# Why is $TΔS$ a good approximation for calculating heat transfer during heating/cooling in real life?

So various books and sources have agreed on defining specific heat capacity (formally) to be:

$$c_v=\frac{T}{n}\left(\frac{\partial S}{\partial T}\right)_V=\frac{1}{n}\left(\frac{\delta Q_{reversible}}{dT}\right)_V$$ $$c_p=\frac{T}{n}\left(\frac{\partial S}{\partial T}\right)_p=\frac{1}{n}\left(\frac{\delta Q_{reversible}}{dT}\right)_p$$

Suppose now one wants to calculate the amount of heat required to heat a certain gas under constant pressure from T1 to T2 (using the specific heat capacity at T1) $$Q_{gas}=mc_{p}(T_2-T_1)=mT_1\Delta S$$

The entire heating process is neither reversible nor quasi-static. Why in real life (or at least in exercise problems), we take the above result $$mT_1ΔS$$ as a good approximation as the heat required?

• There are some incorrect premises here. “The above problem pre-suppose the temperature is undergoing a linear temperature change.” is incorrect; I don’t even know what “linear” is supposed to mean here. A temperature difference is a temperature difference. It sounds like you mean that the heat capacity is assumed to be temperature independent? Commented Mar 26, 2022 at 16:57
• @Chemomechanics Oh, I mean $Q$ is a linear function of $T$ because $c_p$ is assumed to be constant Commented Mar 26, 2022 at 16:59
• The statement “$c_p$ is only a state function when the process is reversible” is also incorrect; it is always a state function. Commented Mar 26, 2022 at 16:59
• Can you clarify “I started wondering when when will the temperature change follow a continuous path”? When does the temperature in a material not follow a continuous path? Commented Mar 26, 2022 at 17:02
• Not for a real material. Commented Mar 26, 2022 at 17:21

## 1 Answer

so in other words, this equation will work no matter what, be it open or closed system, reversible or irreversible process.

No; the equation $$\delta q=mc_XdT$$ describes the temperature change at constant $$X$$ when only heat transfer occurs.

(In contrast, if I were to adiabatically compress the system, for example, then $$\delta q=0$$, but the temperature would still change. Or if I introduced a second substance, a reaction might occur that would change the temperature. So processes other than solely heat transfer need to be excluded.)

The equation essentially defines the heat capacity for a small temperature change when one does nothing but heat or cool a system slightly. That heat transfer can be irreversible.

• About the reversibility of process, take $c_v$ as an example, $c_v=(\frac{dU}{dT})_v=(\frac{TdS}{\delta T})_v$, but only when the process is reversible $TdS=dQ$, so why does it still work for irreversible heat transfer? Commented Mar 26, 2022 at 17:59
• Since you don't have an accepted answer yet - my best guess: $\delta Q = mc dT$ is generally true for reversible and irreversible processes. Indeed, this equation defines the specific heat capacity $c$. The implicit assumption you are making when writing $\Delta Q = \int mc dT$ is not that the process is reversible but rather that it is quasistatic, i.e. we are assuming that the process is so slow that the heat transfer occurs in infinitesimal increments with state variables (e.g. $T$) equalising throughout the system at each step. Commented Mar 26, 2022 at 18:21
• @iyfyts3s I agree that we can use $\delta q=T\,dS$ when all processes are reversible (so that total $dS$ of the universe is zero). In addition, we can use it for a system at internal equilibrium, with no work or mass transfer. In this special case, there's nothing that would alter the system entropy except for heat transfer (reversible or irreversible). Commented Mar 26, 2022 at 18:35
• @ Chemomechanics, I got confused again, for a system at internal equilibrium throughout the process, it should be quasi-static, but when calculating the simple heating/cooling, the process is not even quasi-static. Only at the initial state and final state the system is at internal equilibrium. So in this case, why is $\Delta Q = mc_x \int dT$ still a good approximation? Commented Mar 28, 2022 at 18:03
• The entropy generation that causes the inequality occurs outside the system that's being heated. I'm ignoring this entropy generation (which has no impact on the system) and assuming a reversible path from state variable $T_1$ to state variable $T_2$. The associated heating $Q$ is the same regardless of whether the system is heated reversibly or irreversibly. See also this discussion (and the linked content) on the meaning of the temperature in that inequality. Commented Mar 28, 2022 at 22:29