A bomb calorimeter, or constant volume calorimeter, is a device often used to de

Question

A bomb calorimeter, or constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods. Since the "bomb" itself can absorb energy, a separate experiment is needed to determine the heat capacity of the calorimeter. This is know n as calibrating the calorimeter. In the laboratory a student bums a 0.410-g sample of decanoic acid (C_10 H_20 O_2) in a bomb calorimeter containing 1080 g of water. The temperature increases from 25.30 degree C to 28.00 degree C. The heat capacity of water is 4.184 J g-1 degree C^-1 The molar heat of combustion is 6080. kJ per mole of decanoic acid. C_10 H_20 O_2(s) 9 14 O_2 (g) rightarrow 10 CO_2 (g) + 10 H_2 O(l) + Energy A bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods. In an experiment, a 0.4395 g sample of biphenyl (C_12H_10) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.064 Times 10^3 g of water. During the combustion the temperature increases from 26.33 to 29.52 degree C. The heat capacity of water is 4.184 J g^-1 degree C^-1. The heat capacity of the calorimeter w as determined in a previous experiment to be 946.6 J/degree C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of biphenyl based on these data.

Explanation / Answer

Formula: Heat released by combustion= Heat absorbed by water and calorimeter

Heat of combustion = 6080 kJ/mol decanoic acid

Molar mass of decanoic acid is 172g/mol

Moles of 0.410 g of decanoic acid = (0.410 g/172 g/mol) = 0.002384 mol

Heat released by combustion= (6080 kJ/mol decanoic acid)*(0.002384 mol)

                                                    = 14.49 kJ

                                                     = 14.49*10^3 J

Heat absorbed by water and calorimeter = (mass of water*heat capacity                        water*+heat capacity of calorimeter (C))*temperature change

= (1080g*4.184 J/g oC +C)*(28.00-25.30) oC

= (4518.6 J/ oC + C) 2.7 oC

14.49*10^3 J = (4518.6 J/ oC + C) 2.7 oC

C = (14.49*10^3 J/2.7 oC)-( 4518.6 J/ oC) = 848 J/ oC

Heat capacity of calorimeter is 848 J/ oC.

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