CHEM 150 1st Edition Lecture 14Outline of Last Lecture • PV work• Specific heat• EnthalpyOutline of Current Lecture • Calorimeter• Heat Capacity• Calorimeter constants Current LectureQ: If 4.45kJ of heat are applied to 85.00g of copper at 23.0C, what is the final temperature of the copper? Cs(Cu, solid)= 0.387 J/gECA: q= Cs x m x ΔTΔT = q/Cs x mΔT= 4450J/0.387J/C = 135.2789CTf = Ti + ΔT = 23.0+135.2789C = 158.2789 CQ: If 100.0g of copper at 235.0C are placed in 100.0 mL of water at 25.0C, what is the final temperature of the copper/water system? [assume no loss of water as steam]A: Tf is the same for the Cu and the H2Oand qCu = -q H2OqCu = 0.387J/gC x 100.0g x (Tf-235.0C)qH2O = 4.184J/gC x 100.0g x (Tf-25.0C)0.387J/gC x 100.0g x (Tf-250C) = -4.184J/gC x 100.0g x (Tf-250C)0.387J/gC x Tf - 0.387J/gC x 253.0 = -4184 J/gC x Tf + 4184J/gC x 25.0C Tf = 195.545J/g /4.571J/gC =42.7795C = 42.78CCalorimetry: Measuring ΔH (or ΔE)Bomb calorimeter: Constant volumeif ΔV = 0, then w = PΔV= 0 and ΔE = qQ: A bomb calorimeter has a heat capacity (C) of 4.1356 kJ/C. The combustion of 1.500 g of C6H5COOH raises the temperature of the calorimeter from 22.50 C to 31.69 C. What is ΔE for the combustion of benzoic acid in kJ/mol?A: 1.500g x 1mol/122.118g = 0.0122832molqCal = 4.1356KJ/C (31.69C -22.50C) =38.006164KJqrxn = -qcal = -38.006164KJΔE = Qrxn/n = -38.006164KJ/0.012832mol = -3094.1582 KJ/molΔE= -3.09 x 10^3 KJ/molQ: The heat of combustion of camphor (C10H16O) is -5903.6 kJ/mol. If the combustion of 4.682 g of camphor raises the temperature of a bomb calorimeter from 23.073C to 38.554 EC, what is the heat capacity of the calorimeter?A: 4.682g x 1mol/162.228g = 0.030756495molQrxn = 0.0307565mol x -5903.6 KJ/ 1 mol = -181.574KJQcal = -Qrxn = 181.574KgCp= Qcal/ΔT = 181.574KJ/(38.554C - 23.073C) =
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