What is the change in entropy here?

heat capacity of water = 4.186 J/g/K ...you have written for J/kg/K ...i dont know why ??

dS = Cp ln (T2/T1) + R ln (P1/P2)

since the process takes place at atmosphereic condition so pressure remains constant ...so P1 = P2

so dS = Cp ln (T2/T1) + R ln 1

dS = Cp ln (T2/T1) + 0 = Cp ln (T2/T1)

now calculate entropy change for both 1 kg of water separately using the above formula....in both cases T2 or final temperature will be 323 K

so 1 kg (or 1000 g) of water at 273 K is taken to 323 K

so entropy change for this much water

dS1 = 1000 X 4.186 X ln (323/273) = 4186 X ln 1.183 = 4186 X 0.168 = 703.248 J/K

similarly 1000 g of water at 373 K is taken to 323 K ...so entropy change for this much water :

dS2 = 1000 X 4.186 X ln 323/373 = 4186 X ln 0.866 = 4186 X -0.144 = -602.784 J/K

note that dS2 is -ve ...this is true because water at 373 K has been reudced to come to 323 K and as temperature decreases entropy decreases ....

total entropy change of the system = dS1 + dS2 = 703.248 + (-602.784) = 703.248 - 602.784 = 100.464 J/K

so b) is the answer ....

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