posted by Anonymous .
The enthalpy change for the solution process when solid sodium hydroxide dissolves in water is -44.4 kJ/mol. When a 13.6 g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0 o C to __________ o C. Assume that the solution has the same specific heat as liquid water, 4.18 J/(g oC).
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).