More review since you asked  : )

Key concept: Calorimeters are closed systems. Energy is conserved.

                    Soo.... if the calorimeter cools, that is the amount of energy that the reaction or phase change "ate"

                    If the calorimeter heats up, that energy came from the reaction or phase change.

Formulas: deltaH = m Cp delta T heating and cooling w/out phase change

               deltaH = m Hf  energy required to melt the ice. What we are solving for.

Point: Calorimeter cools and ice goes through two step process. Melting and warming to T2.

Data: Sally Super student conducts the following experiment: She adds 20 g ice to 100 g water in calorimeter and obtains        final temperature of 6.0 C.

        mass water in calorimeter: mwater= 100.0 g

        initial temp of calorimeter:  T1= 22.0 C

        final temp of calorimeter: T2 = 6.00 C

        delta T water = -16.0 C

        mass of ice: mice = 20.0 g.

        delta T melted ice = +6.00 C

1. Heat loss of calorimeter: m Cp delta T

    100 g x 4.18 J/g C x -16.0 C  = -6688 joules

2. Heat gained in phase change (this is a two-step problem: melt ice, warm to T2):  m Hf  ⁺ m Cp delta T

   +6688 joules =  (20.0 g x Hf) + (20.0 g x 4.18 J/g C x  6.00 C)

    6186.4 J = 20.0 Hf

    Hf = 309.32 J

How does this compare to the actual value of Hf?  (309.2 - 333.6/333.6) x 100 = - 7.3% within our normal limits of accuracy.

Now for enthalpies of reaction. Again, remember that the heat (enthalpy) lost by one component is gained by the other. We can measure one, and do minor calculations to figure out the other.

Suppose you dissolve 20.0 grams of NaOH in a calorimeter with 100.0 grams of water. Given the specific heat capacity of water is 4.18 J/g C, what will the end temperature of the calorimeter be if its initial temperature is 17 C? The deltaH of solution of NaOH is -44.4 KJ/mole.

1. How many moles NaOH dissolved?  

 0.500 mole NaOH.

2. Heat released by NaOH:

    0.500 mole x 44.4 kJ/mole= -22.2 kJ is released into water

3. Heat (enthalpy) absorbed by water

    m Cp delta T

22200 J = 100.0 g x 4.18 J/g C x delta T

delta T = +53.1 C

Tf = Tinitial + deltaT

Tf = 70.1 C