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IB Specific Heat Lab

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Introduction

Lab #3

Specific Heat

By: Rendol Williams

Mr. Smith

Irmo High School

4A Physics SL

Wednesday, November 16, 2011

Design

ASPECT 1:

The purpose of this lab is to prove an understanding of the concept of calorimetry, and establish a basic knowledge of applying those skills in the category of specific heat capacity. Through this experiment we will outline a method for determining specific heat capacities experimentally in lab. Specific heat represents the amount of thermal energy required to change the mass of one kilogram of a substance by one Kelvin. We will manipulate the equation Q=mcT to calculate various quantitative data that will inevitably lead to us an experimental specific heat capacity that we could compare to the actual or accepted value. We will also exercise the use of thermal equilibrium as we experiment in moving heat from a warmer body to a colder body in the form of thermal energy. The amount of heat lost must

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Middle

T

        Q= Heat (Joules)                m= mass        c= specific heat capacity (J g-1 K-1)                

T= Change in Temperature

        Q lost= Heat Lost (Metal)                Q gained= Heat Gained (Water)                Q lost = Qgained

Q gained = (4.183 ± 0) (122.6 ± .2) (25.2 ± .1 – 22.5 ± .1)        4.183= c of water

                        Q gained = 1384.7 ± .3

Zinc                Q lost =1384.7 ± .3

                        Q lost (1384.7 ± .3) = (c) (56.25° C ± .05) (100 - 25.2 ± .1)

                                c of zinc = .329 ± .04 J g-1 K-1

Q gained = (4.183 ± 0) (122.6 ± .2) (29.5 ± .1 – 24.5 ± .1)        

                        Q gained = 2564.2 ± .3

Tin                Q lost =2564.2 ± .3

                        Q lost (2564.2 ± .3) = (c) (70.5° C ± .05) (100 – 29.5± .1)

                                c of zinc = .510 ± .04 J g-1 K-1

Q gained = (4.183 ± 0) (122.6 ± .2) (32.0 ± .1 – 29.0 ± .1)        

                        Q gained = 1538.5 ± .3

Copper                Q lost =1538.5 ± .3

                        Q lost (1538.5 ± .3) = (c) (70.25° C ± .05) (100 – 32.0 ± .1)

                                c of zinc = .320 ± .04 J g-1 K-1

Q

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Conclusion

        Evaluation

As the percent error expresses the experiment overall could have been more accurate. Many simple adjustments could be made to result in better quantitative data being recorded. One source of error could be the error of validity of the scale. The triple beam balance is a good tool but if you desire more accurate results the electronic scale with zeroing would be a better alternative. Another source of error was the heat lost from the cup and the air flow coming into the cup. Heat from the cup would have dispersed into the open air since the top was open. A simple way of solving this problem would just be to put a lid over the top of the cup to prevent major heat loss. Some still would be loss but the ratio would be much smaller.

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