Thermodynamics: Lossy Capacitor

[BobNye]

Homework Statement

A capacitor is charged to Q₀ and then isolated from its surroundings thermally, electrically, and mechanically. Over time the charge leaks across the dielectric filling and is neutralized. The Free Energy is given

F=Q²/2C₀ *T/T₀ + κT(1-lnT/T₀)

κ, T₀, and C₀ are constants.

What is the final temperature T₂ in terms of the inital temp. T₁? What is the change in entropy?

Homework Equations

F=U-TS
S=-(dF/dT)_Q
1st and 2nd Law of Thermodynamics, etc.

The Attempt at a Solution

Calculate S=-(dF/dT)_Q
S=-Q²/(2C * T₀) + κ ln(T/T₀)

Observe that U (internal energy) is constant in the initial and final states. Equate U₁=F₁+T₁S₁ and U₂=F₂+T₂S₂

U₁=Q²/2C *T₁/T₀ + κT₁(1-lnT₁/T₀) + T₁ [-Q²/(2C * T₀) + κ ln(T/T₀)]=κT₁

U₂=κT₂(1-lnT₂/T₀) + T₂ [κ ln(T/T₀)]=κT₂

U₁=U₂ -> κT₁=κT₂ -> T₁ = T₂

Basically this implies that there is no energy at all stored in the capacitor, and that when the charge dissipates it does not increase the temperature of the dielectric. This doesn't make sense. Where have I gone wrong? I would appreciate your help greatly.

PS. I calculated the entropy using this method but it seems pointless to post it since this answer is clearly wrong.

 

[mark726]

Dear fellow scientist,

Thank you for your post. I believe the issue with your solution lies in the assumption that the internal energy (U) remains constant in the initial and final states. This may not be the case in this scenario as the charge is leaking across the dielectric and being neutralized, which could potentially lead to a change in internal energy.

Instead, let's consider the fact that the system is isolated from its surroundings, meaning there is no exchange of energy with the environment. This implies that the change in internal energy (ΔU) is equal to 0. Using the first law of thermodynamics, we can write:

ΔU = Q - W = 0

Where Q is the heat added to the system and W is the work done on the system. Since there is no work being done on the system, we can rewrite this equation as:

Q = 0

This means that the change in internal energy is solely due to the change in heat (ΔU = Q). Therefore, we can write:

ΔU = Q = TΔS

Where ΔS is the change in entropy. Rearranging this equation, we get:

ΔS = ΔU/T

Now, to calculate the change in entropy, we need to calculate the change in internal energy (ΔU). From the given equation for Free Energy (F), we can write:

ΔF = F2 - F1 = -Q2/2C0 *T2/T0 + κT2(1-lnT2/T0) - Q1/2C0 *T1/T0 + κT1(1-lnT1/T0)

Since the system is isolated, we can write:

ΔF = ΔU - TΔS = 0 - TΔS = -TΔS

Therefore, we can write:

-TΔS = -Q2/2C0 *T2/T0 + κT2(1-lnT2/T0) - Q1/2C0 *T1/T0 + κT1(1-lnT1/T0)

Rearranging and simplifying, we get:

ΔS = (Q2/2C0 *T2/T0 - Q1/2C0 *T1/T0) - κ ln(T2/T0)

Now, we know that Q1 = Q0 (since the