Understanding Callen's eq. (4.9) and the maximum work theorem

Question

I'm currently reading Herbert B. Callen's Thermodynamics and an Introduction to Thermostatics, II ed, and I don't quite understand the last few propositions made on page 105 (a scan attached below).

Consider a subsystem connected to a Reversible Heat Source (RHS) and a Reversible Work Source (RWS). The subsystem's internal energy changes by $\mathrm{d}U$ in some process, and $\mathrm{d}U<0$. The released energy goes, in the form of heat $\text{đ}Q_{\text{RHS}}>0$, to the RHS, and, in the form of work $\text{đ}W_{\text{RWS}}>0$, to the RWS. So by the conservation of energy $$ -\mathrm{d}U = \text{đ}W_{\text{RWS}} + \text{đ}Q_{\text{RHS}}. $$

Denote by $\mathrm{d}S$ the change of entropy of the subsystem, and by $T_{\text{RHS}}$ the temperature of the RHS. Maximum work theorem suggests that $\text{đ}W_{\text{RWS}}$ is maximal when the process is reversible and the total entropy of the system doesn't change: $$ \frac{\text{đ}Q_{\text{RHS}}}{T_{\text{RHS}}} = -\mathrm{d}S. $$

Then Callen attempts to "calculate the maximum delivered work" and gives the following equation 4.9 $$ \max \text{đ}W_{\text{RWS}} = \frac{T_{\text{RHS}}}{T} \text{đ}Q - \mathrm{d}U = \left(1 - \frac{T_{\text{RHS}}}{T}\right) (- \text{đ}Q) - \mathrm{d}W $$ Here he used $\mathrm{d}U=\text{đ}Q+\text{đ}W$ without clarifying the notation explicitly.

Question: why is $-\text{đ}Q\neq \text{đ}Q_{\text{RHS}}$ and $-\text{đ}W\neq\text{đ}W_{\text{RWS}}$ here? I thought that when the subsystem is connected to the RWS, all the work done by the subsystem is done upon the RWS. Same applies to RHS: where does the subsystem's heat go, if not exactly to RHS? RWS can't exchange heat and RHS can do no work, so the setup doesn't seem to permit any conversion of work into heat and vice versa.

What I noticed so far in my attempts to understand it, is:

1. $-(\text{đ}Q + \text{đ}W) = \text{đ}Q_{\text{RHS}} + \text{đ}W_{\text{RWS}}$ (this follows from equating different expressions for $\mathrm{d}U$)
2. $\text{đ}Q$ cannot be equal to $\text{đ}Q_{\text{RHS}}=\frac{T_{\text{RHS}}}{T} \text{đ}Q$ because we equated the entropy changes and $T_{\text{RHS}}$ doesn't equal the subsystem's temperature $T$. That doesn't answer my question, though: how exactly did the heat get converted into work (or vice versa) here, why didn't it all go to the RHS?
3. If we substitute $T_{\text{RHS}}=T$ into eq4.9 we get $-\text{đ}W=\max\text{đ}W_{\text{RWS}}$, as I would expect.

Answer 1

I believe Callen's point is to contrast reversible and irreversible operation.

Under reversible operation, the system temperature $T$ equals the reversible heat source temperature $T_\mathrm{RHS}$; in addition, the relevant generalized force (the one that drives shifts in the conjugate generalized displacement when transferring work) is identical between the system and the reversible work source.

Only in this way can reversibility be maintained, as entropy is produced any time energy moves down a gradient (in temperature or some generalized force), but reversibility implies zero entropy production.

We encounter the standard paradox of zero gradients: There's nothing to actually drive a process to achieve nonzero $Q_\mathrm{RHS}$ and $W_\mathrm{RWS}$, as all heat transfer is driven by a temperature difference and all work is driven by a generalized force gradient (e.g., a pressure difference).

We remind ourselves that reversibility is an idealization where we bring $T$ arbitrarily close to $T_\mathrm{RHS}$ and still achieve nonzero $Q_\mathrm{RHS}$ (albeit with a vanishing rate as $T\to T_\mathrm{RHS}$). The (unrealizable) limit of the engine operating at $T= T_\mathrm{RHS}$ corresponds to reversibility.

We further recall that all this conceptual effort is justified by the benefit of not having to track entropy production, which would bog us down in the details of actual mechanisms and geometries.

With all that said, it's useful to derive equations that characterize the efficiency for various values of $T$ in real, irreversible engines. Higher values of $T$ would understandably tend to increase $Q$ (because of the larger driving force of a greater temperature difference) and thus decrease $W$ for fixed $\Delta U$. When you don't see the subscripts "RHS" and "RWS" here, Callen appears to be referring to real systems.

None of this means that heat transfer is going anywhere but the heat source or that work is being done on anything but the work source. It just means that the real magnitudes are different from the idealized reversible values: $Q>Q_\text{RHS}$, and therefore (by energy conservation) $W<W_\text{RWS}$.

Edit: The comments indicate that further clarification is desired. Let's examine the case of a real, irreversible system with $Q>Q_\text{RHS}$ and $W<W_\text{RWS}$. Consider, for example, a motor connected to a large, static block of material (the heat reservoir) and a thermally nonconducting shaft leading elsewhere (the work reservoir). The fixed input energy is the electrical power supplied to the motor. The motor can’t do work on the block or heat the shaft. Nevertheless, the motor, if jammed, would irreversibly disperse all the input electrical energy into the heat reservoir, through Joule heating, and none into the work reservoir, as the jammed shaft is motionless.

Answer 2

Actually, the heat extracted from the subsystem does not have to directly go to the heat source (and same for work). As you'll see when reading the Carnot Cycle section in that same textbook, you can employ auxiliary subsystems that sit "between" the primary subsystem and the work/heat sources and can essentially convert heat to work and vice versa. Importantly these auxiliary systems, over the course of the process, do not input/output any net energy or entropy.

As an example, the auxiliary subsystem might simply be a gas, connected to the work source and the primary subsystem. The primary subsystem could give off some heat $dQ$ and cause the gas to expand isothermally; thus overall this heat $dQ$ ended up becoming work entering the work source. (Of course, this example is only one part of an overall process; eventually the auxiliary gas has to come back to its original state).