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Andrew Mason said:<snip>
Example:
Consider 1 kg of water at 0C contained in a thin plastic bag and a lake at 10C (state A). I place the bag in the lake. Heat flows into the bag of water from the lake until the bag of water is 10C, the lake is still at 10C (state B). What is the change in entropy?
The change in entropy of the system (the bag of water) + the surroundings (the lake) is:
\Delta S_{bag} = \int_{A}^{B} dQ_{bag}/T_{bag}
\Delta S_{lake} = \int_{A'}^{B'} dQ_{lake}/T_{lake}
\Delta S_{total} = \Delta S_{bag} + \Delta S_{lake}
In order to determine the change in entropy, you have to find the reversible path from state A to state B for the bag. This would be a Carnot heat engine operating between the bag and the lake in which the work output would be stored and not converted to heat (say by lifting a weight). The total heat transferred to the cold reservoir (the bag of water) is raises the 1kg of water 10 degrees - 10 Kcal.
But in this process, the amount of heat lost by the lake would be greater than the heat gained by the water in the bag, (since some of that heat was used to do work which is stored in lifting the weight). So this cannot be the reversible process for the surroundings in going from state A' to B'. In going from A' to B', the reversible path, again, is a Carnot heat engine between the lake and the bag in which the heat lost by the lake is 10Kcal.
Conclusion: when you determine the change in entropy of the system and the change in entropy of the surroundings resulting from a non-reversible process, the paths for calculating the entropy changes of the system and surroundings are different.
AM
This is a great example, as it highlights a lot of the shortcomings in the way thermodynamics is traditionally taught
First, there's some assumptions which have not been spelled out- the lake is a reservoir, for one- neither the temperature nor the volume of the lake are allowed to change (actually they can, but the analysis then becomes too complicated to justify the increased model accuracy). Second, the bag is present to prevent diffusion of the cold water into the warmer water- this creates mass loss from thermodynamic system to the reservoir, which is also more difficult to model in the traditional picture.
The approach used above then invokes Carnot cycles and machines, neither of which has anything to do with the actual physics of heat and mass diffusion. Reversible processes are not ever observed in this situation.
It's much easier to simply evaluate the integrals: The heat flow Q through the bag depends on the temperature difference and the specific heat of the enclosed water: Q = c m (T_bag-T_lake), or dQ = cmdT_bag. The entropy integral for the bag is then dS = cm Ln(T_lake/T_i), where T_i is the inital temperature of the bag. If you like, the specific heat can be a function of temperature and the integral suitably evaluated.
Keeping ‘c’ constant, this integral evaluates to 35.9 cal/K. Positive, as the heat flows into the bag.
Now, for the lake. Again, the integral can be evaluated very straightforwardly: dS = 10 kcal/283 K = -35.3 cal/K. We have a potential problem as the entropy of the lake is *decreased* because the heat flowed out of the lake into the bag.
However, the entropy of the two processes together is positive: 0.63 cal/K. Which is as it should be. Also note that time is not part of any of this analysis: in that way, the concept of ‘equilibrium’ is retained- the temperature of the bag is assumed to reach a new value and not change from that.
My point is that traditional textbook presentations of thermodynamics tries so hard to force reversible processes and equilibrium onto the student that the essential physics is obscured and overly complicated. I claim that it is far better to start with the acknowledgment that most processes are irreversible, and then introduce equilibrium as a way to restrict the number of variables. In this way, the Gibbs free energy is the natural starting point (also since the Gibbs free energy change is what is measured in calorimetry experiments)