work or a heat interaction? Recall that work can be
described with mechanical or electromagnetic variables.

My opinion is that we should first define the system. If we take the juice as our system, there is no work done by the system; nor on the system since the system boundaries do not move.
There is not heat neither that transverse the system boundary.
Am I correct?

On a second thought, I think that we should define the bottle and its content as the system. Then we have clearly a work done on the system.

Is there any contradiction? What am I missing here?

thermodynamics

share|cite|improve this question

edited 4 hours ago

Dimitris

asked 6 hours ago

DimitrisDimitris

1836

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that the bottle is not absolutely full of liquid, you are doing work on the juice by shaking it, because you are mixing the vapor and liquid phases when you shake the bottle. In short order, this work turns into heat. In my opinion, the question is ambiguous.
  $endgroup$
  – David White
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In introductory themodynamics, we often assume that the system is macroscopically stationary. This is not the case here; there are opportunities to do work on the system other than to compress it.
  $endgroup$
  – Chemomechanics
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidWhite Yeah, it seems like it's confusing at best to try to pin such a complex scenario as either "work or heat" without clearly defining what part of the interaction they mean.
  $endgroup$
  – JMac
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: physics.stackexchange.com/q/431399
  $endgroup$
  – dmckee♦
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is no different from Joule's experiment to demonstrate the mechanical equivalent of heat, except that in Joule's experiment the work was easier to measure.
  $endgroup$
  – alephzero
  6 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

One student of mine has found one question in a book of thermodynamics (1st year university level).

Work or Heat? You grab a bottle of juice and shake it thoroughly.
Is this an example of a work or a heat interaction? Recall that work can be
described with mechanical or electromagnetic variables.

My opinion is that we should first define the system. If we take the juice as our system, there is no work done by the system; nor on the system since the system boundaries do not move.
There is not heat neither that transverse the system boundary.
Am I correct?

On a second thought, I think that we should define the bottle and its content as the system. Then we have clearly a work done on the system.

Is there any contradiction? What am I missing here?

thermodynamics

share|cite|improve this question

edited 4 hours ago

Dimitris

asked 6 hours ago

DimitrisDimitris

1836

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming that the bottle is not absolutely full of liquid, you are doing work on the juice by shaking it, because you are mixing the vapor and liquid phases when you shake the bottle. In short order, this work turns into heat. In my opinion, the question is ambiguous.
  $endgroup$
  – David White
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In introductory themodynamics, we often assume that the system is macroscopically stationary. This is not the case here; there are opportunities to do work on the system other than to compress it.
  $endgroup$
  – Chemomechanics
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidWhite Yeah, it seems like it's confusing at best to try to pin such a complex scenario as either "work or heat" without clearly defining what part of the interaction they mean.
  $endgroup$
  – JMac
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: physics.stackexchange.com/q/431399
  $endgroup$
  – dmckee♦
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is no different from Joule's experiment to demonstrate the mechanical equivalent of heat, except that in Joule's experiment the work was easier to measure.
  $endgroup$
  – alephzero
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

One student of mine has found one question in a book of thermodynamics (1st year university level).

Work or Heat? You grab a bottle of juice and shake it thoroughly.
Is this an example of a work or a heat interaction? Recall that work can be
described with mechanical or electromagnetic variables.

My opinion is that we should first define the system. If we take the juice as our system, there is no work done by the system; nor on the system since the system boundaries do not move.
There is not heat neither that transverse the system boundary.
Am I correct?

On a second thought, I think that we should define the bottle and its content as the system. Then we have clearly a work done on the system.

Is there any contradiction? What am I missing here?

thermodynamics

share|cite|improve this question

edited 4 hours ago

Dimitris

asked 6 hours ago

DimitrisDimitris

1836

$endgroup$

share|cite|improve this question

edited 4 hours ago

Dimitris

asked 6 hours ago

DimitrisDimitris

1836

share|cite|improve this question

edited 4 hours ago

Dimitris

asked 6 hours ago

DimitrisDimitris

1836

asked 6 hours ago

DimitrisDimitris

1836

asked 6 hours ago

DimitrisDimitris

1836

2 Answers
2

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oldest

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2

$begingroup$

You are doing work on the juice.

Let the juice and the bottle be the system. Lets say the bottle is glass (rigid) or if plastic when you shake the bottle you don't squeeze it. There are several types of work that you can do. One is boundary work which expands or contracts the boundary (bottle). This doesn't apply. There is also stirrer work which increases the temperature. This was the basis of the famous Joule experiment that proved the equivalency of heat and work.

Although this normally involves stirring the contents of the bottle with a paddle wheel that is in the bottle increasing the temperature of the contents, shaking the contents also does work on the contents. In addition to what @David White pointed out, shaking the contents creates viscous friction within the juice, raising its temperature.

Of course there is also the possibility of heat transfer from your hand to the juice if the juice is at a lower temperature than your hand.

Hope this helps.

share|cite|improve this answer

answered 6 hours ago

Bob DBob D

3,8432317

$endgroup$

add a comment | 

2

$begingroup$

Even if the system is the juice only, the walls of the bottle are moving and exerting forces on the juice. So you have forces applied through displacements. This is clearly work. The mechanical work that the walls of the bottle do on the juice is soon dissipated by the viscous friction in the fluid, which converts the mechanical work into internal energy of the juice (i.e., small temperature rise). If the bottle is insulated (effectively), there is no heat transfer involved.

share|cite|improve this answer

answered 6 hours ago

Chester MillerChester Miller

15.2k2824

$endgroup$

add a comment | 

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2

$begingroup$

You are doing work on the juice.

Let the juice and the bottle be the system. Lets say the bottle is glass (rigid) or if plastic when you shake the bottle you don't squeeze it. There are several types of work that you can do. One is boundary work which expands or contracts the boundary (bottle). This doesn't apply. There is also stirrer work which increases the temperature. This was the basis of the famous Joule experiment that proved the equivalency of heat and work.

Although this normally involves stirring the contents of the bottle with a paddle wheel that is in the bottle increasing the temperature of the contents, shaking the contents also does work on the contents. In addition to what @David White pointed out, shaking the contents creates viscous friction within the juice, raising its temperature.

Of course there is also the possibility of heat transfer from your hand to the juice if the juice is at a lower temperature than your hand.

Hope this helps.

share|cite|improve this answer

answered 6 hours ago

Bob DBob D

3,8432317

$endgroup$

add a comment | 

2

$begingroup$

You are doing work on the juice.

Let the juice and the bottle be the system. Lets say the bottle is glass (rigid) or if plastic when you shake the bottle you don't squeeze it. There are several types of work that you can do. One is boundary work which expands or contracts the boundary (bottle). This doesn't apply. There is also stirrer work which increases the temperature. This was the basis of the famous Joule experiment that proved the equivalency of heat and work.

Although this normally involves stirring the contents of the bottle with a paddle wheel that is in the bottle increasing the temperature of the contents, shaking the contents also does work on the contents. In addition to what @David White pointed out, shaking the contents creates viscous friction within the juice, raising its temperature.

Of course there is also the possibility of heat transfer from your hand to the juice if the juice is at a lower temperature than your hand.

Hope this helps.

share|cite|improve this answer

answered 6 hours ago

Bob DBob D

3,8432317

$endgroup$

add a comment | 

$begingroup$

You are doing work on the juice.

Let the juice and the bottle be the system. Lets say the bottle is glass (rigid) or if plastic when you shake the bottle you don't squeeze it. There are several types of work that you can do. One is boundary work which expands or contracts the boundary (bottle). This doesn't apply. There is also stirrer work which increases the temperature. This was the basis of the famous Joule experiment that proved the equivalency of heat and work.

Although this normally involves stirring the contents of the bottle with a paddle wheel that is in the bottle increasing the temperature of the contents, shaking the contents also does work on the contents. In addition to what @David White pointed out, shaking the contents creates viscous friction within the juice, raising its temperature.

Of course there is also the possibility of heat transfer from your hand to the juice if the juice is at a lower temperature than your hand.

Hope this helps.

share|cite|improve this answer

answered 6 hours ago

Bob DBob D

3,8432317

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Bob DBob D

3,8432317

answered 6 hours ago

Bob DBob D

3,8432317

answered 6 hours ago

Bob DBob D

3,8432317

2

$begingroup$

Even if the system is the juice only, the walls of the bottle are moving and exerting forces on the juice. So you have forces applied through displacements. This is clearly work. The mechanical work that the walls of the bottle do on the juice is soon dissipated by the viscous friction in the fluid, which converts the mechanical work into internal energy of the juice (i.e., small temperature rise). If the bottle is insulated (effectively), there is no heat transfer involved.

share|cite|improve this answer

answered 6 hours ago

Chester MillerChester Miller

15.2k2824

$endgroup$

add a comment | 

2

$begingroup$

Even if the system is the juice only, the walls of the bottle are moving and exerting forces on the juice. So you have forces applied through displacements. This is clearly work. The mechanical work that the walls of the bottle do on the juice is soon dissipated by the viscous friction in the fluid, which converts the mechanical work into internal energy of the juice (i.e., small temperature rise). If the bottle is insulated (effectively), there is no heat transfer involved.

share|cite|improve this answer

answered 6 hours ago

Chester MillerChester Miller

15.2k2824

$endgroup$

add a comment | 

$begingroup$

Even if the system is the juice only, the walls of the bottle are moving and exerting forces on the juice. So you have forces applied through displacements. This is clearly work. The mechanical work that the walls of the bottle do on the juice is soon dissipated by the viscous friction in the fluid, which converts the mechanical work into internal energy of the juice (i.e., small temperature rise). If the bottle is insulated (effectively), there is no heat transfer involved.

share|cite|improve this answer

answered 6 hours ago

Chester MillerChester Miller

15.2k2824

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Chester MillerChester Miller

15.2k2824

answered 6 hours ago

Chester MillerChester Miller

15.2k2824

answered 6 hours ago

Chester MillerChester Miller

15.2k2824