[purple urkeL]

What is the final temperature in a squeezed cold pack that contains 40.0g NH(4)NO(3) dissolved in 125 mL of water? Assume a specific heat of 4.18 J/g x C for the solution, an initial temperature of 27.5 degrees Celsius, and no heat transfer between the cold pack and the environment. the enthalpy change for the reaction is 25.7 kJ.

So far, I've tried setting it up by using 25700 J = 4.18 x 40.0 (T(final) - 27.5) but when I solve for T(final), I am not getting the right answer. I figure I have to do something with the moles but I am not understanding. I know that the 40 grams dissolved in 125 ml is also 4M solution. But that's all I know. Any help would be greatly appreciated! Thanks!

[Yggdrasil]

The enthalpy change for the reaction is 25.7 kJ per mole of ammonium nitrate disolved.  How many moles of ammonium nitrate are you dissolving?

Also, for your m, you need to consider the mass of the water because that's what's cooling down, not the ammonium nitrate itself.

Finally, since the reaction absorbs heat, your q will be negative (so that T_final < T_initial)