Does the rate of heat addition to a thermodynamic system affect its final state?

Question

I have read that the entropy change of a system is greater than or equal to the integral of the heat added to it divided by its temperature. The case where entropy is equal to the integral is when the heat is reversibly added, i.e., through an infinitesimal temperature difference and by implication, an infinitesimal heat transfer rate. I assume the greater the temperature difference, and thus the greater the heat transfer rate, the greater the entropy change of the system.

Additionally, the heat added at constant pressure is equal to $mC_pdT$, so that for a given amount of heat addition and constant mass, there is a definite temperature change.

Now, consider a cylinder-piston setup. In one case we add heat very slowly (reversibly), and in another we add the heat fast (irreversibly). In both cases, we add the same amount of heat and do so at constant pressure, so that the final temperature and pressure are the same in both cases. The initial and final state variables are therefore the same between the two cases, including the entropy, however in the slow case heat is added reversibly and in the other case irreversibly.

So, I am confused, because I am told that rate of heat addition can affect the entropy of a system, but I cannot see how in my thought experiment. I was hoping someone might help me clear this up. Thanks!

Answer 1

Let's go through some things you say and clean them up:

I assume the greater the temperature difference, and thus the greater the heat transfer rate, the greater the entropy change of the system.

As you say, under reversible conditions the entropy change is $$ \Delta S_{12}=\int_{T_1}^{T_2} \frac{dQ}{T} $$ where $T_1$ and $T_2$ are the initial and final temperatures. We cannot quite relate the magnitude of $\Delta S_{12}$ to the magnitude of $\Delta T$, as we can see from the $TS$ graph below: As you can see, we can construct various paths between the same end temperatures with any $\Delta S$ we wish, positive or negative.

Secondly, the temperature difference $T_2-T_1$ does not say anything about the rate of heating. This rate depends, among many other things, on the difference between the temperature of the system and the temperature of the source that is used to heat up the system.

So, the temperature difference says nothing about the magnitude of the $\Delta S$ and nothing about whether heating is fast or slow.

Now to reversible (slow) versus irreversible (fast) heating: if we heat fast, the pressure of the system will likely increase first before it finally equalizes with the pressure of the surroundings. In this case $Q\neq \Delta H$, because this equation assumes constant pressure, which we don't have. To calculate heat in this case we need more information about the details of the process. Regardless, we can be certain about one thing: the same amount of heat will lead to different final states under slow or fast heating.

Answer 2

Consider heat added to the piston/cylinder system. $Q - W = \Delta U$ where $Q$ is the heat added to the system, $W$ is the work done by the system, and $\Delta U$ is the change in internal energy of the system.

For $Q$ added reversibly (slowly), $W$ is less than for the same amount of heat added irreversibly (rapidly), so the change in internal energy for the irreversible process is greater that the change in internal energy for the reversible process.

If the system contains an ideal gas, the internal energy is a function of temperature alone, so the final state for the irreversible case has a higher temperature than for the reversible case.

The same heat added to the system results in less work done by the system for the irreversible case.

The final entropy of the gas is greater for the irreversible case than for the reversible case for the same heat added. See Difference between reversible and irreversible heat transfer on this exchange.