Justin Germani Specific Heat Capacity January 24, 2012
Data:
Object 1
Object 2
Object 3
For each table above:
∆- represents uncertainty
-mass of Aluminum Calorimeter is 42.5g
Sample Calculations:
Calculate specific heat capacity (c):
Object 1
Q lost by block = Q gained by water + Q gained by Al
0.0623(c)(-73.9) = 0.114(4200)(03.3) + 0.0425(910)(03.3)
c = 371 J/kg/°C (3 significant figures)
Results:
Data:
Freezing Aluminum block
Calculate Initial Temperature of Aluminum block:
Q gained by block = Q lost by water + Q lost by Al calorimeter
0.0623(910)(16.2-T)=.114(4200)(03.5)+.0465(910)(03.5)
918.427-56.963(T) =1675.8 + 148.1025
T = -15.9°C
Percentage difference:
Sample Calculations:
Conclusion:
The specific heat capacity of object one which was Iron was calculated to be 370.9 J/Kg/°C.
The specific heat capacity of object two which was Aluminum was calculated to be 625.6 J/Kg/°C.
The specific heat capacity of object three which was Copper was calculated to be 318.9 J/Kg/°C.
The specific heat capacity of the objects can be calculated ...
This is a preview of the whole essay
Data:
Freezing Aluminum block
Calculate Initial Temperature of Aluminum block:
Q gained by block = Q lost by water + Q lost by Al calorimeter
0.0623(910)(16.2-T)=.114(4200)(03.5)+.0465(910)(03.5)
918.427-56.963(T) =1675.8 + 148.1025
T = -15.9°C
Percentage difference:
Sample Calculations:
Conclusion:
The specific heat capacity of object one which was Iron was calculated to be 370.9 J/Kg/°C.
The specific heat capacity of object two which was Aluminum was calculated to be 625.6 J/Kg/°C.
The specific heat capacity of object three which was Copper was calculated to be 318.9 J/Kg/°C.
The specific heat capacity of the objects can be calculated because the heat gained in the reaction is equal to the heat lost during the reaction. Due to the concept of thermal equilibrium we know that when objects of differing temperatures are in contact with each other, in this case the block, the water and Aluminum calorimeter they will eventually transfer heat to the point at which they are the same temperature. By measuring the temperature before the blocks were added to the water and then again when the temperature remains constant indicating equilibrium temperature the change in temperature was able to be calculated and using this along with the mass of the objects and water and calorimeter the specific heat capacity could be calculated using the Q=mc∆T formula.
For the “How cold is it in the freezer compartment?” question an aluminum block was placed in the freezer and then placed into the water in the calorimeter of which the temperature had been recorded. The temperature of the water with the Aluminum block was recorded until the constant temperature was reached. With this we can calculate the initial temperature of the aluminum block which gives us the temperature of the freezer.
Evaluation:
There were several limitations in the equipment used to carry out the procedure.
The scale used to measure the mass of the aluminum calorimeter and water only measured up to 200g this was a major limitation as it only allowed for a small amount of water to be added to the calorimeter which may have affected the heat transfer as the submerged block may not have been fully submerged therefore some heat may have been lost to the air.
In moving the heated block from the boiling water to the calorimeter the block could have lost heat but this would be minimal due to the relatively high Specific heat capacities of the blocks.
When the block was moved into the calorimeter it was dropped into it and this caused some water to splash out which would decrease the mass of water in the calorimeter which could affect the overall temperature change.
The value for the specific heat capacity (shc) of object one which was Iron was compared to the theoretical value for the shc of Iron and gave a 17.6% difference this value is relatively high and could be due to uncertainties in the measurements and errors produced from the procedure.
The % difference of object two was very large at 31.3% this could be due to random error but more likely due to the block not being fully submerged as I mentioned earlier.
The value of the percentage difference for Object3 which was Copper was relatively high at 17.2%,
In order to improve upon the experiment a scale that measures up to a higher mass such as 500g. Another method to improve the experiment would be to add more water to the calorimeter in order to insure the block was fully submerged. Another improvement that could be made would be to not drop the blocks into the calorimeter and therefore minimize the risk of losing water after the mass was taken.