[doubleU (OP)]

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??


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[RD]

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 is not the same as "heat of solution"]. 

[doubleU (OP)]

i got the final heat to be 38.2

[RD]

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

[lightarrow]

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.

[Nizzle]

Good job doing doubleU's homework lightarrow []

[lightarrow]

Good job doing doubleU's homework lightarrow []

I answered just because I considered it as an interesting problem for all readers. Of course it's not my habit doing others' homeworks  [].