Chemistry Help!?

You will need more information, i.e. you need the heat of solution of NaOH. Assuming that this value is -44.51 (kJ/mol) at 25(C) (from Handbook of Chemistry and Physics) you can calculate the temperature rise as follows:

The molecular weight of NaOH = 23 (g/mol) + 16 (g/mol) + 1(g/mol) = 40 (g/mol)

8.5 (g) is 8.5 (g) / 40 (g/mol) = 0.21 (mol) of NaOH

This will generate 0.21 (mol) * 44.51 (kJ/mol) = 9.5 (kJ)

The temperature rise can be calculated using

Q = m x c x delta T, with Q is the heat, m = mass of water, c = specific heat of water at constant pressure (4.18 (J/g/K)) and delta T is the temperature change.

Rewrite to: delta T = Q / m / c

There is 100 (mL) of water. The density of water is 1.0 (g/mL). (For simplicity we ignore the NaOH mass and the density variation that occurs when dissolving NaOH, just like we ignore the fact that the specific heat of water with NaOH dissolved is not necessarily 4.18 (J/g/K))

Then there is:

m = 100 (mL) * 1.0 (g/mL) = 100 (g) water

Q = 9.5 (kJ) = 9.5 x 10^3 (J)

c = 4.18 (J/g/K)

delta T = 9.5 x 10^3 (J) / 4.18 (J/g/K) / 100 (g) = 23 (K)

There will be a 23 (C) temperature rise.

youre notgiving enough info.. gotta tell what temperature the water is starting at..

unless you just want to know the specific heat for the reaction.. in which case you would need a starting heat (generally in Joules or Kelvin)... so get back to me with that info... then I can help ya out!