4) Fill the calorimeter approximately half full with distilled water at room temperature and record the mass and temperature
5) Add the metal carefully into the beaker and wait at least 10 minutes before doing anything.
6) While the metal is still in the boiling water, measure the temperature of the water with a thermometer and record. It will be assumed that the temperature of the metal is the same as the boiling water.
7) After the metal has been heating 10 minutes, remove the test tube from the beaker. Immediately pour the metal into the calorimeter cup so that the metal is covered by the water. Cover the calorimeter with its cover
9) Insert the thermometer through the small hole in the cover. Stir very slowly with a stirring rod, and record the highest temperature reached by the water.
Follow-Up :
X = unknown specific heat of metal
(152.19 ± 0.05)(4.186)(20.2 ± 0.05 - 17.35 ± 0.05 ) = (57.39 ± 0.05)(X)(20.2 ± 0.05 - 100.8 ± 0.05)
1815.64 ± 0.05 = 4625.63 ± 0.05X
X = 0.393 ± 0.05
Specific heat actual of copper is 0.386
[(0.393 - 0.386)/0.386] X 100 = 1.81% error
Conclusion: Overall I think that 1.81% is not a big margin of error because of several reason which I couldn't really prevent and there were also some limitations while doing the lab. One of the limitations is that when we put the metal in the water and take it out after 10 minutes we just assume that it has reached the point of equilibrium, and this then creates a limitation. Another limitation that I think was present was that when we are transferring the metal into the calorimeter the temperature of the metal might have changed a little even thought we tried to move it as quickly as possible and that could be a limitation. There are also some errors that we are not able to prevent from happening. For example, the electrical thermometer could have a different temperature and it will also change the temperature of the water because it will take some time for the temperature to stay at constant. Even though we try as hard to not have any errors human error will always be present.
