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July 24, 2014

July 24, 2014

Posted by **Anne** on Wednesday, December 12, 2007 at 8:26pm.

Question:

The heat capacity of a bomb calorimeter is 87.5 kJ/K (this value is for the total heat capacity including that of the water jacket around the reaction chamber). If 67.2g of CH4 (g), is combusted under such reaction conditions, what will be the increase in temperature of the calorimeter? Delta Ecombustion for CH4 (g) is -885.4 kJ/mol.

Answer:

drwls, Wednesday, December 12, 2007 at 7:38am

Calculate the heat energy release:

The number of moles of CH4 you are burning is 67.2/16 = 4.20 mol

Multiply that by 885.4 kJ/mol for the heat release. Then divide that by 87.5 kJ/K for the temperature rise.

Or, you could do it all at once

Delta T = (67.2g*885.4kJ/mol)/(16 g/mol*87.5 kJ/K) = ? K

- Chemistry -
**DrBob222**, Wednesday, December 12, 2007 at 8:43pmSo tell us your problem. What is it you don't understand.

- Chemistry -
**Anne**, Wednesday, December 12, 2007 at 9:01pmIs it okay to have a negative temperature (in degrees Celsius) in this case?

- Chemistry -
**DrBob222**, Wednesday, December 12, 2007 at 9:08pmCombustion reactions give off energy, as this one does; therefore, the temperature will rise. The question asks for how much the T will rise. So, no, I wouldn't expect the T to go down.

- Chemistry -
**Anne**, Wednesday, December 12, 2007 at 9:14pmBut I got an answer of 42.5K, which means when I convert it to degrees Celsius I get 42.5K - 273 K = -23.05 degrees Celsius.

Does this sound weird to you then?

- Chemistry -
- Chemistry -
**DrBob222**, Wednesday, December 12, 2007 at 9:29pmYes, it's weird. But your answer is correct; you're just confused on what your answer is. I worked the problem and I have 42.5 degrees BUT THAT IS DELTA T. That IS the rise in degrees C or degrees K (both are equivalent). The problem didn't ask for the T, it asked for delta T and that is what the equation is.

delta H = delta E = Cp*delta T

885.4 kJ/mol* 4.2 mol = 87.5 kJ*delta T.

Solve for delta T = 42.5 degrees K (or 42.5 degrees C.).

Glad I could help.

- Chemistry -
**Anne**, Wednesday, December 12, 2007 at 9:47pmOh, I see now. I really appreciate your help :) Have a wonderful day!

- Chemistry -

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