Question #babca

1 Answer
Dec 18, 2014

A reversible process is one that is carried out sufficiently slowly that the entire system is very nearly at equilibrium during the process.


Suppose you have two 1 kg blocks of copper (specific heat of 385 J/kg-K). Block 1 is initially at 301 K and Block 2 is at 299 K. Now you bring the two blocks into thermal contact with each other. Heat will flow from the first to the second until both blocks are at 300 K. The temperature difference is so small that for all practical purposes the heat transfer is reversible. This can be demonstrated by the fact that the changes in entropy of the two blocks are (nearly) equal in magnitude and opposite in sign.

Block 1: #DeltaS~=(-385J)/(300.5K)=-1.281J/K#
Block 2: #DeltaS~=(385J)/(299.5.5K)=1.285J/K#
(only about 0.3% difference)

But consider another experiment where Block 1 is at 350 K and Block 2 is at 250 K. The blocks are brought into contact momentarily so the temperature change is 1.0 K. The entropy changes are now:

Block 1: #DeltaS~=(-385J)/(349.5K)=-1.102J/K#
Block 2: #DeltaS~=(385J)/(250.5K)=1.537J/K#
(about 28% difference)

This shows that although energy is conserved in both cases, entropy is only conserved in reversible processes. In the first case, the temperature difference between the blocks is so small that (for all practical purposes) the process is reversible and entropy is conserved.

Good explanation, nice graphs can be found here: