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# specific heat of a solid

Extracts from this document...

Introduction

SPECIFIC HEAT OF A SOLID:

TOPIC 3.2 IN THE SYLLABUS – 1.5 HOURS

When the heater is used as shown in the diagram, it provides energy E

E = V x I x t

V being the voltage provided by the power supply (should be constant), I the current (if the voltage is maintained constant, so will be the current) and t the time.

As the voltage (and hence the current) is kept constant, you will register temperature and time first using no thermal insulator and then using it.

Using the previous information about the heater together with the specific heat capacity of the blocks (Aluminium = 878 J K-1 Kg-1; Copper = 361 J K-1 Kg-1; Mild steel = 480 J K-1 Kg-1) do the appropriate graph to obtain the specific heat capacity of the block you worked with.

Criteria to be assessed in this practical are: Data Collection and Processing; Conclusion and Evaluation and Manipulative skills.

Middle

11520

12

4

270

12960

12

4

300

14400

12

4

330

15840

12

4

360

17280

12

4

390

18720

12

4

420

20160

12

4

450

21600

12

4

480

23040

Q=E therefore Q= V·I·t

 Aluminium mass (±0.00001kg) temperature ( ± 0.5 k) ΔT (±0.5 k) mΔT (±0.51kg·k) Q (±0.03J) 1.00156 295 0 0 0 1.00156 295 0 0 1440 1.00156 295 0 0 2880 1.00156 295 0 0 4320 1.00156 296 1 1.00156 5760 1.00156 298 3 3.00468 7200 1.00156 299 4 4.00624 8640 1.00156 301 6 6.00936 10080 1.00156 303 8 8.01248 11520 1.00156 305 11 11.01716 12960 1.00156 307 13 13.02028 14400 1.00156 309 15 15.0234 15840 1.00156 311 17 17.02652 17280 1.00156 313 19 19.02964 18720 1.00156 315 21 21.03276 20160 1.00156 317 23 23.03588 21600 1.00156 319 24 24.03744 23040

% error of c= (real value- experimental value)/ (real value) · 100

%error of c= ((878-784.2)/ 878) · 100 % error of c= 10.7%

% uncertainty of c= %u of Q+ %u of m+ %u of T

Average Q= 11520J

%u of Q= (0.03/11520) · 100  %u of Q= 2.60 ·10^-4%

Average T = 304.3 k

%u of T= (0.5/304.3) ·100  %u of T= 0.164%

Conclusion

Improvements

The readings for the temperature weren’t very accurate as the thermometer did only show whole numbers; therefore the value had to be approximated to the nearest unit.

Use other thermometers, which show not only the whole numbers but also the decimals between them, to have more accurate readings. This will definitely reduce the high systematic error experienced.

When the aluminium block with the insulator was heated, it was still hot because of the heating experienced when the aluminium block was heated by itself. This affected the final results and that’s why it has a higher percentage error.

So that the aluminium block could be cooled down to normal temperature, more time has to be waited. Therefore the readings could be much closer to the ones of the real value.

The power supply didn’t provide a constant supply of voltage, affecting also the constant value for current, affecting the accuracy of the experiment.

Don’t have a power supply that varies much in the flow of power, as the one used in the experiment.

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

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