gotcha. the wording of the question tripped me up, but I suppose to interpret it I would need to conceptualize it the way you're mentioning (?). they made it seem spontaneous and quick rather than gradual.
Awesome, thanks so much for the help.
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"One more question for you. Do you understand how the equation in post #7 was obtained, and, specifically, what reversible path was used to obtain dqrev/TdqrevT? What reversible path was used here?"
I... kinda. An applicable equation is mentioned in my...
ok, so just to back things up a bit and make sure I'm all squared away here...:
ΔS1= Cvln( (T1+δQ/Cv)/T1) (if δQ was positive in system 1)
ΔS2= Cvln( (T2-δQ/Cv)/T2) (if δQ was negative in system 2)
ΔS ≈ CvδQ (1/T1 - 1/T2), and knowing that (1/T1 - 1/T2)δQ > 0, ΔS is therefore > 0
... all good...
if initial T2 > T1 then heat would flow from T2 to T1, which means in system 2 the Q would be negative and it would be positive for 1... right? I'm saying that I think I mixed that up earlier
ok, so the Q should be a positive value in system 1 due to T2 > T1... so, adjusting for that and recalculating, I end up with:
ΔS ≈ CvQ (1/T1 - 1/T2).
One of the relevant equations from my text says ΔS = ΔST2+ΔST1 > (1/T1 - 1/T2)δQ > 0. Do I need to expand any further with more work? also, noob...
≈x-x2/2. so, Cln(1+Q/T2) ≈ C (Q/T2 - (Q/T2)2/2), where that second term would go to zero
so, I think we'd arrive at:
ΔS ≈ C (Q/T2 - Q/T1), or ≈ CQ (1/T2 - 1/T1)]
this seems incorrect... maybe I mixed up the positive and negative Q earlier
"so that some heat flows but the temperature of neither system changes appreciably." so I suppose neither T1 nor T2 change (?) but some heat is exchanged. some volume change then?
I presume this is an irreversible process, but in terms of calculations I'm not entirely sure how to expand. entropy change is positive for spontaneous reactions as far as I know, so I'm not sure what else this question wants me to show.
Homework Statement
Two systems that have the same heat capacity Cv but different initial temperatures T1 and T2 (with T2 > T1) are placed in thermal contact with each other for a brief time, so that some heat flows but the temperature of neither system changes appreciably. Show that there is a...
well, I guess that answers it! I must've been interpreting what the question was asking incorrectly and using an incorrect formula. the one described in that link fits this much better, I think.
I'm using an equation provided in my notes. If the equation is supposed to already include V and μ0 then... well, that would be great, but I need the basis for it, which I'm having trouble figuring out.
And with the negative sign, I think it represents the work performed by the material itself...