It will be great if you searched on the internet then just questioning it on here because on here it takes a bit to get that so might as well google it
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
This type of heat transfer is radiation since we can experience the heat from an object without being in contact with it. Energy is being emitted by electromagnetic waves or in moving subatomic particles. Radiation is a consequence of thermal agitation of the molecules.