## Chapter 9 Problem 43

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

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E_Villavicencio 2N
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### Chapter 9 Problem 43

In the solution of this problem, we have to consider two samples of water that were mixed and obtain the change in entropy. However, when we have to use the formula delta S=nCln(T2/T1) the solution manual uses 75.3 J/K as the heat capacity of water. Can someone explain to me where that value came from? Why don't we use 4.184J/K?

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### Re: Chapter 9 Problem 43

So the simplest solution to your question is that this is the exact same value.

If you take $4.184 \frac{J}{mole\cdot K}$ and multiply it by water's molar mass of 18.02 g/mole, you get $75.3 \frac{J}{mole\cdot K}$

In addition, it is found in Table 9.2 of the book on pg334

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