t Of Cold H20 (gained) + the CalorimeterHeat Of Hot H20 (lost) = SH of H20 x Mass of Water x ∆THeat Of Cold H20 (gained) = SH of H20 x Mass of Water x ∆TSpecific Heat of H20 (SH) = 4.184 J/g Amount of H20 Used = 50 gramsTi Cold = 23.1 Ti Hot = 81.4 Tf = 50.3 ∆TCold = 50.3 C 23.1 = 27.2 ∆THot = 50.3 C 81.4 = -31.1 Heat of Cold H20 = 4.184 J/g x 50 g x 27.2 = 5690.24 JHeat of Hot H20 (lost) = 4.184 J/g x 50 g x -31.1 = 6506.12 J6506.12 J = 5690.24 J + Heat gained by CalorimeterHeat gained by Calorimeter = 815.88 JHeat Capacity = Heat gained by the Calorimeter / ∆TCold= 815.88 J / 27.2 = 30.0 J/ ( 0.03 kJ/ )The two calorimeters used were consistent in their heat absorption which is proven by the results in Table 1. Since calorimeter 1 had the lowest heat capacity, it was the best of the two calorimeters Salts in Water Tests:The first experiment performed was the heats of reaction for salts in water. The results of the salts in water experiment were shown in Table 2. Sodium chloride and barium chloride were picked to conduct the tests for our salts, and preformed in calorimeter one. The initial temperatures of the salts, however, were not factored into our equations. Instead room temperature was used for the initial temperature of the salts; which caused a discrepancy. A sample calculation of ∆H for a salt in water is shown below. These calculations are, also, the same ones done for the following experiments.Sample Calculation for ∆H for a Salt in Water (Calorimeter One)Heat Change in the Reaction = Heat Change of Solution + Heat Change of CalorimeterHeat Change of Solution = Total Mass of Solution * Specific Heat * ∆THeat absorbed by Calorimeter = Heat Capacity * ∆TTotal Mass of Solution = 50 g H2O + 5 g NaCl = 55 gSpecific Heat of H2O = 0.004184 kJ/g Ti = 17.1 Tf = 16.9 ∆T = 16.9 C 17.1 = -0.2 He...
