Capacity of a solid by electrical method

Introduction

Thermal conductivity heat is transferred as a consequence of temperature difference between 2 bodies, heat energy passes form a hotter to the colder body. Specific heat capacity is the amount of heat energy required in joules to raise 1kg of a substance by 1 degree Celsius, different substances absorb heat energy at different rates not all substances require the same amount of heat energy to increase the internal temperature of a substance this depends on the mass and the material it’s self. To demonstrate specific heat capacity of a solid an experiment was carried out using 1kg aluminium mass and 1kg copper mass heat was conducted to the masses by an electrical current set at 2amps, each mass had a heating element inserted into the mass and the temperature was recorded every 30 seconds until a change of 10 degrees Celsius was recorded, then the specific heat capacity can be calculated. See table of results and graph.

Theory

Formula for the specific heat capacity Q=MC∆T

Q= heat energy supplied (J) joules

M= mass (kg) kilo grams

C=specific heat capacity (J/kg °c) joules/ kilo grams/ degrees Celsius ∆T= change in temperature °c degrees Celsius

Formula for energy E= IVT

E= energy

I= amps

T= time

Formula for percentage of error

(estimated-actual)actual X 100

= percentage of error %

Apparatus

Apparatus used in this experiment consisted of,

1. Dc power supply

2. Volt meter

3. Amp meter

4. Heating element

5. Thermometer

6. 1gk copper block

7. 1kg aluminium block

Procedure

This experiment was conducted twice firstly with a 1kg aluminium block and then exactly the same procedure was carried out using a 1kg copper block, both blocks have a hole in the centre to insert the heating element and another hole to insert the thermometer.

1. Connect the DC power supply to the amp meter and then to the volt meter, adjust the output of the amps until it reads 2A on the amp meter this gave a reading of 7V on the volt meter. 2. Insert the thermometer into the block and take a reading (T1). 3. Insert the heating element into the block turn on the DC power supply and start the clock, take a reading every 30 seconds. 4. Continue to take reading’s every 30 seconds until the temperature has risen 10 degrees Celsius, at this point turn off power supply and record total time taken and final temperature (T2).

Time in seconds| Temperature in degrees Celsius Aluminium mass| Temperature in degrees Celsius Copper mass| 30| 19| 20|

60| 19| 21|

90| 19| 22|

120| 19| 23|

150| 19.5| 24|

180| 20| 24.5|

210| 20| 25|

240| 20.5| 26|

270| 21| 26.5|

300| 21| 27|

330| 21.5| 28|

360| 22| 29|

390| 22.5| 30|

420| 23| |

450| 23| |

480| 23.5| |

510| 24| |

540| 24| |

570| 24.5| |

600| 25| |

630| 25| |

660| 26| |

690| 26| |

720| 26.5| |

750| 27| |

780| 28| |

810| 28| |

840| 28.5| |

870| 29| |

Table of results

Calculations

Aluminium block E= IVT

E=2*7*870= 12180 J

Q=MC∆T

Q=12180 J

M= 1kg

C=? J/kg°c

∆T= T2-T1

∆T= 29-19

C= Qm*(T2-T1)

C= 121801*(29-19) = 1218

C= 1218 J/kg°c

Experimental value to raise 1kg of aluminium by 1°c = 1218j Actual value= 900 J

Calculations

Copper block

E= IVT

E= 2*7*390= 5460 J

Q=MC∆T

Q= 5460 J

M= 1 kg

C= ? J/kg°c

∆T= T2-T1

∆T= 30-20

C= Qm*(T2-T1)

C= 54601*(30-20)= 546

C= 546 J/kg°c

Experimental value to raise 1kg of copper by 1°c= 546 J

Actual value= 385 J

Experimental error

Copper

(estimated-actual)actual X 100

= percentage of error %

(546-385)385= 0.42

0.42*100= 42%

Experimental error

Aluminium

(estimated-actual)actual X 100

= percentage of error %

(1218-900)900= 0.35

0.35*100= 35%

Conclusion...

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