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Show heat flows from hot to cold bodies

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Homework Statement


Consider two systems which together comprise an isolated system, but are initially not in equilibrium with each other. The temperatures of the two systems are [itex]T_1[/itex] and [itex]T_2[/itex] and the internal energies are [itex]E_1[/itex] and [itex]E_2[/itex]. The systems are separated by a diathermal wall and only allowed to exchange energy by heat exchange. By writing the entropy as a function of the internal energy and the volume, [itex] S = S(E,V) [/itex] and the fact that energy is conserved, show that energy flows from the hotter to the colder body.


Homework Equations


[itex] dE = TdS - PdV [/itex]

[itex] \Delta S > 0 [/itex] (non-reversible process)

The Attempt at a Solution



I guess to begin with I'm confused about how to actually write [itex] S(E,V) [/itex] without knowing an equation of state. I'm also unsure if the term [itex] PdV [/itex] is zero or not, because I see no reason the systems can't expand, yet they "only exchange energy through heat" which implies there is no work...

I would begin by writing,

[itex] dS = \frac{dE}{T} + \frac{P}{T}dV [/itex] but from there I don't know how to integrate, because presumably [itex] T = T(E,V) [/itex] and [itex] P = P(E,V) [/itex].

If someone could help me get started or provide some hints as to where to go, I'd greatly appreciate it.
 

Answers and Replies

  • #2
TSny
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It is given that the two systems can only exchange energy by heat exchange. So, they don't do any work on each other. That should help you decide if the volume of either system changes.

The total entropy of the system is additive: ##S = S_1(E_1, V_1) + S_2(E_2, V_2)##

Take the differential of both sides of this equation and interpret.
 
  • #3
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Ok, so I could write (where [itex] i, f [/itex] stand for initial and final)

[tex] \Delta S = S_f - S_i \to dS_f - dS_i = dS_f - (\frac{dE_1}{T_1} + \frac{dE_2}{T_2}) > 0 [/tex]

but I'm still unsure how to express [itex] dS_f [/itex]... I'm tempted to write [itex] dS_f = \frac{dE_{\textrm{tot}}}{T_f} = 0 [/itex] but I don't think that's quite right...
 
  • #4
TSny
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Ok, so I could write (where [itex] i, f [/itex] stand for initial and final)

[tex] \Delta S = S_f - S_i \to dS_f - dS_i = dS_f - (\frac{dE_1}{T_1} + \frac{dE_2}{T_2}) > 0 [/tex]
I don't see how you got this.

Since ##S = S_1 + S_2##, the differential would be

##dS = d(S_1 + S_2) = dS_1 + dS_2##

Keeping in mind that ##S_1## is considered to be a function of ##E_1## and ##V_1##, how can you express ##dS_1## in terms ##dE_1## and ##dV_1##?
 
  • #5
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[itex] dS_1 = \frac{1}{T_1}dE_1 + P_1dV_1 [/itex] by the first law, but [itex] PdV = 0 [/itex] so [itex] dS_i = dS_1^{(i)} + dS_2^{(i)} = \frac{1}{T_1}dE_1 + \frac{1}{T_2}dE_2 [/itex].

To relate this to the final energy of each piece of the system, I need to use the fact that change in entropy is positive... which is why I wrote down [itex] dS_f [/itex] above.
 
  • #6
TSny
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There is no "final" ##dS## or "initial" ##dS##. There is only a ##dS## that represents a small change in entropy of the system when the two subsystems exchange a little heat.

That is, due to the small heat exchange, the entropy of the system will change from some initial value ##S_i## to some final value ##S_f##. And the change in entropy of the system is ##dS = S_f - S_i##.

Likewise, for each subsystem, ##dS_1 = S_{1,f} - S_{1,i}## and ##dS_2 = S_{2,f} - S_{2,i}##.

So, anyway, you have ##dS = dS_1 + dS_2## and since the volumes are constant you know you can write this as

##dS = \frac{dE_1}{T_1} + \frac{dE_2}{T_2}##


From energy conservation, how are ##dE_1## and ##dE_2## related? Use this relation to write ##dS## in terms of just ##dE_2##.
 
Last edited:
  • #7
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Ah thank you! I understand now, I think the problem was almost trivially simple which made me over-think it...

However, if I did actually want to calculate what [itex] \Delta S [/itex] was in terms of [itex] E_1, E_2, T_1, T_2, T_f [/itex] and [itex] E_f [/itex] I would then certainly need an equation of state, correct?
 
  • #8
TSny
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Yes, that's right.
 

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