[Whapper2006]

So the specific heat of NaOH is .95 cal/goC. THe change in temperature is 8.2oC. 2.33 grams of NaOH is dissolved on 50 mL of water.
What I did was q=mxspecific heatxchange in temp so
(52.33)(.95)(8.2) = -407.651 calories (negative because the system heats up therefore it is exothermic)

My answer keeps getting marked as wrong though. My change in temp and specific heat are correct. but i have a feeling my mass is wrong?  since its a solution dont I have to use the mass of the water + the NaOH?  and isnt the mass of the water 50grams since 50mL=50grams of water cause density is 1?

[Borek]

Please elaborate - are you solving question or are you trying to calcu;ate results of your experiment?

Where did you get specific heat from?

[Whapper2006]

sorry, here are the details.  It was an lab experiment my results were the change in heat.  The specific heat was given to me in the prelab.  And i put 2.33 grams of NaOH into 50 mL of water to make the solution.  I am trying to find the heat produced in calories of he solution. 

[Borek]

What you did looks OK to me. However, are you asked to calculate amount of heat produced? If so, it should be a positive number.