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Topic: Thermochem  (Read 2409 times)

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Offline pocantheses

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Thermochem
« on: October 13, 2014, 01:54:39 PM »
So I know how to do this problem but every time I attempt it, I get it wrong:

""
5.00 g copper has initial temp of 95.0°C.
25.0 g of a sample of water has initial temp of 22.5°C.
Specific heat copper: 0.385 J/g-°C
Specific heat water: 4.186 J/g-°C

What is final temperature of this mixture?
""
i. So I know that the ΔH of each substance can be found by the equation

ΔH = (m)(Cs)(Tfinal-Tinitial)

ii. And I think that the relationship between the ΔH of each substance can be expressed by:

ΔHwater=-ΔHcopper

iii. So I let x be the final temperature. Plugging in the values given, I set up the equation and simplified as such:

(25.0 g)(4.187 J/g-°C)(x-22.5°C) = - (5.00 g)(.385 J/g-°C)(x-95°C)

iv. And then I some how come up with

Tfinal=23.8 °C


What am I doing/writing wrong?


Offline Borek

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Re: Thermochem
« Reply #1 on: October 13, 2014, 04:16:20 PM »
And why do you think it is wrong?
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Offline pocantheses

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Re: Thermochem
« Reply #2 on: October 13, 2014, 07:11:41 PM »
And why do you think it is wrong?

I don't know. It seems off for some reason. Something that is almost boiling point of water being put into water and hardly changing the temperature doesn't seem right. Is the calculation stuff/relationships used right?

Offline Corribus

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Re: Thermochem
« Reply #3 on: October 13, 2014, 08:27:37 PM »
I didn't check the math, but assuming for a moment your answer is correct - this teaches you something important about the difference between temperature and heat. What determines the final temperature when two objects at different temperatures are brought together is not their respective temperatures, but how much heat is available to transfer from one object to the other.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Borek

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Re: Thermochem
« Reply #4 on: October 14, 2014, 03:25:58 AM »
As far as I can tell you calculated correctly, and you got the right answer.

Note that the amount of copper piece is five times smaller than the amount of water, and that specific heat of copper is more than ten times lower than that of water (actually water has very high specific heat). These things add up, making the final temperature look 'assymetrical'.
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