A 10.0 g sample of propane, C3H8, was combusted in a constant-volume bomb calorimeter. The total heat capacity of the bomb calorimeter and water was 8.0 kJ/°C. The molar heat of combustion of propane is -2 222 KJ/mol. If the starting temperature of the water was 20 °C, what will be the final temperature of the bomb calorimeter?

83ºC

Explanation:

A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.

The reaction of combustion of propane is:

C₃H₈ +  5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol

1 mole of propane release 2222kJ

10.0g of propane (Molar mass: 44.1g/mol).

10.0g ₓ (1mol/ 44.1g) = 0.227 moles of C₃H₈

If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):

0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.

Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:

504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.

As initial temperature was 20ºC, final temperature will be:

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