## 9.43 - Where do we get Cp of H2O (l) = 75.3 J/Kmol?

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

Michael_Johanis_2K
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### 9.43 - Where do we get Cp of H2O (l) = 75.3 J/Kmol?

Suppose that 50 g of H2O(l) at 20 degrees C is mixed with 65 g of H2O(l) at 50 degrees C at a constant pressure in a thermally insulated vessel. Calculate delta S and delta S total for the process.

I understand that we find the sum of two entropies, one for each water sample.
We use ChangeEntropy = nCln(Tf/Ti). However, the solutions manual says that Cp,m = 75.3 J/Kmol for water. How do we find this number?

Chem_Mod
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### Re: 9.43 - Where do we get Cp of H2O (l) = 75.3 J/Kmol?

The 75.3 J/K.mol is the molar heat capacity for water under conditions of constant pressure. This is a constant, not a value you are supposed to solve for. All constants and formulas will be provided to you on the formula sheet.

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