# Chemistry

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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

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Jen, Sally, Jason, or whoever -- you must be undergoing an identity crisis. Please use the same name for your posts on Jiskha.

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I worked this for David earlier with slightly different numbers but I'm sure you can transpose as needed.
http://www.jiskha.com/display.cgi?id=1328853263

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Do I use the 293,000 j or 293 kj?

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I suppose it depends upon what you use for specific heat. If you use 4.184 J/g*C then 293,000 is the one to use (remember the 293,000 is just an approximation).

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I still don't get how to do it. I did what you said and it keeps marking me incorrect :(

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If you will post your work I'll try to find the error. You're probably just making an algebra error. What answers have you posted.

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I have this same question on my hw and it is marking me wrong. Can someone please say what the right answer is or confirm it?

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my work:

243.228= 18.0153 * 4.184* (Tfinal-Tinitial)

1825.228=75.376Tfinal

Tfinal= 24.2149

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I am still having trouble as well...

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You have two errors.
First, You have kept the 243,228 in kJ and used J in specific heat. Change that to J (but I would redo that if I were you. It is 6542 x (5.800/2*molar mass C6H6) = I used 78.114 for molar mass benzene. Also, you don't have it in your post but I thing that 6542 is kJ. Anyway, I obtained 242,873.2 J for the combustion of 5.800 g benzene.
Your second error is the 18.015. That is the molar mass of H2O (I suppose) but that is not what goes there. What goes there is the mass of water. From the problem that is 5691g. Outside of those two errors your other numbers are ok. you want to subsitute 21 for Tinitial. I went through all of this with Akle above and I came out with 31,1999 which I would round to 31.20 to 4 s.f. but you need to confirm that See if this doesn't help you.

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Yes it does. Thanks.

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I am having trouble too:
If the heat from burning 6.000 g of
C6H6 2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(l) + 6542 kJ is added to 5691 g of water at 21 °C, what is the final temperature of the water?
Please tell me what I am doing wrong:
293,551= 5691*4.2*(Tf-21C)
293551= 23902.2Tf - 50946.2
343946.2= 23902.2Tf
14.4=Tf

The homework program is telling me it's wrong.

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The final temperature is 35.6