At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

A 3.627g sample of a new organic material is combusted in a bomb calorimeter. The temp of the calorimeter and contents increases from 25.39 Celcius to 30.14 Celcius. Heat capacity (calorimeter constant) is 45.35 KJ/degree C. What is the heat of combusion per gram of the material.

I tried to work this out and need to know if this is correct:
I'm assuming -Qrxn =C(calorimeter) x Delta T
45.35J/degree C x 4.75 (which is change of T) = -20.67KJ since exothermic.
THen Delta Urxn= Qrxn/g of organic material.: -20.57KJ/g sample x gram sample/3.62g
This would be -5.699KJ/g.
Please let me know if this is correct or where I have made an error.