Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: doubleU on 16/10/2009 22:44:26

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 THREADTITLES ARE PHRASED AS QUESTIONS, WHICH IS OUR FORUM POLICY. THANKS. CHRIS]

You need the "Heat of Solution" for NaOH to predict the final temperature ...
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 (http://en.wikipedia.org/wiki/Specific_heat_capacity)is not the same as "heat of solution"].

i got the final heat to be 38.2

A similar problem ... http://www.physicsforums.com/showthread.php?t=161238

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 25°C as well, we have:
moles of solute = 5g/(40g/mol) = 0.125 mol NaOH
molar heat of solution: ΔH_{solution}/(moles of solute) = 44.2kJ/mol
heat of solution: ΔH_{solution} = 44.2kJ/mol • 0.125 mol = 5.525kJ
mass_{solution} = 5g + 100g = 105g = 0.105kg
ΔT_{solution} = ΔH_{solution}/(mass_{solution} • specific heat_{solution}) ≈
≈ ΔH_{solution}/(mass_{solution} • specific heat_{water}) =
= 5.525kJ/(0.105kg • 4.184kJ/kg°C) = 12.58°C.
So the final temperature would be: 25°C + 12.58°C = 37.58°C.

Good job doing doubleU's homework lightarrow [:P]

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 [;)].