Engineering Board Exam Reviewer · Thermodynamics · hard
Air changes from 100 kPa, 300 K to 500 kPa, 500 K. What is the change in specific entropy? (cp = 1.0 kJ/kg· K, R = 0.287 kJ/kg· K)
- 0.5108 kJ/kg· K
- 0.0489 kJ/kg· K
- 0.9727 kJ/kg· K
- -0.0948 kJ/kg· K
Answer
B. 0.0489 kJ/kg· K
Solution
s = cp (T2/T1) - R (P2/P1) = 1.0 (5/3) - 0.287 5 = 0.5108 - 0.4619 = 0.0489 kJ/kg· K.
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