Entropy and heat bath/reservoir

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dRic2
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If I have and object at a different temperature than the thermal/heat reservoir (whatever it's called) an heat flow will take place. If I write the entropy balance for the thermal reservoir it will be:

##\frac {dS} {dt} = \frac {\dot Q} T + \dot S_{gen}##

Now I remember something my professor told me a year ago (I could never fully understand it and it keeps coming back to mind),

from his book (I tried to translate it correctly):
the rate at which entropy is generated within the system (##S_{gen}##) is always zero because, since the temperature is uniform ad constant, the effect of the heat flow is the same for every temperature of the object it's exchanging heat with.

can someone please explain it to me?

Thanks
Ric
 

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  • #2
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If I have and object at a different temperature than the thermal/heat reservoir (whatever it's called) an heat flow will take place. If I write the entropy balance for the thermal reservoir it will be:

##\frac {dS} {dt} = \frac {\dot Q} T + \dot S_{gen}##

Now I remember something my professor told me a year ago (I could never fully understand it and it keeps coming back to mind),

from his book (I tried to translate it correctly):


can someone please explain it to me?

Thanks
Ric
If this is the entropy balance on the ideal reservoir, then there is no entropy generated within the reservoir, since its thermal conductivity is assumed to be infinite, and there are negligible temperature gradients within the reservoir (associated with entropy generation). Also, the heat capacity of the reservoir is assumed to be infinite, so that its temperature remains constant, and all the entropy transferred between the system and the reservoir takes place by heat flow at the reservoir temperature T. If you had assumed that the reservoir had a large, but finite, heat capacity, the change in entropy of the reservoir would have been:
$$\Delta S=mC\ln(T_f/T_i)$$with $$Q=mC(T_f-T_i)$$
If we combine these two equations, we obtain $$\Delta S=mC\ln{\left(1+\frac{Q}{mCT_i}\right)}$$If we take the limit of this as mC becomes infinite, we obtain: $$\Delta S=\frac{Q}{T_i}$$
So an ideal thermal reservoir is one in which no entropy is generated, and, for which the amount of entropy transferred from the system to the ideal reservoir is given by the equation ##\Delta S=\frac{Q}{T_R}##, where ##T_R## is the (constant) reservoir temperature. In other words, an ideal reservoir is always assumed to transfer heat reversibly at the constant reservoir temperature.
 
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dRic2
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Thank you for the clear explanation, but I don't understand the reason for this:

since its thermal conductivity is assumed to be infinite, and there are negligible temperature gradients within the reservoir (associated with entropy generation)

Also can you suggest me some books where entropy balance is carefully explained? In all the books I have read about thermodynamics there is a lot of stuff about the energy balance, but not very much about the entropy balance.

Ric
 
  • #4
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Thank you for the clear explanation, but I don't understand the reason for this:



Also can you suggest me some books where entropy balance is carefully explained? In all the books I have read about thermodynamics there is a lot of stuff about the energy balance, but not very much about the entropy balance.

Ric
A book that I like quite a bit is Fundamentals of Engineering Thermodynamics by Moran et al.
 
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dRic2
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Thank you. A few more questions:

1) If thermal conductivity is infinite, heat (kinetic energy) is transferred entirely from a molecule to an other with no loss (thus no entropy generated). Right?

2) A liquid/solid/gas during a state transition can be assumed as a thermal reservoir? Dose it means it will exchange heat reversibly?

3) Even if the thermal reservoir exchanges heat reversibly, the other system can undergo an irreversibly transformation. Right?
 
  • #6
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Thank you. A few more questions:

1) If thermal conductivity is infinite, heat (kinetic energy) is transferred entirely from a molecule to an other with no loss (thus no entropy generated). Right?
I don't know much about molecular interactions, because I'm a continuum mechanics guy. But I do know that the local rate of entropy generation per unit volume is proportional to the local heat flux squared divided by the thermal conductivity.
2) A liquid/solid/gas during a state transition can be assumed as a thermal reservoir? Dose it means it will exchange heat reversibly?
If you have ice cubes at 0C floating around in liquid water, and heat is introduced at a boundary far from the ice cubes, you can have temperature gradients (and entropy generation) within the liquid water in the vicinity of the heat transfer boundary. But, if the heat is introduced slowly enough at the boundary, the temperature gradients (and entropy generation) will be negligible. The changes very close to the surfaces of the ice cubes will, of course, typically take place with minimal temperature gradients and minimal entropy generation. So it is worthwhile starting to think about the spatial variations of the transient changes taking place in a system in which there is a possibility of irreversibility.
3) Even if the thermal reservoir exchanges heat reversibly, the other system can undergo an irreversibly transformation. Right?
Yes. In this case, all the irreversibility is forced to occur in the "system."
 
  • #7
dRic2
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But I do know that the local rate of entropy generation per unit volume is proportional to the local heat flux squared divided by the thermal conductivity.

Any reference for that? Is this in the book you suggested before? because I give a quick look but I didn't find it. I also took a look at the local entropy balance but I found a very complicated formula for ##\sigma## (or ##S_{gen}##)
 
  • #8
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Any reference for that? Is this in the book you suggested before? because I give a quick look but I didn't find it. I also took a look at the local entropy balance but I found a very complicated formula for ##\sigma## (or ##S_{gen}##)
They have a great derivation in Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 11, Problem 11.D.1 Equation of change for entropy
 
  • #9
dRic2
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Noooooo, I've just returned that book to the library... :( :(
 
  • #10
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Noooooo, I've just returned that book to the library... :( :(
Buy yourself a personal copy. This is the one book that I used more than all the others combined during my 35 year career in industry.
 
  • #11
dRic2
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I will!
 

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