# In search for a physical model for $\Delta S$ for conducting objects

The situation is familiar: A cold object is in contact with a hot object and there is an energy transfer in between them, $Q$.

I know that $Q_{hot} = -Q_{cold}$ and that, because this is a spontaneous process, that $\Delta S > 0$. Therefore, $|\Delta S_{hot}| < |\Delta S_{cold}|$.

This is where I am presented with the equation $$\Delta S = \frac{Q}{T}$$ leading to

$$\frac{Q_{hot}}{T_{hot}} + \frac{Q_{cold}}{T_{cold}} > 0$$ The hot object has a higher temperature and, thus, the "hot" object will have a smaller $\Delta S$ than the "cold" object, leading to an overall positive $\Delta S$.

There are two things that I don't understand.

1. While I can see that $T$ must be inversely related to $\Delta S$, it doesn't make sense to me from the standpoint of a physical model. In my own (fairly basic, mind you) understanding, a larger amount of energy in a system (i.e. the system is at a higher temperature) should lead to a larger number of ways that the energy can be arranged (i.e. greater entropy). I am certain that my reasoning is wrong, but how?

2. This is certainly not an isothermal change. Why then are we able to use single temperature values for the two objects rather than some change in temperature?

• A system presenting a constant temperature boundary condition, a thermostat, does NOT have finite energy, just as a constant voltage source (an ideal battery) or constant current source are idealizations of "large" sources. By large I mean relative to the load they drive. – hyportnex Nov 7 '16 at 21:06
• Which makes a bigger difference in your spending habits: if you have \$0 and I give you \$5 or if you have \$100 and I give you \$5? Here is a better analogy – pentane Nov 7 '16 at 21:12
• @pentane I agree that it is bigger proportionally, but this doesn't seem right to me because changes in entropy are additive. – Christian Wimber Nov 8 '16 at 13:31

Here's how to do the calculation. Let $T_H$ and $T_C$ be the initial temperatures of the hot and cold objects, respectively, and let $(MC)_H$ and $(MC)_C$ be the product of mass and heat capacity. Then, from the first law, $$(MC)_H(T_H-T_F)=(MC)_C(T_F-T_C)=Q$$where $T_F$ is the final temperature of both objects and Q is the total amount of heat transferred. So, $$T_F=\frac{(MC)_HT_H+(MC)_CT_C}{(MC)_H+(MC)_C}$$
The next step is to devise a reversible path between the initial state of the hot object at $T_H$ and the final state at $T_F$, and calculate the integral of dQ/T for that path. This can be done by placing the initially hot object in contact with a continuous sequence of constant temperature reservoirs at slightly different temperatures running from $T=T_H$ to $T=T_F$. For each step along this path, $dQ=-(MC)_HdT$ and $dS=-(MC)_HdT$. So, for the hot reservoir, $$(\Delta S)_H=-\int_{T_F}^{T_H}{\frac{(MC)_H}{T}dT}=-(MC)_H\ln{\frac{T_H}{T_F}}=-Q\frac{\ln{(T_H/T_F)}}{(T_H-T_F)}$$Similarly, for the initially cold object,$$(\Delta S)_C=Q\frac{\ln{(T_F/T_C)}}{(T_F-T_C)}$$ Note that, in both these equations, Q is not divided by the initial temperature of each object, but rather by the so-called log-mean temperature, which is a weighted average of the initial temperature and the final temperature. This answers your second question.
The total change in entropy for both objects is thus given by:$$\Delta S=Q\frac{\ln{(T_F/T_C)}}{(T_F-T_C)}-Q\frac{\ln{(T_H/T_F)}}{(T_H-T_F)}$$ Mathematically, this entropy change is greater or equal to zero irrespective of the masses, heat capacities, and initial temperatures of the objects.