Posted by Anonymous on Monday, November 15, 2010 at 2:25pm.
Enthalpy change is -44,400 J/40 grams (40 grams is 1 mole). Therefore, 13.6 g will change how much?
-44,400 J x 13.6/40 = about -15,000 J but you need to do it more accurately.
q = mass water x specific heat water x (Tfinal-Tinitial)
Substitute +15,000 J (actually the number you get when you do the math) since you are adding the heat to the water, plug in mass water and sp.h.water and solve for Tfinal. I get something like 37 C (again that's approximate).
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