The combustion of 1.014g of benzoic acid brings a temp change of 22.2 to 24.86^0C

During the process the fuse of the bcalorimeter melts from 0.12g to0.010g .with delta H=5857j/g
Find the energy change for the overall process
Combustion of Benzoic acid is given as
C6H5COOH+CO2____CO2+H20 deltaH =26.456kj?g

q(v) = E = [(26,456 kJ/g x 1.014 g)(24,86-22.2)] + [(0.12-0.10)(5.857 Kj/g)(24.86-22.2)] = ?

For the second part do we use delta T, as it is melting and during melting temperature is constant

To find the energy change for the overall process, you need to calculate the energy change for both the combustion of benzoic acid and the melting of the fuse, and then add them together.

1. Energy change for the combustion of benzoic acid:
First, calculate the moles of benzoic acid used in the combustion:

Molar mass of benzoic acid (C6H5COOH) = 122.12 g/mol
Moles of benzoic acid = mass / molar mass
= 1.014 g / 122.12 g/mol
= 0.0083039 mol (approximately)

Now, calculate the energy change for the combustion of benzoic acid using the given delta H value:

Energy change = moles of benzoic acid * delta H
= 0.0083039 mol * 26.456 kJ/mol
= 0.219271 kJ (approximately)

2. Energy change for the melting of the fuse:
Given delta H for the melting of the fuse = 5857 J/g
Mass of the fuse = 0.12 g - 0.010 g = 0.11 g

Energy change = mass of the fuse * delta H
= 0.11 g * 5857 J/g
= 644.27 J

3. Total energy change for the overall process:
Add the energy changes for the combustion and the melting of the fuse together:

Total energy change = energy change for the combustion + energy change for melting of the fuse
= 0.219271 kJ + 644.27 J
= 0.219271 kJ + 0.64427 kJ
= 0.863541 kJ (approximately)

Therefore, the energy change for the overall process is approximately 0.863541 kJ.