• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determining the specific heat of three metals

Extracts from this document...

Introduction

Physics portfolio

Determining the specific heat of three metals

Day to start:Thursday, September 06,2007

                    Due day: Thursday, September 06,2007

Albert.liu

                       The kilmore international school

                       11 B

Method of mixtures:

 When different parts of an isolated system are at different temperatures, heat will flow from the part of higher temperature to the part of lower temperature.

Therefore, from conservation of energy:

 Heat lost by hot object = Heat gained by cold object.

Aim:

This experiment is conducted in order to determine the specific heats of three different metals using the method of mixtures.

Method of this experiment :

  1. Measure the mass of the three different supplied cubic metals. (copper ,Lead ,Aluminum).
  2. Heat the supplied cubic masses of metal in boiling water and record the temperature.
  3. Place some cold water into a separate container of known mass, so that this is enough to just cover the cubic masses. Record the mass and temperature of the water.
...read more.

Middle

The average specific heat of

zinc =(270.833+283.375+295.556)/3=283.255image01.pngimage02.png

image03.png

So the percentage of error is:|(283.255-380)/380|×100%=25.46%

Aluminum:

Trial 1

Trial 2

Trial 3

Mass of Al(m1)    ±0.001kg

0.020

0.020

0.020

Mass of empty cup(m2) ±0.001kg

0.002

0.002

0.002

Mass of cup with water(m3) ±0.001kg

0.046

0.042

0.043

Temperature of cold water(T1) ±0.1℃

18.0

21.0

17.0

Temperature of metal with object(T2) ±0.1℃

84.0

87.0

74.0

Temperature

in  equilibrium (T3) ±0.1℃

23.0

30.0

24.0

the Aluminum : cm(T3-T2) = Cold water: cm(T3-T1)

Trial 1:

c(water)=4200image01.pngimage02.png

M(water)=M3-M2=0.046-0.002=0.044kg (±0.001kg)

C(Al)×0.020×(84.0-23.0)=4200×(0.046-0.002)×(23.0-18.0)

C(Al)=924 image01.pngimage02.png

Trial 2:

c(water)=4200image01.pngimage02.png

m(water)= 0.042-0.002=0.040kg (±0.001kg)

C(Al)×0.020×(87.0-30.0)=4200×0.040×(30.0-21.0)

C(Al)=1326image01.pngimage02.png

Trial 3:

M(water)=m3-m2=0.041 kg (±0.

...read more.

Conclusion

when we use thermometer sometimes it touched the bottom of the beaker. So the temperature of metal will be hotter.

Error:

We got the Parallax error during measure the temperature of using thermometer. Because if we not parallel with the top of the mercury in the thermometer, it will makes the result to be a little bit different. But not affect a lot.

Improvement:

1. When you put the hot metal into the container of the cold water, you must be careful not to let the metal touch the container and make the container be occluded immediately so that the heat can be transfer into the cold water.

2 The big problem is the heat losing on the way that the metal is transferred into the container of the cold water. But I think if we can do this experiment in vacuum the heat lost will be decrease and the data we collected will more and more close to the data in theory.

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Specific heat capacity of an unknown metal

    ] [ (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)

  2. Specific latent heat of fusion of ice

    ] = [ 0.0644 * 385 * (31 � 0.32%) ] = [ (768.614 � 0.32%) Joules ] = [ (768.614 � 2.46) Joules ] = [ (768. 614 � 2. 46) Joules ] is the amount of heat gained by the copper calorimeter.

  1. IB Specific Heat Capacity Lab

    (2) - (1) (see data table) = 103.75 - 41.35 = 62.40 g � (0.01 + 0.01) = � 0.02 g = (0.02 � 62.40) � 100 = 0.032 % Temperature change of water + Calorimeter (?T) (7) - (5) (see data table)

  2. specific heat of a solid

    Is true that it stopped some energy to go to the environment, but in the other hand it took more heat than the one that was lost to the surroundings when using only the aluminium block. This can be proven with the value for the percentage error of the specific

  1. How does the number of holes in a plastic cup affect the time it ...

    for the water to drain whereas it only took the cup 9.96 s (� .01 s) for the water to drain in the cup with 5 holes. The graph is at a negative slope. This is due to the physical barrier of the paper that prevents water from dripping thus as the number of holes increased, the drainage time decreased.

  2. Finding the latent heat of fusion

    Merging equations 2 and 3 we get a new formula to find the latent heat of fusion of ice. 5. From the graph it is possible to take the gradient which is mass per seconds. However, the inverse of the gradient will be seconds per mass which will be a portion on the equation in equation 4.

  1. HL Physics Revision Notes

    that is sealed off from the rest of the environment Control rods ? A material that can absorb excess neutrons whenever this is necessary. How does neutron capture of uranium-238 lead to plutonium-239: 1. Neutron collides with U-238 atom. 2.

  2. Experiment to compare the radiation of heat from different objects.

    surface area of the cans, let?s assume it to be 1 m^2. T = temperature Let?s take the value of the temperature as the Final reading of their temperatures at 180 seconds. e = emissivity ? = Stefan-Boltzmann constant = 1x1x5.67x10^-8x352.78^4 = 878.211 W Energy emitted by the shining can

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work