DIAGRAM:
PROCEDURE:
- First I measured and noted down the mass of the unknown metal block, the empty calorimeter and the stirrer provided by using an electronic balance.
- Then I filled the calorimeter with a certain volume of water, and measured the new mass. I then calculated the mass of water by subtracting the two values (mass of water + calorimeter) – (mass of empty calorimeter).
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Then I measured and noted down the temperature of the cold water in the calorimeter as T1.
- Then I half filled the 250 ml beaker with water and heated it. Then I inserted the metal block with the thread into the beaker being heated and waited for the water to come to a certain temperature which in this case will also be the temperature of the unknown metal block before being transferred to the calorimeter.
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Then I measured and noted down the temperature of the water being heated as T2 which was also the temperature of the metal block and then I removed the metal block and immediately immersed it into the water in the calorimeter and straight away covered the calorimeter in order to avoid any heat loss to the atmosphere.
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Then I left the metal block in the calorimeter and stirred while constantly looking at the thermometer until the water inside the calorimeter reached a constant temperature which was the final temperature of both the metal block and the water. I measured and noted down this final temperature as T3.
RESULTS
Data Collection:
Mass of the unknown metal block = 75.5 g = 0. 0755 kg
Mass of the empty calorimeter and stirrer = 64.43 g = 0. 06443 kg
Mass of calorimeter and stirrer and water = 141.93 g = 0. 14193 kg
Mass of water = 0.14193 – 0.06443 = 0. 0775 kg
Temperature of the cold water – T1 = (30 ± 0. 05) oC
Temperature of the hot metal block – T2 = (95 ± 0. 05) oC
Final Temperature – T3 = (35 ± 0. 05) oC
Specific Heat Capacity of water = 4200 JKg-1K-1
Specific Heat Capacity of copper (Copper Calorimeter) = 385 JKg-1K-1
Data Processing and Presentation:
- Heat lost by hot object = Heat gained by cold object. Therefore:
Total Heat lost by the metal block = Total Heat gained by water + Total Heat gained by
the copper calorimeter
mmcm ∆Tm = mwcw ∆Tw + mccc ∆Tc
mmcm ∆Tm = [ 0.0755 x c x ((95 ± 0.05) – (35 ± 0.05))]
= [0.0755 x c x (60 ± 0.1)] = [0.0745 x c x (60 ± 0.17%) ]
= [(4. 53 ± 0.17%) c] = [(4. 53 ± 0.0077) c] Joules is the amount of heat
lost by the metal block.
mwcw ∆Tw = [ 0.0775 x 4200 x ((35 ± 0.05) – (30 ± 0.05))]
= [0.0775 x 4200 x (5 ± 0.1)] = [0.0775 * 4200 * (5 ± 2%)]
= [ (1627.5 ± 2%) c] = [ (1629.18 ± 32.6)]
= (1627. 5 ± 32. 6) Joules is the amount of heat gained by the cold water.
mccc ∆Tc = [0.06443 x 385 x ((35 ± 0.05) – (30 ± 0.05))]
= [0.06443 x 385 x (5 ± 0.1)] = [0.06443 x 385 x (5 ± 2%) ]
= (124.03 ± 2%) c = [ (124.03 ± 2.48)]
= (124. 03 ± 2. 48) Joules is the amount of heat gained by the Copper
Calorimeter.
Therefore, using the above calculations, we can now calculate the specific heat capacity of the unknown metal block:
[ (4.53 ± 0.0077) c] Joules = [ (1627.5 ± 32.6) ] Joules + [ (124.03 ± 2.48) ] Joules
[ (4.43 ± 0.0077) c] Joules = [ (1751.53 ± 35.08) ] Joules
c = [ (1751.53 ± 35.08) ] = [ (1751.53 ± 2.0%) ] = [ (386.7 ± 2.17%) ]
[ (4.53 ± 0.0077) ] [ (4.43 ± 0.17%) ]
c = [ (386.7 ± 2.17%) ] = [ (386.7 ± 8.4) ] JKg-1K-1
c = (386.7 ± 8.4) JKg-1K-1
Conclusion:
After conducting the above mentioned experiment and after carrying out all the calculations outlined above and in the previous pages, I hereby conclude that the specific heat capacity of the unknown metal block provided is (386.7 ± 8.4) JKg-1K-1. Therefore, the specific heat capacity of the unknown metal block is 386.7 JKg-1K-1 with an error of 2.17%. Thus the specific heat capacity ranges between 378. 3 JKg-1K-1 and 395.1 JKg-1K-1. Copper has a specific heat capacity of 385 JKg-1K-1 which is exist in the range of my reading. Hence the unknown metal provided was copper.
Evaluation:
The experiment was conducted successfully but because I did not get the exact answer some of the errors that might have occurred and some improvements that can be done include:
- If there was more time the experiment could have been repeated and conducted more than once, in order to obtain a better reading which could lead to more appropriate results and more accurate values and it would help to reduce random errors which affected the results obtained.
- There might have been significant heat loss to the surrounding while transferring the hot metal object from the beaker into the calorimeter and this heat loss might have affected the values and results.
- The use of an electric balance to measure the masses of the substances used in the experiment helped to avoid the uncertainties and errors related to the mass values.
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While conducting the experiment, I could not manage to heat the metal object up to the boiling point, as the metal object was removed when it was significantly hot due to the loosening and breaking of the cotton thread, nonetheless a higher temperature such as the boiling point (100oC) could have resulted into more accurate readings and results.
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The main source of error in this experiment was due to the limitation in the choice of apparatus used, the use of better and more accurate equipment would lead to more accurate results (results without errors). For example, rather than using a normal laboratory thermometer which has an error of ± 0.05 oC, it is better to use a digital thermometer which is more exact leading to more accurate temperature readings and hence, results.
