Calorimetry final temp homework help

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Discussion Overview

The discussion revolves around a calorimetry homework problem involving the mixing of nitric acid and barium hydroxide solutions. Participants are trying to determine the final temperature of the resulting solution after a reaction, using given data and previous example results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a calorimetry problem involving the mixing of 0.600 M HNO3 and 0.300 M Ba(OH)2, with initial temperatures provided.
  • Another participant questions the mass calculation, asking for clarification on how the mass of the solutions was derived from their volumes.
  • Some participants clarify that the mass should be calculated from the molarity and volume of the solutions, leading to specific mass values for HNO3 and Ba(OH)2.
  • There is a discussion about the assumption that the specific heats of the solutions are the same as that of water.
  • One participant calculates the final temperature using the derived mass and specific heat, but finds a discrepancy with the answer provided in the textbook.
  • Another participant suggests the possibility of a typo in the textbook answer, while others express uncertainty about how to apply the heat of neutralization value given in a previous question.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final temperature calculation, as there are conflicting results and uncertainty about the application of the heat of neutralization. Some participants agree on the calculations made, while others express confusion and suggest potential errors in the textbook.

Contextual Notes

There are unresolved issues regarding the application of the heat of neutralization and discrepancies in calculated versus expected final temperatures. The discussion also highlights the importance of correctly converting volumes to masses and the implications of specific heat assumptions.

Mitchtwitchita
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can anybody here help me out with this problem?

A quantity of 4.00 x 10^2 mL of 0.600 M HNO3 is mixed with 4.00 x 10^2 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46 degrees C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation).

Results for previous example: qrxn = -2.81 KJ.

qrxn + qsoln = 0
qrxn = -qsoln
therefore, qsoln = 2.81

qsoln = ms(deltaT)
2.81 = (400 g + 400 g)(4.184 KJ/g * degrees C)(x - 18.46 degrees C)
x = [2.81/(800 g)(4.184 KJ/g * degrees C)] + 18.46 degrees C
=18.46 degrees C?

Can anybody please show me where I'm going wrong?
 
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How did you get 400g?

Please be more clear with your units. Which information refers to Nitric acid or Barium Hydroxide?
 
I thought the 400g was in relation to the 400 mL of nitric acid and that the 400 g was in relation to the barium hydroxide.
 
No that's the volume, you want the mass, in grams. So solve for grams of each and we'll go from there.
 
I think HNO3 = 0.400 L x 0.600 mol/1L
=0.24 mol x 63.02 g/1 mol
=15.12 g HNO3

and Ba(OH)3 = 0.400 L x 0.300 mol/1 L
=0.12 mol x 188.3 g/1 mol
=22.60 g Ba(OH)3

15.12g + 22.60g = 37.72g solution?
 
-q_{rxn}=q_{soln}

2.81kJ=m_{a}c_{a}\Delta{T}_{a}+m_{b}c_{b}\Delta{T}_{b}

\Delta{T}_a=\Delta{T}_b=\Delta{T}

2.81kJ=\Delta{T}(m_{a}c_{a}+m_{b}c_{b})

a = Nitric acid
b = Barium hydroxide

You can't just add their weights because they have different specific heats. Are you able to find their specific heats?
 
Last edited:
Yeah, the assumption is that the specific heats are the same as water.
 
Mitchtwitchita said:
Yeah, the assumption is that the specific heats are the same as water.
Oh ok. I was wondering why you were just straight up adding them from your original post.

Anyways, yeah the equation simplifies nicely. Just do what you did originally using the actual mass.

2.81\times 10^3 J=m_{soln}c\Delta T
 
Last edited:
2810 J = (37.7 g)(4.184 J/g * degrees C)(x - 18.46 degrees C)
x = [2810 J/(37.7 g)(4.184 J/g * degrees C)] + 18.46 degrees C
=36.27 degrees C

However, this still doesn't match the answer at the back of the book (22.49 degrees C). Is there something wrong with my math?
 
  • #10
Mitchtwitchita said:
2810 J = (37.7 g)(4.184 J/g * degrees C)(x - 18.46 degrees C)
x = [2810 J/(37.7 g)(4.184 J/g * degrees C)] + 18.46 degrees C
=36.27 degrees C

However, this still doesn't match the answer at the back of the book (22.49 degrees C). Is there something wrong with my math?
I don't know what else to tell you, I got the same thing. Sorry, you'll have to wait for someone else to clean up my mess :p I'll try one more time to see if I can figure it out.
 
  • #11
Oh, I think I got it. Let me finish working it out b4 I type it up.
 
  • #12
Thanks a lot rocophysics! you've been a great help. I can't see any error that I could be making now. Maybe the book has a typo which tends to happen.
 
  • #13
Oh my god, the previous question also says that the heat of neutralization is -56.2 KJ/mol. But that still doesn't help me out because I don't know how to apply it or if it even pertains to this question.
 
  • #14
Mitchtwitchita said:
Oh my god, the previous question also says that the heat of neutralization is -56.2 KJ/mol. But that still doesn't help me out because I don't know how to apply it or if it even pertains to this question.
Ha! No worries. Do you mind typing up the whole problem? I'm getting mad about this! Want to solveee ittt. :-]
 
  • #15
A quantity of 4.00 x 10^2 mL of 0.600 M HNO3 is mixed with 4.00 x 10^2 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46 degrees C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation).

Results for previous example: qrxn = -2.81 KJ.

qrxn + qsoln = 0
qrxn = -qsoln
therefore, qsoln = 2.81

HNO3 = 0.400 L x 0.600 mol/1L
=0.24 mol x 63.02 g/1 mol
=15.12 g HNO3

and Ba(OH)3 = 0.400 L x 0.300 mol/1 L
=0.12 mol x 188.3 g/1 mol
=22.60 g Ba(OH)3

15.12g + 22.60g = 37.72g solution?

2810 J = (37.7 g)(4.184 J/g * degrees C)(x - 18.46 degrees C)
x = [2810 J/(37.7 g)(4.184 J/g * degrees C)] + 18.46 degrees C
=36.27 degrees C

The final answer is 22.49 degrees C
and the heat of neutralization is -56.2 KJ/mol
 

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