- #1
domephilis
- 54
- 6
- 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?
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?
