to determine the value of the final temperature of the calorimeter

Question

A 6.4g sample of aniline (C 6H 5NH 2, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.25 kJ/°C. If the initial temperature was 32.9°C, use the information below to determine the value of the final temperature of the calorimeter.

4 C 6H 5NH 2(l) + 35 O 2(g) → 24 CO 2(g) + 14 H 2O(g) + 4 NO 2(g)

 

 ΔH° rxn= -1.28 x 10 4 kJ

Answer 1

For a bomb calorimeter we can use the equation q = Ccal∆T

q = heat = 1.28x10⁴ kJ for 4 moles 6H5N2

moles C6H5N2 used = 6.4 g x 1 mol/93.13 g = 0.0687 moles

q = 0.0687 mol x 1.28x104 kJ/mol = 880 kJ

Ccal = calorimeter constant = 14.25 kJ/º

∆T = ?

Solving for ∆T we have...

∆T = q / Ccal = 880 kJ / 14.25 kJ/º

∆T = 61.8º

Final temperature = 32.9º + 61.8º = 94.7ºC (note: we add the ∆T because the ∆Hºrxn is negative meaning heat is generated by the reaction and given off to the surroundings)