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

Solution:

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:

$\Delta s = \frac{-Q}{T_{H}}+\frac{Q}{T_{L}}$

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

$\Delta s = \frac{-100(600-1200)}{1200\times 600}$

[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.