During the isothermal heat addition process of a carnot cycle, 900 kj of heat is added to the working fluid from a source at 400oc. 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? [

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 molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

answer carbon and three oxygen atoms;