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Topic: Bomb Calorimeter  (Read 5038 times)

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Offline agentsmith1011

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Bomb Calorimeter
« on: October 21, 2006, 07:30:57 PM »
A bomb calorimeter (Kc = 775 J/K) containing 1150 g of water (c = 4.18 J/g C) is at equilibrium at 20.5 C.  If the heat of combustion for C8H8 is -4560 kJ/mole, determine the equilibrium temperature of the system following the complete combustion of 1.25 g of C8H18.

My Attempt:

n, C8H18 = 1.25g/(114.224 g/mol) = 0.0109 mol

q c + q H2O + q C8H18 = 0
775[ ( Tf - 20.5) + 273 ] + 1150*4.18*(Tf - 20.5) + 0.0109*(-4560*10^3 J/mol)=0
Tf = -8.98 C

Note: I tried adding 273 for the temperatue in q c so all my temperatures would be in celcius.  In q C8H18, I multipled -4560 by 10^3 to convert kJ to J.

I got 2 out of 10 points for this problem.  Can someone point out my mistakes?

Offline moussa

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Re: Bomb Calorimeter
« Reply #1 on: October 22, 2006, 06:36:16 AM »
first you didn't use the KC ??? why is it given then???
water is a product not a reactant >:( how did you put ti=25?? it was not found at all in the begining
2nd 775[ ( Tf - 20.5) + 273 ] + 1150*4.18*(Tf - 20.5) + 0.0109*(-4560*10^3 J/mol)=0
you added 273 in left only??both temperatures must be in the same unit
it has many mistakes!!!
check your teacher for complete solution
earth is larger than a molecule

Offline agentsmith1011

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Re: Bomb Calorimeter
« Reply #2 on: October 22, 2006, 10:11:34 PM »
"first you didn't use the KC Huh why is it given then???"

Actually I did.
qc + qH2O+ qC8H18 = 0
qc = Kc * T = (775 J/K)*(Tf - Ti) k = (775 J/K)* etc...

"water is a product not a reactant Angry how did you put ti=25?? it was not found at all in the begining"

I thought about it for a long time.  I figured the initial temperature for the water and bomb calorimeter must be the equilibrium temperature 20.5 Celcius.

"you added 273 in left only??both temperatures must be in the same unit"

I did that because Kc temperature is in kelvin whereas the others are in celcius.  I figured I had to add 273 to convert the kelvin to celcius.

"check your teacher for complete solution"

The solution isn't posted.  He briefly went over it in class but that day I was late.  Besides, I like the explanations I get here. :)

Offline agentsmith1011

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Re: Bomb Calorimeter
« Reply #3 on: November 05, 2006, 07:18:04 PM »
So I am waiting for an answer for this post?  Clearly if I still need the answer now, I have already tried every other way of getting it.

Offline kamitengo

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Re: Bomb Calorimeter
« Reply #4 on: May 08, 2007, 03:41:23 AM »
Tf = -8.98 C??
lower temperature after combustion of C8H18??
it's unreasonable!

water will became ice? did you consider about heat of freezing?

"I did that because Kc temperature is in kelvin whereas the others are in celcius.  I figured I had to add 273 to convert the kelvin to celcius."

you should use [(Tf+273)-(20.5+273)]
the equilibrium temperature of the system is 29.04 C

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