Understanding Dissipation and Entropy in Newtonian Dampeners"

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In a Newtonian dampener scenario, the mechanical work input is entirely dissipated as heat, leading to an increase in the bath's entropy. The dampener's entropy remains constant as long as its temperature does not rise, meaning the total entropy increase is solely attributed to the bath. The second law of thermodynamics dictates that the overall entropy of the system must increase, which is satisfied by the heat exchange with the bath. There is no universal conservation of entropy law akin to energy conservation; instead, entropy can increase through energy distribution changes. Understanding these principles clarifies the relationship between dissipation, entropy, and thermodynamic systems.
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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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