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We will begin lab in the amphitheatre of DBH
When two objects at different temperatures are brought into physical contact, thermal energy will spontaneously transfer from the warmer object to the colder object until both objects have achieved the same temperature. Assuming the two objects are thermally insulated from their surroundings, the heat lost by the warm object is identical to the heat gained by the cold object. This is a manifestation of the Law of Conservation of Energy.
The heat transfer, q, is a function of the mass of the object (m), the change in temperature undergone by the object ( ) and the object's specific heat ( ). This statement can be expressed mathematically as
q =
Temperature change is always defined as - , which means that q for the hotter object ( ) is negative and q for the colder object ( ) is positive. If energy is conserved, then
+ = 0
and
( ) + ( ) = 0
Now consider dropping an ice cube into water just warm enough to melt the ice cube but not warm enough to further heat the water from the cube. The observation is that the ice cube melts and the warm water cools to C. It is important to recognize that during the phase change, the temperature of the ice cube does not change. Therefore, it is not possible to use the preceding equation to determine the heat transferred. Rather, the energy transferred to the ice cube from the warm water affects the phase change. The energy equation is now adjusted to incorporate the enthalpy required to melt the ice cube, (where f stands for fusion):
( ) + = 0
It is also possible to have thermal energy when chemical reactions occur. The amount and direction of heat flow is dependant on the chemicals reacting. Using a calorimeter, it is possible to experimentally determine the heat of reaction.
In the technique known as constant-pressure calorimetry, enthalpies of phase changes or chemical reactions are determined indirectly by measuring temperature (at constant pressure) changes in a medium, most often water, surrounding the materials undergoing the change. That is, by measuring of the water one can use the preceding equation to calculate for the process of interest. Of course, this means one must know the mass of the water used and water's specific heat: = 4.18 J/(gK).
Today in Part I, you will add a strong base to a strong acid, measure the temperature change in the water as the two react, and use that information to calculate the heat of reaction per gram of NaOH. Then convert your experimental value into an enthalpy in kJ/mol (of NaOH).
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