The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be −3226.7 kJ/mol. When 2.8161 g of benzoic acid are burned in a calorimeter, the temperature rises from 21.84°C to 24.67°C. What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly 2250 g.)

n = moles of benzoic acid = [tex]\frac{\text{Mass of benzoic acid}}{\text{Molar mass of benzoic acid}}=\frac{2.8161g}{122.122g/mole}=0.02306mole[/tex]

[tex]\Delta H[/tex] = enthalpy of combustion = 3226.7 kJ/mole

Now put all the given values in the above formula, get:

[tex]q=(0.02306mole)\times (3226.7kJ/mole)=74.407kJ[/tex]

Now we have to calculate the heat capacity of the bomb calorimeter.

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

[tex]q=[q_1+q_2][/tex]

[tex]q=[c_1\times \Delta T+m_2\times c_2\times \Delta T][/tex]

where,

q = heat released by the reaction = 74.4077 kJ = 74407.7 J

[tex]q_1[/tex] = heat absorbed by the calorimeter

[tex]q_2[/tex] = heat absorbed by the water

[tex]c_1[/tex] = specific heat of calorimeter = ?

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_2[/tex] = mass of water = 2550 g

[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=24.67-21.84=2.83^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]74407J=[(c_1\times 2.83^oC)+(2550g\times 4.18J/g^oC\times 2.83^oC)][/tex]

[tex]c_1=15633.226J/^oC=15.633kJ/^oC[/tex]

Therefore, the heat capacity of the bomb calorimeter [tex]15.633kJ/^oC[/tex]