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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 known as calibrating the calorimeter.
In the laboratory a student burns a 0.714-g sample of β-D-fructose(C6H12O6) in a bomb calorimeter containing 1020. g of water. The temperature increases from 25.60 °C to 27.80 °C. The heat capacity of water is 4.184 J g-1°C-1.
The molar heat of combustion is −2810. kJ per mole of β-D-fructose.
C6H12O6(s) + 6 O2(g) Arrow.gif6 CO2(g) + 6 H2O(l) + Energy
Calculate the heat capacity of the calorimeter.
heat capacity of calorimeter = ___________ J/°C

Respuesta :

Akinny

Answer:

798.8351 J/ °C

Explanation:

Number of moles of  fructose burnt (n)= Mass (m)/ Molar mass(M)

n=  m/M................................................ (1)

m= 0.714 g ;

M = [6 (12) + 12(1) +6(16)]

   = 72 +12+96

   = 180 g

n= 0.714/180

 = 0.003966 moles

The molar heat of combustion for fructose is 2810 kJ  per mole. This implies that the heat released in the experiment is:

2810 000 J x 0.003966

=  11,146.3333J

According the energy conservation, assuming no heat loss, the heat energy released from the combustion of fructose will be transferred to the calorimeter and the water according to the equation below:

Heat from Combustion of fructose = mwCwΔФ + HΔФ...................... (2)

Where mw= mass of water in calorimeter, mw =1020 g

           Cw = Specific heat capacity of water, Cw =4.184 J g-1°C-1

            ΔФ= temperature change , ΔФ =(27.80 -25.60) =2.20 °C

            H = Heat Capacity of Calorimeter, J/ °C

Substituting into equation  (2)

11,146.3333 = (1020 x  4.184 x 2.20) + ( H x 2.20)

11, 146.3333 =    9,388.896 + 2.2 H          

11, 146.3333 - 9,388.896 =2.2 H

1,757.4373  = 2.2 H

H = 1,757.4373/2.2

   =  798.8351 J/ °C

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