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).

To find the final temperature, we need to use the equation:

q = m * c * ΔT

Where:
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature

Given:
The mass of water, m = 250.0 g
The specific heat of water, c = 4.18 J/(g oC)
The initial temperature, T1 = 23.0 oC
The enthalpy change, ΔH = -44.4 kJ/mol

First, let's calculate the number of moles of NaOH that dissolve:

The molar mass of NaOH = 22.99 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol ≈ 40.0 g/mol

Number of moles of NaOH = mass / molar mass
= 13.6 g / 40.0 g/mol
≈ 0.34 mol

Now, let's calculate the heat released by the dissolution of NaOH:

ΔH = q / n
-44.4 kJ/mol = q / 0.34 mol

q = -44.4 kJ/mol * 0.34 mol
q ≈ -15.1 kJ

Since the heat released by the reaction is equal to the heat gained by the water, we have:

q = m * c * ΔT

-15.1 kJ = (250.0 g + 13.6 g) * 4.18 J/(g oC) * (Tf - 23.0 oC)

Simplifying the equation, we get:

-15.1 kJ = 263.6 g * 4.18 J/(g oC) * (Tf - 23.0 oC)
-15.1 kJ = 1100.45 (Tf - 23.0)
Tf - 23.0 = -15.1 kJ / 1100.45
Tf - 23.0 ≈ -13.7

Solving for Tf:

Tf ≈ -13.7 + 23.0
Tf ≈ 9.3 oC

Therefore, the final temperature is approximately 9.3 oC.