Question

8–21 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

Entropy change of hot and cold reservoir are -0.08333 kJ/k and 0.16666 kJ/k respectively.

Explanation:

Given:

Temperature at hot reservoir,$T_{H}=1200 \mathrm{K}$

Temperature at cold reservoir,$T_{L}=600 \mathrm{K}$

Amount oh heat transferred,$Q=100 \mathrm{kJ}$

Entropy change at hot reservoir,$\Delta S_{H}=\frac{Q_{H}}{T_{H}}$

$\Delta S_{H}=\frac{-100}{1200}$

$\Delta S_{H}=-0.083333 \frac{\mathrm{kJ}}{\mathrm{k}}$

Entropy change at cold reservoir,$\Delta S_{L}=\frac{Q_{L}}{T_{L}}$

$\\\Delta S_{L}=\frac{100}{600}$

$\Delta S_{L}=0.166666 \frac{\mathrm{kJ}}{\mathrm{k}}$

Total entropy change,

$\Delta S=\Delta S_{n}+\Delta S_{L}$

$\Delta S=-0.083333+0.166666$

$\Delta S=0.083333 \frac{\mathrm{kJ}}{\mathrm{k}}$

$\Delta S$ is not less than zero.Hence,it fulfills increase of entropy principle.