Answer:
(a) 0.699 kJ/K
(b) -0.671 kJ/K
(c) 0.028 kJ/K
Explanation:
The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).
(a) The entropy change of the refrigerant (ΔS) = Q/T
Q = 180 kJ
T = -15.64 + 273.15 = 257.51 K
ΔS = Q/T
= 180/257.51 = 0.699 kJ/K
(b) The entropy change (ΔS) of the cooled space (ΔS
) = -Q/T
Q = -180 kJ
T = -5 + 273.15 = 268.15 K
ΔS = Q/T
= -180/268.15 = -0.671 kJ/K
(c) The total entropy change for this process (ΔS) = ΔS
+ ΔS
= 0.699 - 0.671 = 0.028 kJ/K