0.22 Entropy and the second law of thermodynamics  (Page 8/9)

All that has happened is the transfer of heat from one piece of metal to the other. No chemical reactions or phase changes have occurred and there has been no rearrangement of the atoms in the metal samples. Why did this heat transfer increase the entropy? We know from our previous study that raising the temperature of a sample by adding heat increases the entropy of the sample. Let’s say that the amount of heat transferred into sample B was q . The energy of sample B goes up by q during the heat transfer. And therefore the entropy of B must go up.

However, by conservation of energy, this same amount of heat q must also transfer out of sample A. The energy of sample A goes down by q during the heat transfer. And therefore the entropy of A must go down. Since the heat transfer is the same and the types and masses of the metal samples are the same, it would seem that the entropy increase of B should be exactly equal to the entropy decrease of A, in which case the total entropy change should be zero. But we know that this is not true, because the transfer of heat is a spontaneous process in an isolated system, so ∆S>0.

This can only mean that the entropy increase of the cold sample B must be larger than the entropy decrease of the hot sample A as they approach thermal equilibrium. This tells us that the quantity of heat alone is not enough to predict the entropy change associated with heat transfer. We must also know the temperature to predict the entropy change.

Our observations also tell us that the entropy change is greater for the lower temperature sample. Entropy change and temperature are therefore inversely related to each other. Adding heat to a cold sample produces a greater entropy change than adding heat to a hot sample. We can conclude all of this based just upon observing that heat spontaneously flows from a hot piece of metal to a cold piece of metal. What is harder to conclude is what the exact inverse relationship is. With much more effort and mathematical reasoning, we can show that the correct inverse relationship between entropy change and temperature is the simplest inverse relationship, namely an inverse proportion:

∆S = q/T for transfer of heat q at temperature T

This equation works when the temperature is held constant during the heating process. If the temperature is not held constant, we have to do an integral of 1/T from the initial temperature to the final temperature to calculate ∆S. That is not important for our purposes here. But it is interesting to ask how we can heat a body without changing its temperature. We will discuss this in the next section.

Observation 5: heat transfer during chemical or physical processes

The results of the previous observations and reasoning are important in all cases where heat is transferred, not just in heat transfers involving substances at two different temperatures. Let’s think back to our goal here. How can the Second Law be applied to a process in a system that is not isolated? We would like to understand how it is possible for a process in a non-isolated system to be spontaneous when ∆S<0 for the system in that process. A good example is the freezing of water at temperatures below 0 ºC. What is so special about 0 ºC? Why is it that below this temperature, the freezing process becomes spontaneous even though ∆S<0?