Decrease a substance's temp by stirring

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Homework Help Overview

The discussion revolves around the effects of stirring on the temperature of water in a sealed and insulated container. Participants explore whether stirring causes the temperature to increase, decrease, or remain constant, referencing principles of thermodynamics and energy transfer.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants suggest that stirring adds energy to the water, potentially increasing its temperature, while others question how this interacts with heat loss mechanisms in different scenarios. The original poster expresses confusion about the relationship between stirring and temperature changes.

Discussion Status

Participants are actively engaging with the question, offering various perspectives on the thermodynamic principles involved. Some have provided insights into how stirring affects energy distribution and heat loss, while others are still seeking clarity on the fundamental concepts.

Contextual Notes

There is an emphasis on the ideal conditions of the system being discussed, such as the effects of insulation and sealing on heat transfer, which may not reflect real-world scenarios like a cup of coffee.

misskitty
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This question is really bugging me.

If water is in a sealed and insulated container and is stirred will the temperature slightly increase, slightly decrease or stay the same? Explain your answer.

I'm really stuck. I was going to say slighty increase, but I'm not sure because can you increase the temperture of something by stirring it? I know you can decrease a substances temperature by stirring. Ahh, help please, I'm really confused. :frown:
 
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Since the stirring is transferring work to the water, the temperature would increase slightly.

Using the first law of thermodynamics:

[tex]E_2-E_1 = Q - W[/tex]

Since it is adiabatic, Q=0. Since work is being done on the system, the work would be negative, which makes...

[tex]E_2-E_1 = 0-(-W)[/tex]

[tex]E_2-E_1=\Delta KE+\Delta PE+\Delta U = W[/tex]

Assuming there is no change in the kinetic or potential energy of the system, the only change is in internal energy.
 
misskitty said:
This question is really bugging me.

If water is in a sealed and insulated container and is stirred will the temperature slightly increase, slightly decrease or stay the same? Explain your answer.

I'm really stuck. I was going to say slighty increase, but I'm not sure because can you increase the temperture of something by stirring it? I know you can decrease a substances temperature by stirring. Ahh, help please, I'm really confused. :frown:
This is exactly how Joule discovered that heat was a form of energy. Since, in stirring, one has applies a force over a distance, work is being done. The work adds kinetic energy to the water as evidenced by its motion. However, eventually the water stops moving due to friction, and that kinetic energy is dispersed (into a more random kinetic energy of individual molecules of water) and increases the temperature of the water.

AM
 
Well, by stirring something, you're putting energy in...
 
Hot water normally cools by conduction of heat through the container wall and by evaporation of the hottest molecules. Stirring the water encourages evaporation and also brings fresh hot water into contact with the container wall. So it encourages both types of heat loss. However, if the container is completely insulated to prevent conduction loss, and completely sealed to prevent evaporation loss, then the energy put into stirring will cause it to heat slightly.
 
So why does stirring a hot cup of coffee cool the coffee? Does the heat energy transfer to the spoon?
 
See rdt2's explanation...the cup of coffee model is not an ideal, closed system. To summarize what rdt2 said, heat is transferred away from the coffee through the mug and into the air. Although stirring adds energy (and heat) to the system, it also encourages the cooling process through the mug and into the air. It encourages a faster LOSS of heat than it causes an INCREASE in heat, so you get a net loss of heat.
 
Ok. Thanks. :biggrin: That helps...now can someone explain to me why I can do all the really complicated stuff no problem and get it while the rest of my peers have their mouths hanging open out of sheer confusion...yet I can't get something SIMPLE like THIS? :devil:
 
No, I can't really explain that...but take solace knowing that everyone needs a little help now and then.
:smile:
 
  • #10
Thats good to know. Sometimes I feel like I'm the only one. Kinda makes me feel a bit stupid when I have to ask for help...::redface:
 

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