I am trying to understand the derivation of a general energy balance in battery thermodynamics. The following relation is frequently found to determine the heat generation of a battery:
$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0)$
where $\dot{Q}$ is the heat generation, $I$ current, $T$ temperature and $E$ cell voltage. The index $0$ denotes the open circuit voltage. Furthermore, $\Delta S = zF\,\frac{\mathrm dE_0}{\mathrm dT}$, with $F$ as Faraday constant and $z$ as number of exchanged electrons.
The derivation is as follows:
First law of thermodynamics: $\mathrm dU = \mathrm dQ - \mathrm dW \tag{1}$ with $\mathrm dW = p\,\mathrm dV + \mathrm dW_\mathrm{el}$
$\mathrm dH = \mathrm dU + p\,\mathrm dV + V\,\mathrm dp \tag{2}$
Substituting (2) in (1):
$\mathrm dH - p\,\mathrm dV -V\,\mathrm dp = \mathrm dQ - p\,\mathrm dV - \mathrm dW_\mathrm{el} \tag{3}$
with $\mathrm dp=0$ follows:
$\mathrm dH = \mathrm dQ -\mathrm dW_\mathrm{el}$, differentiation with respect to time:
$$\frac{\mathrm dH}{\mathrm dt} = \frac{\mathrm dQ}{\mathrm dt} - \frac{\mathrm dW_\mathrm{el}}{\mathrm dt} = \dot{Q} - EI$$
So far, everything is fine. But now:
$$H = G + TS \rightarrow \frac{\mathrm dH}{\mathrm dt} = \frac{\mathrm dG}{\mathrm dt} + T\,\frac{\mathrm dS}{\mathrm dt} + \frac{\mathrm dT}{\mathrm dt}S$$
Here my first problem in understanding arises: In general $\frac{\mathrm dG}{\mathrm dt}$ and $\frac{\mathrm dT}{\mathrm dt}S$ cancels each other, since, $S=-\frac{\mathrm dG}{\mathrm dT}$ and $\frac{\mathrm dG}{\mathrm dt}$ can be expanded with $\mathrm dT$: $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$. This means, that the influence of the open circuit voltage in the given energy balance would vanish.
I assume, that this is prevented by simply stating isothermal conditions: $\frac{\mathrm dT}{\mathrm dt}S = 0$. But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell. Can somebody explain this to me?
Now if $\mathrm dT=0$ the derivation would proceed like this:
$$\frac{\mathrm dG}{\mathrm dt} + T\,\frac{\mathrm dS}{\mathrm dt} = \dot{Q} - EI$$
$$\frac{\mathrm d(-znFE_0)}{\mathrm dt} + T\,\frac{\mathrm d\left(znF\,\frac{\mathrm dE_0}{\mathrm dT}\right)}{\mathrm dt} = \dot{Q} - EI$$ where $znF=C$ is the charge and $\frac{\mathrm dC}{\mathrm dt} = I$
And here my second problem arises: to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $\frac{\mathrm dE_0}{\mathrm dT}$ constant, so that the chain rule does not apply. And I do not understand why this should be valid?
Can anybody help me with this?