Understanding Dissipation and Entropy in Newtonian Dampeners"

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SUMMARY

The discussion focuses on the relationship between dissipation and entropy in a Newtonian dampener characterized by a viscosity coefficient η, subjected to a fixed rate e' in a thermal bath at temperature T. The mechanical work input, represented as dW = ηe'*e'dt, is entirely converted into heat, leading to an increase in the bath's entropy of -dW/T. The participants clarify that while the dampener exchanges heat with the bath, its entropy remains constant as long as its temperature does not rise, emphasizing that the total entropy increase is solely attributed to the bath, in accordance with the second law of thermodynamics.

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  • Study the principles of thermodynamics, focusing on the second law and entropy changes
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muzialis
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Hello there,

I have a question on the dissipation and entropy.

Let us consider a Newtonian dampener with viscosity coefficient η, pulled at a fixed rate e', immersed in an infinite bath at temperature T.
The mechanical work input in time dt is then dW = ηe'*e'dt, and is all dissipated into heat.
The bath will see an increase of its entropy of -dW/T.
And the dampener? Some might argue that it will exchange heat with the bath, but its entropy gain cannot be opposite and equal the entropy gain of the bath, as the second law prescribes the entropy for the overall dissipative sistem has to increase.
So how do I compute the total change in entropy? Where is the "additional" entropy coming from, to satisfy the second law?
I am so puzzled, hop somebody can relieve me!

Thanks
 
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As long as the dampener's temperature does not increase, it's entropy remains constant. The overall increase of entropy is entirely due to the increase of entropy of the bath.
 
Please note there is no universal "conservation of entropy law" as there is with energy.

Entropy can increase because processes rearrange the distribution of energy, although the energy remains constant.
 

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