Answer:
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
<em>1 mole of propane release 2222kJ</em>
10.0g of propane (Molar mass: 44.1g/mol).
10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>
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:
<h2>83ºC</h2>
