Answer:
The heat absorbed by the bomb Calorimeter Q = 5.07 [tex]\frac{KJ}{mole}[/tex]
Explanation:
Given data
Heat capacity of calorimeter [tex]C_{cal}[/tex] = 6.97 [tex]\frac{KJ}{c}[/tex]
Temperature Rise ΔT = 32 °c
Therefore the heat absorbed by the bomb calorimeter
[tex]Q = C_{cal}[/tex] ΔT
Q = 6.97 × 32
Q = 223.04 KJ
Since 1 mole of propane weighs = 44 gm , So
[tex]Q = \frac{223.04}{44}[/tex]
Q = 5.07 [tex]\frac{KJ}{mole}[/tex]
Therefore the heat absorbed by the bomb calorimeter Q = 5.07 [tex]\frac{KJ}{mole}[/tex]
