Recent content by TristanJones

ok, I think I get it. cheers

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. --- "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

Ok, so ΔS = Cln(1-Q/CT1) + Cln(1+Q/T2) is where we're at?

I suppose: S1= Cvln( (T1-Q/Cv)/T1) I take it I do the same thing for system 2 and then add? S2= Cvln( (T2+Q/Cv)/T2)

ok, so for system 1 would it be -Q/Cv + T1 and for system 2: +Q/Cv + T2? (I appreciate your patience; apologies for implying doubt)

"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...