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Industrial chemistry - HSN forum

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Industrial chemistry


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#1 floss

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Posted 06 March 2007 - 10:41 AM

This is prob a really easy question but I've got a complete blank on it!

After mixing 50cm cubed NaOH(aq)degrees celsius with 50cm cubed HCl(aq) at 17 degrees celsius the temperature rose to 25 degrees c.
What was the temperature change caused by the reaction?
A - 2 degrees c
B - 5 degrees c
C - 8 degrees c
D - 9 degrees c


I know the answer is 5 but I don't understand why...could someone please explain it to me? dry.gif



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#2 Nathan

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Posted 06 March 2007 - 08:30 PM

Are you sure it's 5? I would have said that it was 8, tbh. Because it starts at 17, rises 8 degrees and finishes at 25. No? unsure.gif

#3 floss

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Posted 06 March 2007 - 08:49 PM

Ah!! perhaps the marking scheme was wrong biggrin.gif
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#4 dfx

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Posted 07 March 2007 - 08:42 AM

QUOTE(Fiona Jane @ Mar 6 2007, 10:41 AM) View Post
[color=#9932CC][i]After mixing 50cm cubed NaOH(aq)degrees celsius


At how many degrees celsius?


#5 Nathan

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Posted 07 March 2007 - 08:06 PM

Oh, yeah, I never even noticed that unsure.gif. You would need to take the average temperature of both liquids. I'm assuming if the marking scheme states the answer as 5 then the starting temperature for the NaOH would be given as 23 degrees C in the question.

#6 dfx

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Posted 07 March 2007 - 11:35 PM

QUOTE(Nathan @ Mar 7 2007, 08:06 PM) View Post
Oh, yeah, I never even noticed that unsure.gif.


The devil is in the detail. wink.gif aka typo.

#7 floss

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Posted 11 March 2007 - 12:33 PM

QUOTE(dfx @ Mar 7 2007, 11:35 PM) View Post
QUOTE(Nathan @ Mar 7 2007, 08:06 PM) View Post
Oh, yeah, I never even noticed that unsure.gif.


The devil is in the detail. wink.gif aka typo.


I'm not very good at typing, but don't mock the afflicted! the starting temperature was 23 degrees... blink.gif



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#8 Nathan

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Posted 11 March 2007 - 02:40 PM

QUOTE(Fiona Jane @ Mar 11 2007, 12:33 PM) View Post
I'm not very good at typing, but don't mock the afflicted! the starting temperature was 23 degrees... blink.gif


Do you understand now why the answer's 5? smile.gif

#9 floss

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Posted 11 March 2007 - 08:23 PM

QUOTE(Nathan @ Mar 11 2007, 02:40 PM) View Post
Do you understand now why the answer's 5? smile.gif


the short answer is no... unfortunately! sad.gif
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#10 Nathan

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Posted 11 March 2007 - 08:40 PM

QUOTE(Fiona Jane @ Mar 11 2007, 08:23 PM) View Post
the short answer is no... unfortunately! sad.gif


No worries. Basically, when you're working out delta.gifT when you've got liquids, you need to take the temperature of both liquids before mixing them. In general, you're going to get two different temperatures. So that you can (fairly) accurately calculate delta.gifT, you need to take the average starting temperature. Then your final temperature is easy enough, it's just the temperature of the mixture at its hottest/coldest, depending on whether the reaction is exo- or endo- thermic.

Taking your example,

QUOTE
After mixing 50cm cubed NaOH(aq)at 23 degrees celsius with 50cm cubed HCl(aq) at 17 degrees celsius the temperature rose to 25 degrees c.
What was the temperature change caused by the reaction?


So;
Starting temperature of NaOH = 23 deg.C
Starting temperature of HCl = 17 deg.C
Average temperature = (23+17)/2 = 40/2 = 20 deg.C

Final temperature of NaOH + HCl mixture = 25 deg.C

delta.gifT = 25-20 = +5 deg.C

Does that make more sense now?

#11 floss

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Posted 11 March 2007 - 08:52 PM

QUOTE(Nathan @ Mar 11 2007, 08:40 PM) View Post
QUOTE(Fiona Jane @ Mar 11 2007, 08:23 PM) View Post
the short answer is no... unfortunately! sad.gif


No worries. Basically, when you're working out delta.gifT when you've got liquids, you need to take the temperature of both liquids before mixing them. In general, you're going to get two different temperatures. So that you can (fairly) accurately calculate delta.gifT, you need to take the average starting temperature. Then your final temperature is easy enough, it's just the temperature of the mixture at its hottest/coldest, depending on whether the reaction is exo- or endo- thermic.

Taking your example,

QUOTE
After mixing 50cm cubed NaOH(aq)at 23 degrees celsius with 50cm cubed HCl(aq) at 17 degrees celsius the temperature rose to 25 degrees c.
What was the temperature change caused by the reaction?
So;
Starting temperature of NaOH = 23 deg.C
Starting temperature of HCl = 17 deg.C
Average temperature = (23+17)/2 = 40/2 = 20 deg.C

Final temperature of NaOH + HCl mixture = 25 deg.C

delta.gifT = 25-20 = +5 deg.C

Does that make more sense now?



Thanks, I missed that bit in class you see and just couldn't get my head around it, but yes, I get it now! smile.gif
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#12 Nathan

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Posted 11 March 2007 - 08:57 PM

Yay! biggrin.gif

#13 floss

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Posted 11 March 2007 - 09:01 PM

QUOTE(Nathan @ Mar 11 2007, 08:57 PM) View Post
Yay! biggrin.gif


Very few people have successfully managed to teach me chemistry things - you deserve a medal!! hehe x

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