Change in Entropy When Mixing Water at Different Temperatures

AI Thread Summary
The discussion focuses on calculating the change in entropy when mixing water at different temperatures. The correct approach involves a quasi-static process where both water sources reach a final temperature through a series of reservoirs, resulting in a calculated entropy change of 3.2 cal/K. An alternative method was attempted, involving the gradual mixing of hot and cold water, which yielded a higher entropy change of 14.77 cal/K. The confusion arises from the assumption that the process is reversible; however, pouring the mixed water back does not return it to its original state, thus violating the conditions for reversibility. The conclusion confirms that the initial method is valid while the alternative approach is flawed due to the irreversible nature of the mixing process.
domephilis
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Homework Statement
Exercise 11.30 Find the change in entropy if 500g of water at 80C is added to 300g of water at 20C. (Fundamentals of Physics I by R. Shankar)
Relevant Equations
$$dS = \frac{\Delta Q}{T}$$ $$\Delta Q = mc \Delta T$$
After re-reading the book, I did figure out what I was supposed to do. Take both waters through a series of reservoirs to bring them down to their final temperature while allowing for a quasi-static process. Thus, $$\Delta S = m_1c \int_{T_1}^{T*} \frac{dT}{T} + m_2c \int_{T_2}^{T*} \frac{dT}{T}$$. And I did get the correct answer, 3.2 cal/K.

But, before that, I tried a different method; and, my question is why I'm wrong. The method is as follows. I also thought that I needed a reversible (or quasi-static) process. So I imagined the hotter water slowly dripping into the colder water while allowing equilibrium at almost every moment. Then, $$\frac{\Delta Q}{T} = \frac{c(T_h-T(m))dm}{T(m)}$$. T(m) can be easily determined through some calorimetric work. $$T(m) = \frac{T_hm+300T_c}{m+300}$$ We can then integrate both sides to get the change in entropy. $$\Delta S_c = T_h\int_{0}^{500}\frac{dm}{T(m)}-500$$ (Units are in calories, grams, and Kelvin. I have omitted the specific heat because it is one for water.) Hence, $$\Delta S_c = T_h\frac{300}{T_h}(1-\frac{T_c}{T_h})\ln(1+\frac{5}{3}\frac{T_h}{T_c})$$. A similar logic applies to the change in entropy of the hotter water being cooled down by the cold water. Here, I assume that entropy is additive (which was given in the book). The final answer I got was 14.77 cal/K. Setting aside the gory details of the calculations, is there anything wrong with the logic?
 
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Your alternative path is not reversible. Can you see why?
 
That’s my confusion. I added the water in slowly allowing for equilibrium at almost every moment. If I wanted to reverse a step, I can just pour that drop back in. I am mimicking the quasi-static process described in the book for a gas (i.e. slowly take out grains of sand which compresses a piston while being at equilibrium after each step …).
 
domephilis said:
That’s my confusion. I added the water in slowly allowing for equilibrium at almost every moment. If I wanted to reverse a step, I can just pour that drop back in. I am mimicking the quasi-static process described in the book for a gas (i.e. slowly take out grains of sand which compresses a piston while being at equilibrium after each step …).
I’m not sure this is right, but here’s my guess. When I pour the hot to the cold, I heat the cold water somewhat and gets warmer water. When I try to reverse it by pouring the water back, that bit back is not the 80C that it used to be and will cool the hotter water. Hence, we end up with a result different than our original. Does that sound right?
 
domephilis said:
I’m not sure this is right, but here’s my guess. When I pour the hot to the cold, I heat the cold water somewhat and gets warmer water. When I try to reverse it by pouring the water back, that bit back is not the 80C that it used to be and will cool the hotter water. Hence, we end up with a result different than our original. Does that sound right?
Yes. Very good.
 
Chestermiller said:
Yes. Very good.
I see. Thank you very much. Have a nice day.
 
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