Question

Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1320 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. (Round the final answers to four decimal places.)

Answer 1

Answer:0.0909 kJ/K

Explanation:

Given

Temperature of hot Reservoir $T_h=1320 K$

Temperature of cold Reservoir $T_l=600 K$

Heat of 100 kJ is transferred form hot reservoir to cold reservoir

Hot Reservoir is Rejecting heat therefore $Q_1=-100 kJ$

Heat is added to Reservoir therefore $Q_2=100 kJ$

Entropy change for system

$\Delta s=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}$

$\Delta s=\frac{-100}{1320}+\frac{100}{600}$

$\Delta s=-0.0757+0.1666=0.0909 kJ/K$

As entropy change is Positive therefore entropy Principle is satisfied