Answer:
The value of specific entropy = 0.847 [tex]\frac{KJ}{kg K}[/tex]
Explanation:
Constant pressure = 120 K pa
From the property tables of R - 134 a at P = 120 K pa
Temperature T = - 22.5 ° c
Value of heat transfer = 212.32 [tex]\frac{KJ}{kg}[/tex]
Thus entropy change ΔS = [tex]\frac{dq}{T}[/tex]
⇒ ΔS = [tex]\frac{212.32}{250.5}[/tex]
⇒ ΔS = 0.847 [tex]\frac{KJ}{kg K}[/tex]
Therefore the value of specific entropy = 0.847 [tex]\frac{KJ}{kg K}[/tex]
