Respuesta :
Answer:
the entropy change of the fluid during the process process is is 1.337 kJ/K, the change for the source is -1.337 kJ/K and the total entropy change is 0
Explanation:
since the Carnot cycle is a reversible cycle, the entropy change is related with the heat exchanged through:
ΔS =∫dQ/T
since the temperature remains constant
ΔS =∫dQ/T=(1/T)*∫dQ = Q/T
Q= heat added to the system
T= absolute temperature = 400°C= 673 K
therefore
ΔS = Q/T = 900 kJ/ 673 K = 1.337 kJ/K
ΔS working fluid = 1.337 kJ/K
since the process is reversible, the entropy change of the universe (total entropy change) is 0 (there is no entropy generation). thus
ΔS universe = ΔS working fluid + ΔS source = 0
ΔS source= -ΔS working fluid = -1.337 kJ/K
The entropy change of the fluid is -1.337kJ/K, the entropy change of the source is -1.337kJ/K and the entropy change of the system is 0J/K
Data;
- Q = 900kJ
- T = 400°C
Entropy of a System
This is a rate of disorderliness of a system.
The entropy of the fluid is calculated as
[tex]\delta S_f = \frac{Q}{T}\\ \delta S_f = \frac{900}{673} \\\delta S_f = 1.337 kJ/K[/tex]
The entropy change of the fluid is -1.337kJ/K
b)
The entropy change of the source
This is calculated as
[tex]\delta S_s = -\frac{Q}{T} \\\delta S_s = -\frac{900}{673} = -1.337kJ/K[/tex]
The entropy change of the source is -1.337kJ/K
c)
Total entropy change of the process is the sum of the entropy change of the fluid and entropy change of the source
[tex]\delta S = \delta S_f + \delta S_s\\\delta S = -1.337 + 1.337 = 0[/tex]
The entropy change of the system is 0J/K
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