Respuesta :
Answer:
Heat Capacity of calorimeter is 6.842 kJ/°C
Explanation:
4 C6H5NH2(l) + 35 O2(g) → 24 CO2(g) + 14 H2O(g) + 4 NO2(g)
ΔH°rxn = -1.28 x 10⁴ kJ
4 moles of aniline gives 12800 kJ of heat,
1 mole of aniline will give 12800/4 kJ amount of heat = 3200 kJ
ΔH°rxn for 1 mole of aniline = -3200 kJ
1 mole of aniline weighs 93.13 g/mol
6.55 g of aniline has 6.55/93.13 moles = 0.0703 moles
And this would give off 0.0703 × 3200J amount of heat off
0.0703 × 3200 = 225.1 kJ.
Heat lost by aniline = Heat gained by calorimeter,
Heat lost by aniline = 225.1 kJ
Heat gained by calorimeter = C ΔT = C (32.9)
225.1 = C (32.9)
C = 6.842 kJ/°C
The heat capacity of the calorimeter has been found to be 125.53 J/kg.K.
To calculate the heat capacity of the calorimeter, the heat gained by the calorimeter has to be calculated. The heated aniline results in the transfer of heat to the calorimeter.
Thus, the heat gained by the calorimeter = heat lost by aniline.
Heat lost by the aniline = mc[tex]\Delta[/tex]T
m = mass = 6.55 g = 0.00655 kg
c = specific heat = 2060.6 J/ kg. K
[tex]\Delta[/tex]T = change in temperature = 32.9[tex]\rm ^\circ C[/tex] = 306.05 K
Substituting the values in the equation,
Q = 0.00655 [tex]\times[/tex] 2060.6 [tex]\rm[/tex][tex]\times[/tex] 306.05
Q = 4,130.73 Joules
Q = 4.13 kJ
The heat lost by aniline = heat gained by calorimeter
Heat lost by aniline = 4.13 kJ
Heat gained by calorimeter = 4.13 kJ
Heat capacity [tex]\times[/tex] change in temperature = Heat gained by calorimeter
Heat capacity = [tex]\rm \dfrac{heat\;gained}{change\;in\;temperature}[/tex]
Heat capacity of calorimeter = [tex]\rm \dfrac{4.13}{32.9}[/tex]
Heat capacity of calorimeter = 0.1255 kJ/kg.K
Heat capacity of calorimeter = 125.53 J/kg.K
The heat capacity of the calorimeter has been found to be 125.53 J/kg.K.
For more information about the heat capacity, refer to the link:
https://brainly.com/question/11194034
