Question

A 3.917 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 23.13 ∘ C to 29.28 ∘ C. The heat capacity (calorimeter constant) of the calorimeter is 44.51 kJ / ∘ C, what is the heat of combustion per gram of the material?

Answer 1

Answer: The heat of combustion of the organic material is -69.88 kJ/g

Step-by-step explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

`q=c\Delta T`

where,

q = heat absorbed

c = heat capacity of calorimeter = 44.51 kJ/°C

`\Delta T` = change in temperature =
`T_2-T_1=(29.28-23.13)^oC=6.15^oC`

Putting values in above equation, we get:

`q=44.51kJ/^oC* 6.15^oC=273.74kJ`

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the enthalpy change of the reaction, we use the equation:

`\Delta H_(rxn)=(q)/(m)`

where,

q = amount of heat released = -273.74 kJ

m = mass of organic material = 3.917 g

`\Delta H_(rxn)` = enthalpy change of the reaction

Putting values in above equation, we get:

`\Delta H_(rxn)=(-273.74kJ)/(3.917g)=-69.88kJ/g`

Hence, the heat of combustion of the organic material is -69.88 kJ/g