**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.