# 7 ºc table 2 temperature change of heated water when

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O: 23.7 ºC Table 2. Temperature Change of Heated Water When Placed into Calorimeter Time (min) Temperature ( o C) 0 43.2 1 42.8 2 42.4 3 42.1 4 41.6 5 41.4 Table 3. Temperature Change of Heated & Cool Water Mixture Time (min) Temperature ( o C) 1 30.5 2 30.5 3 30.5 4 30.4 5 30.4 6 30.4 7 30.3 8 30.3 9 30.3 10 30.2

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Part C Concentration of HCL: 1.0520 M Temperature of HCl before mixing: 21.2 ºC Temperature of NaOH before mixing: 21.7 ºC Table 4. Temperature Change of HCl (aq) with the Addition of NaOH (aq) Time (s) Temperature ( o C) 5 26.8 10 26.9 15 26.8 20 26.8 25 26.8 30 26.8 35 26.8 40 26.8 45 26.8 50 26.8 55 26.8 60 26.8 90 26.7 120 26.7 150 26.7 180 26.7 210 26.7 240 26.6 270 26.6 300 26.6 330 26.6 360 26.6 390 26.6 420 26.6 450 26.6 480 26.6 510 26.5 540 26.5 570 26.5 600 26.5
Results and Calculations Part A 0 200 400 600 800 1000 1200 1400 1600 23 26 29 32 35 38 41 44 47 Figure 1. Cooling Curve of Temperature when NaOH and H2O are combined Time (s) Temperature (ºC ) ΔT = 23.0 ºC

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Part A Calculations Calculating molar enthalpy for NaOH: m water V water = 250.0g m soln = m water + m NaOH = 250.0g + 10.01g = 260.01g n = m/M = 10.01 g/ 39.998 g/mol = 0.25 mol NaOH Heat of solution for NaOH per mole of NaOH assuming no heat lost to calorimeter: C cal =0 because no heat was lost to the calorimeter q rxn = - (m soln * S soln ) (T f - T i ) = -(260.01 g) (4.184 J/g* º C) (23.0 º C) = -25021.3 J ∆H soln = q rxn / n = -25021.3 J/0.25 mol = -100085.1 J/mol Therefore, the molar enthalpy of the solution without heat lost to the calorimeter is -100085.1 J/mol. Heat of solution of NaOH per mole of NaOH including heat lost to Calorimeter: q rxn = - (C cal + m soln * S soln ) (Tf - Ti) = - (14.2 J/ C + 260.01g * 4.184 J/g* C) (23.0 ̊ ̊ º C) = -25347.9 J ∆H soln = -25347.9 J/0.25 mol = -101391.5 J/mol Therefore, the molar enthalpy of the solution with health lost to calorimeter is -101391.5 J/mol.
Part B -8 -6 -4 -2 0 2 4 6 8 10 12 23 26 29 32 35 38 41 44 Figure 2. Temperature of Water prior to and after Addition of Cool Water Time (min) Temperature (ºC )

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Part B Calculations T i for cool water = 23.7 C ̊ T i for heated water = 40.9 C ̊ T f for cool water = 30.7 C ̊ T f for heated water = 30.7 C ̊ ∆T heated water = -10.2 C ∆T cool water = 7 C ̊ ̊ mhot water = mcal with hot water - mcal = 139.74 g – 58.53 g = 81.21 g mcold water = mcal with cold & hot water – mhot water - mcal = 261.98 g – 81.21 g – 58.53 g = 122.24 g Calculating heat capacity of calorimeter: q hot water = - (q cold water + q calorimeter ) which means q cal = -q hot water – q cold water C cal * ΔT = - (m*s* ΔT) hot water – (m*s* ΔT) cold water = -81.21g * 4.184 J/g* C * (-10.2 C) – 122.24g * 4.184 J/g* C * 7 C = -144.38 J ̊ ̊ ̊ ̊ C cal = -144.38 J / (-10.2 C ̊ ) = 14.2 J/ C = 0.0142 KJ/ C ̊ ̊ Therefore, the calorimeter constant is 0.0142 kJ/ºC.

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