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I was given a question,

"A thermos bottle containing milk is shaken vigorously. Consider the milk as the system.

(a) Will the temperature rise as a result of the shaking?

(b) has heat been added to the system?

(c) Has work been done on the system?

(d) Has the system's internal energy changed?".

I know that the shaking, which can be viewed more simply as stirring, does increase the temperature of the system, and work has been done on the system, but does shaking/stirring add heat $Q$? If yes, could someone explain how? And how would that effect the internal energy (U)?

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When you shake the thermos, you're adding kinetic energy (motion) to to milk. That kinetic energy quickly dissipates to heat due to fluidic friction, thus heating the milk. So:

a) Yes, the temperature of the milk will rise.

b) No. Kinetic energy has been added to the system, which dissipated to heat within the system.

c) Yes. Work was done to accelerate the milk (though not in a single direction).

d) Yes. By the conservation of energy, the energy added from shaking is added to the system, first resulting in fluidic motion, and then dissipating to heat because of fluidic friction.

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    $\begingroup$ I would not use the word "heat" with regard to viscous friction (even though it is sometime imprecisely referred to as "viscous heating"). In thermodynamics, heat is reserved exclusively for thermal energy transferred between the surroundings and the system. (through the interface between system and surroundings). $\endgroup$ – Chet Miller Sep 29 '18 at 0:54
  • $\begingroup$ So for heat to be added to the system, would that only be the case if the thermos' contents increased in temperature due to the direct transfer of energy from something that was higher in temperature? Your answer to B was very helpful to understand the increase in temperature, but I don't quite understand why the kinetic energy resulting from the shaking dissipating as heat through the system does not qualify as heat being added to the system by shaking. $\endgroup$ – Ben Sep 29 '18 at 0:59
  • $\begingroup$ @Ben, not only are we not adding heat, in the real life case we actually remove heat from the system. When the thermos is shaken, we perform work on it, and as a result of this, the temperature increases. As it equilibrates again, the heat is slowly transferred out of the thermos and into the environment. $\endgroup$ – Al Nejati Sep 29 '18 at 1:05
  • $\begingroup$ @ChesterMiller Good distinction! $\endgroup$ – Stuart Van Horne Sep 29 '18 at 1:13
  • $\begingroup$ @Ben Yes, heat is added if it is placed in contact with something hotter, resulting in heat transfer, as Chester described above. I suppose you could say in a non-technical sense that heat was added to the system indirectly by the shaking, but I believe that most physicists would describe it as a 2 step process. $\endgroup$ – Stuart Van Horne Sep 29 '18 at 1:18
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The point here turns on the distinction between different ways of understanding the transfer of energy to or from a system.

Because "heat" and "work" are both measures of how much energy is transferred, and the difference between them is a semantic distinction imposed from the outside, rather than something with a rigorous physical basis.

  • work is energy transferred by means that can be characterized by macroscopic variables.

  • heat is energy transferred by means that can only be characterized by understanding the microscopic detail of the interaction.

So, can you figure out the energy increase of the system by knowing the force applied by the arm as a function of time (i.e. macroscopic variables characterizing the interaction)?


It is, of course, worth noting that Joule's paddle wheel experiment worked because the macroscopic work done by the falling weight was sufficient to characterize the energy transfer.

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    $\begingroup$ dmckee, that was a subtle and very good answer ... hence the upvote. $\endgroup$ – David White Sep 29 '18 at 1:11

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