Problem 5.15 of Statistical Physics by Reif ( Calculation of entropy)

[nuclear_dog]

Homework Statement

1 kg of water with specific heat (C) of 4180 Joules /kg/degree is given at 0°C. It is taken to 100°C by two methods :-
(i) by bringing it in contact with a reservoir at 100°C.
(ii) by bringing it in contact with a reservoir at 50°C , and then with another reservoir at 100°C.

Calculate the entropy change of water and the reservoir in both cases.

Homework Equations

The Attempt at a Solution

Well, for a small amount of heat transferred (dQ), the entropy change is given by dS = \frac{dQ}{T} . dQ can be calculated by the formula dQ = C*m*dT . Then ΔS is obtained by integrating between Tf and Ti . So ΔS comes out to be C*m*ln(Tf/Ti). By this method I get the same answer in both the cases.

 

[rude man]

You have (correctly) computed the change in entropy of the water but what happened to the changes in entropy of the three reservoirs?

 

[nuclear_dog]

Thanks for the reply.
What I wanted to know was, whether the entropy change of only the water in the two cases would be same or not. By the method I have used, I get the same entropy change for water in both the cases.

 

[rude man]

But the problem asked for the entropy changes of the water AND THE RESERVOIRS. So you haven't answered the problem.

Yes, the change in entropy of the water is the same for both processes. That's because entropy is a state function. ΔS is a function of the beginning (A) and end (B) states only. It doesn't matter by what process you get from A to B, not even if it's irreversible (as is the case here). Your water starts at A = 0C and ends at B = 100C, all at the same pressure.