Ans. Given, heat of combustion of glucose = - 2.81 x 103 kJ/mol. The -ve sign indicates that heat is being released during combustion.
Moles of glucose combusted in bomb calorimeter = Mass of glucose sample / Molar mass of glucose
= 1.94 g / 180.16 g mol-1
= 0.01076821 moles
Amount of heat produced from combustion of 0.01076821 moles glucose =
Molar heat of combustion of glucose x moles of glucose
= - 2.81 x 103 kJ/mol x 0.01076821 moles
= - 0.0303 x 103 kJ
= - 30.3 kJ
Thus, combustion of given glucose sample releases 30.3 kJ heat
#1. Heat absorbed by calorimeter.
In calorimeter, the heat released by combustion of a sample (here, glucose) is absorbed by the calorimeter to increase its temperature.
Thus, heat absorbed by calorimeter = + 30.3 kJ. The +ve sign indicates that heat is being absorbed by the calorimeter.
#2. Let the initial temperature of calorimeter = T
Now, q = C x dT - equation 2
where, q = heat gained by calorimeter = 30.3 kJ
C = heat capacity of calorimeter system = 5.25 kJ/0C
dT = Final – initial temperature = 36.00C – T
Putting the values in above equation-
30.3 kJ = 5.25 kJ/0C x (36.00C – T)
Or, 36.00C – T = 30.3 kJ / (5.25 kJ/0C) = 5.770C
Or, -T = 5.770C - 36.00C = -30.23
So, T = 30.230C
Thus, initial temperature of calorimeter = 30.230C
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