Entropy change of hot and cold reservoir are -0.08333 kJ/k and 0.16666 kJ/k respectively.
Explanation:
Given:
Temperature at hot reservoir,[tex]T_{H}=1200 \mathrm{K}[/tex]
Temperature at cold reservoir,[tex]T_{L}=600 \mathrm{K}[/tex]
Amount oh heat transferred,[tex]Q=100 \mathrm{kJ}[/tex]
Entropy change at hot reservoir,[tex]\Delta S_{H}=\frac{Q_{H}}{T_{H}}[/tex]
[tex]\Delta S_{H}=\frac{-100}{1200}[/tex]
[tex]\Delta S_{H}=-0.083333 \frac{\mathrm{kJ}}{\mathrm{k}}[/tex]
Entropy change at cold reservoir,[tex]\Delta S_{L}=\frac{Q_{L}}{T_{L}}[/tex]
[tex]\\\Delta S_{L}=\frac{100}{600}\\[/tex]
[tex]\Delta S_{L}=0.166666 \frac{\mathrm{kJ}}{\mathrm{k}}[/tex]
Total entropy change,
[tex]\Delta S=\Delta S_{n}+\Delta S_{L}[/tex]
[tex]\Delta S=-0.083333+0.166666[/tex]
[tex]\Delta S=0.083333 \frac{\mathrm{kJ}}{\mathrm{k}}[/tex]
[tex]\Delta S[/tex] is not less than zero.Hence,it fulfills increase of entropy principle.
