posted by Akle .
DrBob222, I got an answer of 33 and 24.23 as the temperature but it is incorrect. I looked at your response to David and unfortunately it did not help me. Would you please take a look at this question and help me out. Thanks.
If the heat from burning 5.800 g of C6H6 is added to 5691 g of water at 21 °C, what is the final temperature of the water?
2C6H6(l) +15O2 (g)----> 12CO2(g)+6H2O(l) +6542
heat from burning 5.8 g benzene is
6542 x (5.800/2*78.114) = 242,873.2 J.
242,873.2 = 5691 g x 4.184 J/g x (Tf-21)
242,873.2 = 23,811.14T - 500,034.02
T = (500,034.02 + 242,873.2)/23,811.14
T = about 31.19999 which I would round to 31.20 (which is consistent with my estimate with David of 33. I did make a typo on his post which I hope he caught. That is not 293,000, it is closer to 243,000. I hit a 9 instead of a 4. I assume the 21 C is 21.00. There are 4 s.f. everywhere else in the problem, including the 5.800 so 31.20 should do it. You should go through this carefully and confirm all of the numbers. The method is ok but I could have made a calculator error. Check carefully for significant figure issues. I not that many students on this board have the right method and numbers BUT they fail to type in the right number of s.f. Those data bases are so unforgiving. Hope this helps (and works).
Thank you! It did and i found my error a minute before you posted this. Once again, I always appreciate your help DrBob222 :)