[baggravation]

NaOH + HCl ------> NaCl + HOH

I need to figure out the quantity of heat in this reaction. The volume of the solution is 150 mL (NaCl + HOH). The inital temperature is 26 degrees celcius and the final temperature is 23 degrees celcius. That's all I know. How would I go about figuring this out?

[eunChae]

First i gotta say that i couldnt find out the result just with the given values (and also i dont believe it can be calculated ,thus we re gonna make some necessary assumptions:
*the solution volume is exactly 150 ml
*The NaCl(aq) is sufficiently dilute that its 'density' and 'specific heat' are about the same as those values for pure water: 1.00 g/mL and 4.18 J/(g*C), respectively.
*The system is completely isolated:No heat escapes from the calorimeter.
*The heat required to warm any part of the calorimeter other than the NaCl(aq) is negligible.

The heat produced by the reaction and retained in the calorimeter is
     
     q= mass * specific heat * (delta)T
 
     q = 150 mL * 1.00 g/mL * 4.18 J/(g*C) * (26-23)C = 1881 J

[eunChae]

I should also note that, as i know, the reaction of NaOH and HCl is exothermic, i.e. the temperature of the final solution should be risen up instead of being cooled...

[lancenti]

First way to go about this is to use eunChae's method, but since your data's messed up that's probably not going to work.

Second way is to consider the number of moles of water formed, and look up the value of something called the Standard Enthalpy Change of Neutralization for SASB reactions. If my memory serves me right, this is 56 kJ/mol.

This method is better in the sense that it is the 'ideal', no need to account for heat lost to the environment and all.