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s.h_math s.h_math
wrote...
Posts: 293
Rep: 3 0
2 months ago Edited: 2 months ago, bio_man


Ive attempted this question 4 times now & this is my most confident answer do you possibly think you can let me know where I am going wrong? Thanks!

A bomb calorimeter has a heat capacity of 4.76 kJ/0C. When 2.047 g of ethanol, C2H5 OH(I), is combusted in the bomb
calorimeter, the temperature of the calorimeter and its contents increased from 22.72 oc to 35.48 oc. Calculate the change in
internal energy, AU, for the combustion of 2.047 g of ethanol in the bomb calorimeter.
-1367.32
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wrote...
Educator
2 months ago
Will figure out when I get home!
s.h_math Author
wrote...
2 months ago
I have been waiting for your response for quite some time.

please let me know, Thanks!
wrote...
Educator
2 months ago
I'm not sure how to do it, but I asked a colleague. Will update by noon!
s.h_math Author
wrote...
2 months ago
Alrighty!

Thanks!!
wrote...
Educator
2 months ago
Try this method:


s.h_math Author
wrote...
2 months ago
Unfortunately, still incorrect.

The relationship between the change in internal energy  (Δ𝑈),
  heat  (𝑞),
  and work  (𝑤)
  is given by the following equation.
Δ𝑈=𝑞+𝑤
 
Work is related to pressure  (𝑃)
  and the change in volume  (Δ𝑉)
  through the following equation.
𝑤=−𝑃Δ𝑉
 
A bomb calorimeter has a constant volume, so  Δ𝑉=0
  for any reaction carried out in a bomb calorimeter. Therefore,  𝑤=−𝑃Δ𝑉=0
  for a reaction carried out in a bomb calorimeter, and the equation for the change in internal energy simplifies to  Δ𝑈=𝑞.
 
For a reaction carried out in a bomb calorimeter, the heat absorbed by the solution and calorimeter  (𝑞soln)
  is equal in magnitude but opposite in sign to the heat released by the reaction  (𝑞v).
 
𝑞soln=−𝑞v
 
The heat absorbed by the solution and calorimeter is also related to the heat capacity of the calorimeter  (𝐶cal)
  and the change in temperature  (Δ𝑇)
  through the following equation.
𝑞soln=𝐶calΔ𝑇
 
Combine these equations to get an expression for  Δ𝑈
  in terms of  𝐶cal
  and  Δ𝑇.
 

feedback^^^
wrote...
Educator
2 months ago Edited: 2 months ago, bio_man
I think the issue with your answer are the significant figures. Does the question tell you what to round to?

Here's my attempt. I get the same as you but we're off by 0.06.



I used our tutorial found here:



If it is a rounding/significant figure situation, the final answer to four significant digits is -1367. If it is to three significant figures, we write it in scientific notation as:

\[-1.37 \times 10^3\]

or simply:

\[-1370\]

Both show 3 significant figures.
s.h_math Author
wrote...
2 months ago
Unfortunately, no. I just gave it a try! its marked it as incorrect again!
wrote...
Educator
2 months ago
Hi

I haven't looked at this since, but did you end up getting the right answer?
s.h_math Author
wrote...
2 months ago
haha no!

It was such a weird question for some reason.

I understand you have been busy as well! its all good bioman I know you honestly tried your best to figure out the correct answer.
wrote...
Educator
2 months ago
No problem.

If you do get the answer, please update the thread
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