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.

Step-by-step 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)=(Q_(H))/(T_(H))`

`\Delta S_(H)=(-100)/(1200)`

`\Delta S_(H)=-0.083333 \frac{\mathrm{kJ}}{\mathrm{k}}`

Entropy change at cold reservoir,
`\Delta S_(L)=(Q_(L))/(T_(L))`

`\\\Delta S_(L)=(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.