Lab Partner: DATA: Atmospheric pressure o Part A: Calibration of Thermometers ma

Question

Lab Partner: DATA: Atmospheric pressure o Part A: Calibration of Thermometers marks) (2 Thermometer 1 Water Ice Bath Temperature (TC Water Boiling Temperature C Thermometer 2 Water Ice Bath Temperature ("C Water Boil Temperature (C Part B: Heat Capacity of the Calorimeter (5 marks) Mass of Calorimeter (g) A1 o. Mass of Calorimeter Cold Water (g) A2: 2 Mass of Cold Water (g) (A2-A1) Temperature of Cold Watertrc) Thermometer 1) measured: corrected Temperature of Hot Water rmometer 2) measured corrected: Final Temperature of Water(eC) (Thermometer 1) measured 39 2 Mass of Calorimeter Cold Water Hot Water (g) A3 Z2.q Mass of Hot Water (g) (A3-A2)

Explanation / Answer

Part A : Heat capacity of calorimeter

m(hot water) = 103.3 g

m(cold water) = 49.0 g

let 1.7 oC be the correction in thermometer (this is rough number, you may feed your number here) then,

dTb(hot water) = (47.5 - 39.2) - 1.7 = 6.6 oC

dTb(cold water) = (39.2 - 22.5) - 1.7 = 15 oC

Q(hot) = 103.3 x 4.18 x 6.6 = 2849.84 J

Q(cold) = 49 x 4.18 x 15 = 3072.3 J

heat gained by calorimeter = 222.46 J

Heat capacity of calorimeter = 222.46/15 = 14.83 J/oC

Part C) Molar heat of neutralization

Trial 1

Mass of solution = HCl + NaOH = 39.9 g

average temperature of solution = [(HCl + NaOH)/2] - 1.7 = 20.8 oC

[1.7 oC is taken as the correction in temperature, a rought value]

dTb = (30 - 1.5) - 20.8 = 7.5 oC

dHsoln = mCpdT

            = 39.9 x 4.18 x 7.5

            = 1250.865 J

molar heat of neutralization = 1250.865 J/1 M x 0.0196 L x 1000

                                             = 63.82 kJ/mol

Similarly other values for Trials may be calculated

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