When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the calorimeter is 69.6 J∘C and it is submerged in 944mL of water, how much heat (in kJ) was produced by the hydrocarbon combustion?

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

c: specific heat of water = 4.186 J/gºC

ΔT : change in temperature = 2.06 ºC

so solving for q:

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC

ΔT : change in temperature = 2.06 ºC

q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J + 143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ