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The heat capacity of a bomb calorimeter is found t
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Answer #1

Ans. Given, heat of combustion of glucose = - 2.81 x 103 kJ/mol. The -ve sign indicates that heat is being released during combustion.

Moles of glucose combusted in bomb calorimeter = Mass of glucose sample / Molar mass of glucose

                                    = 1.94 g / 180.16 g mol-1

                                    = 0.01076821 moles

Amount of heat produced from combustion of 0.01076821 moles glucose =

                        Molar heat of combustion of glucose x moles of glucose

                        = - 2.81 x 103 kJ/mol x 0.01076821 moles

                        = - 0.0303 x 103 kJ

                        = - 30.3 kJ

Thus, combustion of given glucose sample releases 30.3 kJ heat

#1. Heat absorbed by calorimeter.

In calorimeter, the heat released by combustion of a sample (here, glucose) is absorbed by the calorimeter to increase its temperature.

Thus, heat absorbed by calorimeter = + 30.3 kJ. The +ve sign indicates that heat is being absorbed by the calorimeter.

#2. Let the initial temperature of calorimeter = T

Now,    q = C x dT        - equation 2

            where, q = heat gained by calorimeter = 30.3 kJ

C = heat capacity of calorimeter system = 5.25 kJ/0C

                        dT = Final – initial temperature = 36.00C – T

Putting the values in above equation-

            30.3 kJ = 5.25 kJ/0C x (36.00C – T)

            Or, 36.00C – T = 30.3 kJ / (5.25 kJ/0C) = 5.770C

            Or, -T = 5.770C - 36.00C = -30.23

            So, T = 30.230C

Thus, initial temperature of calorimeter = 30.230C

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