dS = dq/T
tea pot before heating
tea pot being heated and S is increasing
Start adding heat
Entropy is S1
Entropy is S2 so that S2 > S1
Input of q increase overall S, but there is
turbulence
Entropy is S3 and S3 > S2
Thus, at this point not all of the heat has been changed into an entropy
term. Rather some of the heat was use to create some internal energy.
Assume no time passes for the following operation.
Turning off heat and isolating the system -
Entropy is S4 and S4
= S3
The system is allowed to settle isolated
Entropy is S5 and S5 > S4
Notice that in the last step the entropy increases without any external influence. Notice also that this process will not reverse, i. e. the internal temperature will not spontaneously drop with an increase in turbulence.