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.

Question

07-10-2022
Chemistry

Answered

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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.

Sagot :

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