ThermodynamicsHard
Question
A reversible heat engine absorbs 40 kJ of heat at 500 K and performs 10 kJ of work rejecting the remaining amount to the sink at 300 K. The entropy change for the universe is
Options
A.−80 J/K
B.100 J/K
C.20 J/K
D.180 J/K
Solution
$\Delta S_{unit} = \Delta S_{source} + \Delta S_{Heatengine} + \Delta S_{\text{sink}}$
$= - \frac{40 \times 10^{3}}{500} + 0 + \frac{30 \times 10^{3}}{300} = + 20\text{ J/K}$
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