1. a) 50ml of water at 46.9°C were mixed with 50ml of water at 25.1°C in a calorimeter also at 25.1°C. The final temperature was 30.1°C. Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter. (density of water = 1.00g/ml, specific heat capacity = 4.18 j/gk
b) when 5.00g of NaOH (s) are added to 100g of water (using the same calorimeter as in part a), the temperature rises from 25.0°C to 37.5°C. calculate the molar heat of solution, i.e. delta H for the process. assume that the specific heat of water is 4.18 j/gk and that of the NaOH (aq) solution is the same.
chemistry - DrBob222, Monday, November 12, 2012 at 4:18pm
[mass warm H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] ]+ [Ccal x (Tfinal-Tinitial)] = 0
Solve for Ccal.
[mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal x (Tfinal-Tintial)] = q = dHsoln in joules.
Convert 5.00 g NaOH to mols. mols = grams/molar mass.
Then dHsoln = q/mol NaOH = ?J/mol. These are most often quoted in kJ/mol and you can make that conversion; divide J/mol by 1000 for kJ/mol