Let me start by saying I agree with what both @Barbaud Julien and @Chester Miller said. So my answer is only intended to provide a different perspective.
First, it is true that the entropy is a state function of a system, and therefore the change of entropy of the system between any two given equilibrium states will be the same whether or not the process between the states is reversible. However, the change in entropy of the system plus surroundings will depend on the process. It will be zero if the process is reversible, and greater than zero if irreversible.
Consider a cycle consisting of two isothermal and two adiabatic processes. If the processes were all reversible, we would obviously have a Carnot cycle. Since your question focuses on the effect of heat transfer on entropy, let’s assume the adiabatic processes are reversible adiabatic (isentropic) and consider only the impact of the heat transfer processes being reversible or irreversible.
Let’s take the example of a heat addition expansion process of an ideal gas between two equilibrium states. Let the initial and final temperatures be equal. Consequently, from the ideal gas law, the initial and final pressure-volume products are equal. Clearly the process between the states may be a reversible isothermal process. However, it also need not be. We’ll consider both.
Let the temperature of the system be $T_{sys}$ and the temperature of the surroundings be $T_{sys}+dT$. Consider the system and surroundings as thermal reservoirs, i.e., heat transfer between them does not alter their temperatures so that the heat transfer occurs isothermally.
Let a specific quantity of heat, $Q$, transfer from the surroundings to the system. The resulting entropy changes are:
$$\Delta S_{sys}=\frac{+Q}{T_{sys}}$$
$$\Delta S_{surr}=\frac{-Q}{T_{sys}+dT}$$
The net change in entropy (system + surroundings) is thus:
$$\Delta S_{net}=\frac{+Q}{T_{sys}}+\frac{-Q}{T_{sys}+dT}$$
Now, note that if $dT\to 0$, then $\Delta S_{net}\to 0$ and the process is a reversible isothermal process.
However, for any finite temperature difference, $dT>0$, $\Delta S_{net}>0$ and the process is irreversible.
For both the reversible and irreversible processes, the change in entropy of the system is the same. However, for the irreversible process the change in entropy of the system is greater than the change in entropy of the surroundings. The excess entropy that is created in the irreversible expansion process means more heat must be rejected to the cold temperature reservoir (surroundings) during the isothermal compression in order for the cycle entropy to be zero. That results in less energy to do work in the cycle.
Although in this example the same amount of heat is transferred reversibly and irreversibly, clearly the rate of heat transfer will be greater for the irreversible than the reversible process owing to the finite temperature differential for the irreversible process. So for the amount of heat transfer to be the same, the product of the very slow heat transfer rate times the very long time for the reversible process would need to equal the product of the higher heat transfer rate times the shorter time duration for the irreversible process.
Hope this helps.
Understanding entropy change is much simpler than it seems from your description of Callen. Here are the basics:
Entropy is a physical (state) property of the material(s) comprising a system at thermodynamic equilibrium, and the entropy change between two thermodynamics equilibrium states of a system depends only on the two end states (and not on any specific process path between the two end states).
For a closed system, there are only two ways that the entropy of the system can change:
(a) by heat flow across the system boundary with its surroundings at the temperature present at the boundary $T_B$; this is equal to the integral from the initial state to the final state of $dQ/T_B$ along whatever path is taken between the two end states.
(b) by entropy generation within the system as a result of irreversibility; this is equal to the integral from the initial state to the final state of $d\sigma$ along whatever path is taken between the two end states, where $d\sigma$ is the differential amount of entropy generated along the path..
Contribution (a) is present both for reversible and irreversible paths. Contribution (b) is positive for irreversible paths and approaches zero for reversible paths. For any arbitrary path between the two end states, the two contributions add linearly: $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$
Determining the amount of entropy generation along an irreversible path is very complicated so, to determine the entropy change for a system between any initial and final thermodynamic equilibrium states, we are forced to choose only from the set of possible paths that are reversible in applying our equation. The reversible path we choose does not have to bear any resemblance to the actual path for the process of interest. All reversible paths with give the same result, and will also provide the entropy change for any of the irreversible paths.
Armed only with these basics, one can determine the change in entropy for a closed system experiencing any process, provided that application of the 1st law of thermodynamics is sufficient to establish the final thermodynamic equilibrium state.
Best Answer
When we plot a process as a continuous line, curve, etc., of states using thermodynamic variables we are assuming that each point on the chart represents an equilibrium state, i.e., the process taking place is a series of “quasi-static” or “quasi-equilibrium” states. For example, refer to the T-S diagram below for a Rankine refrigeration cycle. The compressor, condenser, and evaporator processes are each reversible. The expansion valve (throttling) process is not. Note how it is represented by a series of hashed lines implying that the states between 3 and 4 are not equilibrium states. The enthalpy at state 3 and state 4 are the same but it is not a constant enthalpy process. Although Zemansky singles out the T-S diagram, the same holds true for other graphs of processes. The following is from an MIT web course on thermodynamics concerning processes.
“…properties define a state only when a system is in equilibrium. If a process involves finite, unbalanced forces, the system can pass through non-equilibrium states, which we cannot treat. An extremely useful idealization, however, is that only infinitesimal unbalanced forces exist, so that the process can be viewed as taking place in a series of quasi-equilibrium states”.
Hope this helps