The heat of combustion of benzene, C6H6, is -41.74 kJ/g. Combustion of 3.65 of benzene causes a temperature rise of 4.41 C^o in a certain bomb calorimeter. What is the heat capacity of this bomb calorimeter? (kJ/C^o)

Question

So, I what I did was I multiplied -41.74 kJ/g and 3.65 to get the kJ alone. This gave me -152.351. Then, I divided that number by -4.41 (negative because it goes down by that amount) to get kJ/C^o. I got 34.5 kJ/C^o as my answer. Apparently, this is not the answer. Help?

Answer 1

If you carry that out to one more place MAYBE it is
(41.74*3.65/4.41) = 34.5467 and if I round that to 34.55 and round that to 34.6 to three significant figures; otherwise, your procedure looks ok to me.

Answer 2

yesh