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Author Topic: What temperature change occurs when solid sodium hydroxide dissolves in water?  (Read 20509 times)

Offline doubleU

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if you put 5.00g of solid NaOH in 100ml of water at the temp of 25 C, what is the final temp of the system?you need the specific heat of NaOH? specific heat= ?
the solution will heat up. Do we plug in 25 C for both ?
is the equation: msΔt NaOH = msΔT water??


[MOD EDIT - PLEASE ENSURE THAT YOUR THREAD-TITLES ARE PHRASED AS QUESTIONS, WHICH IS OUR FORUM POLICY. THANKS. CHRIS]
« Last Edit: 17/10/2009 10:46:01 by chris »


 

Offline RD

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You need the "Heat of Solution" for NaOH to predict the final temperature ...

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For a given solute, the heat of solution is the change in energy that occurs as one mole of the solute dissolves in water.
http://pulse.pharmacy.arizona.edu/resources/heatofsolution.pdf


["Specific heat" capacity is not the same as "heat of solution"]. 
« Last Edit: 16/10/2009 23:44:17 by RD »
 

Offline doubleU

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i got the final heat to be 38.2
 

Offline RD

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Offline lightarrow

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if you put 5.00g of solid NaOH in 100ml of water at the temp of 25 C, what is the final temp of the system?you need the specific heat of NaOH? specific heat= ?
the solution will heat up. Do we plug in 25 C for both ?
is the equation: msΔt NaOH = msΔT water??
From the .pdf file provided from RD:
http://pulse.pharmacy.arizona.edu/resources/heatofsolution.pdf

in the simplistic hypothesys that the specific heat of the solution is the same as that of water, and assuming that the initial temperature of the solid NaOH is 25C as well, we have:

moles of solute = 5g/(40g/mol) = 0.125 mol NaOH

molar heat of solution: ΔHsolution/(moles of solute) = -44.2kJ/mol

heat of solution: ΔHsolution = -44.2kJ/mol 0.125 mol = -5.525kJ

masssolution = 5g + 100g = 105g = 0.105kg

ΔTsolution = -ΔHsolution/(masssolution specific heatsolution) ≈

≈ -ΔHsolution/(masssolution specific heatwater) =

= 5.525kJ/(0.105kg 4.184kJ/kgC) = 12.58C.

So the final temperature would be: 25C + 12.58C = 37.58C.
« Last Edit: 17/10/2009 12:00:54 by lightarrow »
 

Offline Nizzle

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Good job doing doubleU's homework lightarrow :P
 

Offline lightarrow

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Good job doing doubleU's homework lightarrow :P
I answered just because I considered it as an interesting problem for all readers. Of course it's not my habit doing others' homeworks  ;).
 

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