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inna [77]
3 years ago
12

During the isothermal heat addition process of a Carnot cycle, 900 kJ of heat is added to the working fluid from a source at 400

oC. What is the entropy change of the working fluid? What is the entropy change of the source? What is the total entropy change for the process? [
Chemistry
1 answer:
Alinara [238K]3 years ago
7 0

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

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<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

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