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Offline hillkr

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Calirometry
« on: October 17, 2014, 04:33:50 PM »
Problem:
Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHcomb for dimethylphthalate (C10H10O4) is 4685 kJ/mol. Assume 0.930 g of dimethylphthalate is combusted in a calorimeter whose heat capacity (Ccalorimeter) is 6.894 kJ/°C at 21.447°C. What is the final temperature of the calorimeter?
Scratch work:
C10H10O4 = 194.075 g/mol 0.930 g = 0.00479 g/mol
ΔHcomb = 4685 kJ/mol
Cp = 6.894 kJ/°C
initial heat = 21.447°C
heat released = 4685 × 0.00479 = 22.45 kJ
What I did and got it wrong:n·Cp·ΔT
0.00479·6.894·(22.45-21.447) = 0.033121

How do I work this problem?
« Last Edit: October 17, 2014, 04:49:27 PM by Borek »

Offline Borek

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Re: Calirometry
« Reply #1 on: October 17, 2014, 04:53:09 PM »
C10H10O4 = 194.075 g/mol 0.930 g = 0.00479 g/mol

Nothing makes sense here, but surprisingly the number is the one needed later.

Quote
heat released = 4685 × 0.00479 = 22.45 kJ

OK

Quote
n·Cp·ΔT

Nope, that's not the correct equation. What is the definition of the heat capacity? How does it differ from the specific heat capacity? When do we use one, when the other?
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Offline hillkr

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Re: Calirometry
« Reply #2 on: October 17, 2014, 05:29:58 PM »
For C10H10O4 194.075 is the molecular weight and the problem said 0.930 g of C10H10O4 so i divided 0.930 by 194.075 to get the mols for the problem which was equal to 0.00479g/mol.

Heat capacity (Cp) is the quantity of energy needed to raise the temperature of an object by 1°C.
Specific heat (cs) is the energy required to raise the temperature of one gram of a substance by 1°C
So heat capacity is raising the temperature of the whole thing and specific heat is 1 gram.
I assumed I would use the heat capacity equation because the problem gave the heat capacity. My teacher did not go over one of these problems in class so i really do not know how to do it. 

Offline mjc123

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Re: Calirometry
« Reply #3 on: October 17, 2014, 06:11:54 PM »
Quote
For C10H10O4 194.075 is the molecular weight and the problem said 0.930 g of C10H10O4 so i divided 0.930 by 194.075 to get the mols for the problem which was equal to 0.00479g/mol.
The unit for moles is moles, not g/mol, which is the unit of molecular weight. g divided by g/mol gives mol. But the number is correct.
Quote
Heat capacity (Cp) is the quantity of energy needed to raise the temperature of an object by 1°C
Correct. An object. Not a mole of objects, or a mole of something else, like dimethyl phthalate. So why multiply by 0.00479?
Quote
0.00479·6.894·(22.45-21.447) = 0.033121
Why do you subtract a quantity in °C from a quantity in kJ? The final temperature is an unknown, it's what you're trying to find. You know the heat, you know the heat capacity. So...

Offline hillkr

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Re: Calirometry
« Reply #4 on: October 17, 2014, 06:51:04 PM »
So it is heat/heat capacity to get the temperature change then you add that to the temperature given.
22.45 ÷ 6.894 = 3.256
21.447 + 3.256 = 24.70°C

Thank you so much for helping me.

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