Question

Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600 K. Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.

Answer 1

Given:

`T_(H)` = 1200 K

`T_(L)` = 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:

`\Delta s = (-Q)/(T_(H))+(Q)/(T_(L))`

`\Delta s = \frac{Q(T_(L)-T_{_(H)})}{T_(H)T_(L)}`

`\Delta s = (-100(600-1200))/(1200* 600)`

[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.