Re: heat of solution
- From: Bob <bbx107@xxxxxxxxxxxxxxx>
- Date: Tue, 09 Aug 2005 20:33:18 -0700
On Tue, 09 Aug 2005 17:15:51 GMT, "John Doe III"
<john_doeIII@xxxxxxxxx> wrote:
>Thanks everyone, I know the answer. I know their solution. I just wanted to
>know if anyone here can come up with the same answer. A lot of times it's
>easy to describe what needs to be calculated but hard to find the correct
>answer. Can anyone tell, at least, if answer is A, B, C or D? A couple of
>days ago I asked about the polarity of CH3Cl, CH2Cl2, CHCl3, CCl4. Many
>responders could describe polarity and could tell what needed to be done to
>find the correct answer but surprisingly came up with a wrong order. For
>this particular question, I am curious to know if any of you can come up
>with answer. If you don't want to tell me your solution: fine, I'll post
>their solution but please at least tell if you think answer is A, B, C or D.
>
>20 grams of NaCl is poured into a coffee cup calorimeter containing 250 ml
>of water. If the temperature inside the calorimeter drops 1 C by the time
>NaCl is totally dissolved, what is the heat of solution for NaCl and water?
>(specific heat of water is 4.18 J/gC)
Eyeballing it, it is D. (+3)
Hey, I'll even explain what I did... The specific heat is 4.18
kJ/kg-deg = 4.18 kJ/L-deg. Multiply the given amts by 4 --> 80 g NaCl
in a liter. 80 g is about 1.3 mol. 4.18/1.3 is about 3. Did I do
something foolish?
But, with respect, I do not think much of your approach as stated
above -- and judging by the other replies, I am not alone.
You are the one doing these problems. Is there a sign issue here? If
so, say so. Why should we all have to get out calculators and work
thru all the details, just cuz you don?t want to post what the real Q
is.
We have had a lot of interesting discussions from your Q. I think it
is clear we do not think too much of the question sets (or at least of
a few of some individual questions; I suspect you post only those few
where this is some issue.) It's fine to ask about those, and question
their answers and/or your work. But why make us all do it all?
bob
.
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- From: John Doe III
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