Respuesta :
Answer:
-6.517 kJ is the heat of combustion for this sample that is 0.2301 grams of ethanol.
Explanation:
Mass of bomb calorimeter = 2.000 kg = 2.000 × 1000 g = 2000. g
1 kg = 1000 g
Specific heat of bomb calorimeter = c = 2.45 J/g°C
Change in temperature = [tex]\Delta T=1.33^oC[/tex]
Heat absorbed by the bomb claorimeter = Q
[tex]Q=m\times c\times \Delta T[/tex]
[tex]=2000 g\times 2.45 J/g^oC\times 1.33^oC=6,517 J[/tex]
Heat energy released on combustion of 0.2301 grams of ethanol = Q'= -Q
Q' = -Q = -6,517 J = -6.517 kJ
1 J = 0.001 kJ
-6.517 kJ is the heat of combustion for this sample that is 0.2301 grams of ethanol.