Entropy change in an RC circuit

[yamata1]

Hello,I would like some help for a problem

Homework Statement

Initially:At t=0 [/B]the cylindrical capacitor of capacitance c=\frac{\epsilon s}{d} (d the distance between the 2 electrodes and s their surface; \epsilon = \epsilon(T) is the dielectric permittivity) is discharged and we close the circuit and charge isothermally the cylindrical capacitor until it has a c*v charge.
The current is i=\frac{dq}{dt} and the voltage is v(t), the electric charge of the capacitor is q(t)=c*v(t).
We have an RC circuit receiving work W from a generator of voltage v and energy Q_{exchanged} from a thermal reservoir (thermostat) of T_0 Kelvin,the circuit has its entropy change\Delta S_{in} during the charging.
We have C_q the heat capacity for a constant charge q and \lambda= \frac{qT \epsilon'}{\epsilon c} another coefficient in the equation :
TdS_{in}= C_q dT + \lambda dq (1)

What is the entropy of the universe \Delta S_{univ} equal to ?
Does C_q depend on q ?

Homework Equations

\Delta S_{thermostat}=\frac{Q_{exchanged}}{T_0}

The Attempt at a Solution

If we integrate C_q over T ,I think q does appear in the formula for C_q but it's not a variable I think.To find \Delta S_{in} we divide the equation (1) by T then integrate \frac{\lambda}{T} over q from 0 to q=cv.
\Delta S_{univ}=\Delta S_{thermostat}+\Delta S_{in}
I would like to know how to express \Delta S_{thermostat} and it's sign. Thank you.

 

[Gene Naden]

What physical physical picture you are trying to work with? That is, what phenomenon are you interested in? It seems you have the capacitor bathed in a temperature reservoir. Why?

 

[yamata1]

What physical physical picture you are trying to work with? That is, what phenomenon are you interested in? It seems you have the capacitor bathed in a temperature reservoir. Why?

I want to know how to express the entropy change , I think the entropy change of the universe is 0 but I'm not sure what the equation is.The RC circuit is getting heat from the reservoir and work from the generator,I'm guessing it's to make it a reversible process.

 

[Gene Naden]

I am not sure, but I suspect you are confused. Ordinarily one does not consider heat when analyzing an RC circuit, and the capacitance of a capacitor does not vary much with temperature. I am not so sure that the net change in entropy of the universe is zero; For irreversible processes, entropy increases.

Generally the flow of current through a circuit with resistance is not a reversible process. The resistors give off heat but never absorb heat.

 

[rude man]

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 = \lambda dq/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_dielectric = (ε'/εc) ∫q dq from q=0 to q = Cv.
Then the total change in entropy of the universe is just ΔS_dielectric - ΔS_thermostat.

Sorry, not very definite; hope others will do better.

 

[yamata1]

I think I have an explanation :

The generator gives the system {resitance,capacitor} the electric energy :

The capacitor charges itself by en absorbing half ( the work $W=\int vdq$). The difference represents the lost energy through the Joule effect :

This energy is received through heat by the heat reservoir at fixed temperature ,which increases the entropy by :

Accouting for the resistance, we have :
\Delta S_{univ}=\frac{\varepsilon'}{2\varepsilon.c}\cdot q_{f}^{2}-\frac{\varepsilon'}{2\varepsilon.c}\cdot q_{f}^{2}+\frac{q_{f}^{2}}{2c.T_{0}}=\frac{q_{f}^{2}}{2c.T_{0}}

there is a creation of entropy caused by the Joulet effect which is an irreversible phenomenon as Gene Naden wrote

I am not sure, but I suspect you are confused. Ordinarily one does not consider heat when analyzing an RC circuit, and the capacitance of a capacitor does not vary much with temperature. I am not so sure that the net change in entropy of the universe is zero; For irreversible processes, entropy increases.

Generally the flow of current through a circuit with resistance is not a reversible process. The resistors give off heat but never absorb heat.

.
The power received by the heat reservoir is the power produced by the Joule effect :