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Measuring the Heat capacity of a Metal Free essay! Download now

Home > GCSE > Physics > Measuring the Heat capacity of a Metal

Measuring the Heat capacity of a Metal

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Downloads to date: N/A | Words: 650 | Submitted: 24-Jan-2012
Spelling accuracy: 98.5% | Number of pages: 6 | Filetype: Word .doc


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Measuring the Heat capacity of a Metal

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Measuring the Heat capacity of a Metal
Background theory
This experiment gives the value for specific thermal capacity of a metal in this case iron. The specific heat capacity is the amount of heat energy needed to raise a specific amount of a substance.

Method
Measure and record the mass of the metal block in kg and Set up the circuit and set the power supply to 12V.




The ammeter A measures the current through the heating coil. The voltmeter V measures the voltage across the heating coil. That will enable to measure the energy input into the heater. Measure the temperature of the block using the mercury in glass thermometer provided. Allow the thermometer to reach thermal equilibrium and then write down the temperature. Turn on the current recording the temperature of the block every 30s for 10 minutes. After 10 minutes turn the heater off and continue recording the temperature of the block every 30s for a further 5 minutes.




Results
Table showing the voltage across the heater and the current through it at the start, middle and end of the heating.
Time (seconds)
Temperature (C)
Power (Voltage)
Current (Ampere)
30
16
60
16.4
90
17
120
18
150
19
180
20
210
20.5
240
21.5
270
22
6.8
2.53
300
23
330
24
360
25
390
26
420
27
450
28
480
29
510
29.5
540
30.5
570
31
6.82
2.53
600
32
6.84
2.54



Table showing temperature of the block every 30s for 10 minutes.

Time (seconds)
Temperature (C)
30
16
60
16.4
90
17
120
18
150
19
180
20
210
20.5
240
21.5
270
22
300
23
330
24
360
25
390
26
420
27
450
28
480
29
510
29.5
540
30.5
570
31
600
32
630
33
660
34
690
34.5
720
35
750
35
780
36
810
36
840
36
870
36
900
36







Graph showing temperature against time for the full 15 minutes


7. (650-200) / (32.5-20)
= (450) / (12.5)
=36C/s

8. 6.84 x 2.54 = 17.3736 J/ s

9. Use the energy supplied in one second i.e. the power (Question8) and the rise of temperature in one second (Question7) to calculate the specific heat capacity of the metal using the equation:
Power = mass x specific heat capacity x temperature rise per second
- When the equation is rearranged to get C it will be:
q = m x c x ?t
=17.3736/10184 x 36
= 0.061 J/g C/s
10. The graph has a linear shop which shows that there is a steady and constant rate at which temperature rises. in the last few minutes graph is no longer liber and when it reaches a certain point no changes occurs for few minutes as the heat is turned off so the temperature neither increases or decreases for a while. Even though the temperature did rise even after the heater was turned off, the metal stayed at the same temperature hence no power was passing through. When it reaches 36C it will start to fall again until it hits the room temperature.
11.


Conclusion
Main issue with this experiment is the energy losses. The electrical energy measured going into the heater does not correspond exactly to the thermal energy delivered to the block which in turn result in less increase of temperature. For improvement
Energy losses are a problem in this experiment. The electrical energy measured going into the heater does not correspond exactly to the thermal energy actually delivered to ...

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