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Entropy change in an RC circuit
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[QUOTE="rude man, post: 6013210, member: 350494"] This is pretty advanced thermodynamics. I can only give you some ideas: Since the capacitor is being charged isothermally, dT = 0 and you are left with dS = [itex]\lambda dq[/itex]/T = (ε'/εc)q dq. You didn't say what ε' and c are. But your eq. (1) is of course a form of the Maxwell "2nd T dS equation" so there must exist a coefficient representing the change of polarization with T ( related by the state equation). So maybe ε' = dε/dT and ΔS[SUB]dielectric[/SUB] =[SIZE=5][SIZE=4] (ε'/εc) [SIZE=5]∫[/SIZE]q dq from q=0 to q = Cv.[/SIZE][/SIZE] Then the total [URL='https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/']change in entropy[/URL] of the universe is just ΔS[SUB]dielectric[/SUB] - ΔS[SUB]thermostat[/SUB]. Sorry, not very definite; hope others will do better. [/QUOTE]
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Entropy change in an RC circuit
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