Lab - Thermochemistry
Uploaded by erobinson on Apr 30, 2004
Thermochemistry
Introduction:
This lab will teach us some of the basics of thermochemistry, such as temperature, heat, and heat capacity through creating our own simple calorimeter.
Data:
[img:4c4688b9b8]http://www.collegepimp.com/echeat/lab6.gif[/img:4c4688b9b8]
Discussion:
The purpose of this lab was to learn how to determine the specific heat of metal, a solution, neutralization and the heat capacity of a calorimeter. To determine the heat of the calorimeter we added hot water to cold water contained inside our calorimeter and observed the change in temperature inside the calorimeter. Knowing that the heat of gained by cold water and the calorimeter should be exactly equal to the heat lost by the hot water, we are able to solve for the heat capacity of the calorimeter itself. We found the heat capacity of our calorimeter to be 25.97kJ.
Our next task was to determine the heat capacity of an unknown metal. Using the same idea from the previous experiment, we knew that the heat lost by the metal would be exactly equal to the heat gained by the water and calorimeter. Since we now know the heat capacity of the calorimeter we can solve for the heat capacity (Cmetal) of our metal. Both my lab partner and I found the specific heat capacity of the unknown metal to be 0.129 J/Kg. This corresponds most “reasonably” with lead, even though according the chart given, it would be gold. This result does not seem reasonable however because I would not expect us to be handling lead with our bare hands. We are unable to determine the cause for this potential error.
To determine the heat of the solution we multiplied the heat capacity of the calorimeter by the change in temperature and added the mass of the solution times the specific heat capacity of the solution and the change in temperature, then set the entire quantity negative, resulting in 633 joules. From this we could determine the molar heat of the solution by dividing by moles of NH4NO3. ΔHsolution was found to be 25.3kK/mol.
Lastly, we were tasked with determining the heat of neutralization. This was found by using the same equation as in part 3 of our experiment but changing ΔT and the mass of our solution. We found the heat of the neutralization for NaOH & HCl to be -4424J and NaOH & CH3COOH to be -4443J. It is interesting that both values were similar; I believe this means that both acetic acid and hydrochloric acid have...