Solution:
Given:
[tex]T_{H}[/tex] = 1200 K
[tex]T_{L}[/tex] = 600 K
Q = 100 kJ
The Entropy change of the two reservoirs is given by the sum of entropy change of each reservoir system and is given by the formula:
[tex]\Delta s = \frac{-Q}{T_{H}}+\frac{Q}{T_{L}}[/tex]
[tex]\Delta s = \frac{Q(T_{L}-T_{_{H}})}{T_{H}T_{L}}[/tex]
[tex]\Delta s = \frac{-100(600-1200)}{1200\times 600}[/tex]
[tex]\Delta s = 0.0833kJ/K
Since, the change in entropy is positive and according to the Increase in entropy principle, for any process the total change in entropy of a system is always greater than or equal to zero (with its enclosing adiabatic surrounding).
Therefore, the entropy principle is satisfied.
