J.R. S. answered 02/12/22
Ph.D. University Professor with 10+ years Tutoring Experience
So, Lauren K, you asked this same question the other day, and I submitted an answer. Was there something wrong with that answer? If so, instead of posting the question again, you could leave a comment so I could respond. Otherwise, I'll answer it again making the same error (if there was an error). It helps everyone on this site if you participate in feedback. Simply posting the question again, will only get you the same answer. See below.
Heat generated by combustion of 3.450 g naphthalene:
3.450 g x 40.1 kJ / g = 138 kJ
This heat will be absorbed by both the water and the calorimeter to raise the temperature of both.
q = mC∆T + Ccal∆T
q = heat = 138 kJ
m = mass of water = 1.000 kg
C = specific heat of water = 4.184 kJ/kgº
∆T = change in temperature = 4.345º
Ccal = calorimeter constant
Solve the equation for Ccal...
138 kJ = (1.000 kg)(4.184 kJ/kgº)(4.345º) + (Ccal)(4.345º)
138 kJ = 18.18 + 4.345Ccal
Ccal = 27.6 kJ/º

J.R. S.
02/12/22
Lauren K.
Okay thank you, I'm so sorry I didn't even notice I posted the same question twice! It's been so hectic this week with homework/assessments I feel like ive been all over the place lol, thank you for helping!02/12/22