Question

A 1.82 g sample of solid N(CH3)4Cl (s) is dissolved in 148 mL of water in a coffee cup calorimeter. Once all of the N(CH3)4Cl (s) is dissolved in the water, the final temperature of the solution is found to be 12.46°C. If the initial temperature of the water in the calorimeter was 21.36 °C, calculate the calorimeter constant (in J/K) for the coffee cup calorimeter. Report your answer to three significant figures. The heat of solvation of N(CH3)4Cl (s) is 4.08 kJ/mol.

Answer 1

Answer – We are given the, mass of solid N(CH₃)₄Cl = 1.82 g , volume = 148 mL , tf = 12.46^oC, ti = 21.36^oC, ∆Hfus = 4.08 kJ/mol

Heat loss by the water = Heat gain by the N(CH₃)₄Cl(s) + heat of calorimeter

Heat loss by the water, q = m*C*∆t

                                        = 148 g * 4.184 J/mol.^oC * (12.46^oC-21.36^oC)

                                       = -5511.2 J

Heat gain by the N(CH₃)₄Cl(s)

q = m*∆Hfus

moles of N(CH₃)₄Cl = 1.82 g /109.60 g.mol^-1

                                 = 0.0166 moles

So, q = 0.0166 moles * 4080 J/mol

          = 67.65 J

So, heat of calorimeter = 5511.2 J -67.65

                                     = 5443.4 J

So, ∆H/∆t = C calorimeter constant

So, calorimeter constant, C = 5443.4 J /(12.46^oC-21.36^oC)

                                            = 612 J/K