A 0.3423 g sample of pentane, C5H12, was burned in a bomb (constant-volume) calorimeter. The temperature of the calorimeter and the 1.000 kg of water contained therein rose from 20.22°C to 22.82°C. The heat capacity of the calorimeter is 2210J/°C. The specific heat of water is 4.184 J/g·°C. Calculate the heat of combustion of pentane in kJ/g.

Respuesta :

Answer:

16.82 kJ/g

Explanation:

The Heat energy generated by the pentance combusion caused the temperature increasing of the caloriemeter and the water inside it.

Heat energy of a substace can be calculated using

Q = mcθ

C= mc

where Q = Heat energy

          m =mass

          c = specific heat capacity

          C = heat capacity

          θ =temperature difference

(Temperature difference = final temperature - initial temperature.)

heat absorbed by water = mcθ

                                         = 1  * 4.184 * (22.82-20.22)

                                         = 10.88 J --------------(1)

heat absorbed by calorimeter = Cθ

                                                  = 2210 *(22.82-20.22)

                                                  = 5746 J ---------------(2)

Combustion energy = heat absorbed by water + heat absorbed by                                                  calorimeter

Combustion energy =  10.88+ 5746 = 5756.88 J

                                 = 5.757 kJ

heat of combustion of pentane = Combustion energy / mass

                                                    = 5.757 / 0.3423

                                                    = 16.82 kJ/g

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