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a mixture of 1.78kg of water and 262g of ice at 0 degree celcius is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0 degree celcius. (1) calculate the entropy change of the system during this process. (2) the system is then returned to the first equilibrium state, but in an irreversible way (by using av bunsen burner, for instance) calculate the entropy change of the system during this process. (3) show that your answer is consistent with the second law of themodynamics.

a mixture of 1.78kg of water and 262g of ice at 0 degree celcius is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0 degree celcius.
(1) calculate the entropy change of the system during this process.
(2) the system is then returned to the first equilibrium state, but in an irreversible way (by using av bunsen burner, for instance) calculate the entropy change of the system during this process.
(3) show that your answer is consistent with the second law of themodynamics.
 
 

Grade:6

1 Answers

Arun
25763 Points
2 years ago
233-179_1.PNG J/kg )/(273 K)= -927 J/KFrom the above observation we conclude that, the change in entropy ΔS of the system during this process will be -927 J/K.(b) Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy ΔS of the system during this process is equal to the negative of previous case.So, ΔS = -(- 927 J/K)= 927 J/KFrom the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.(c) In accordance to second law of thermodynamics, entropy change ΔS is always zero.The total change in entropy will be,ΔS = (-927 J/K) + (927 J/K)= 0From the above observation we conclude that, our answer is consistent with the second law of thermodynamics. 3= -mL/T,ΔS = -mL/T= -(0.76 kg) (333×10
 
Hope it helps
 
Regards
Arun (askIITians forum expert)

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