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CHE 101 - Applied Chemistry

Mike Lawliss

A RUBRIC & SAMPLE LAB REPORT

 Formal Lab Reports must be written within the following format:

RUBRIC:

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SUNY_RUBRIC

 Specific Heat

Joe Student

Instructor:  Michael Lawliss

Class:  Chem 101 Spring 2005

 

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Aim:

             The Aim of this Lab Exercise is to show that the composition of a metal cylinder can be established using a laboratory derived specific heat.

 

Introduction: 

        The cylinder’s specific heat was determined indirectly by observing the temperature change of the water that the cylinder was submerged in.  Submerging the cylinder was necessary due to the fact that we could not measure the temp of the metal directly.  To calculate a specific heat we used the Heat Equation ( Heat = (mass) (spec heat) (∆T).  The calculated value of Specific Heat was then compared to a chart of known Specific Heats.  Using that information and other observable traits of the metal cylinder a guesstimate was formed for the composition of the cylinder.  
       This exercise will use the principles of thermodynamics, when a metal object is submerged in a sample of water, any heat exchanged from the metal to the water will result in a temperature change for the water until both water and metal arrive at equilibrium.  When this is done inside an insulated container the heat loss to the environment is kept to a minimum. 

 

Method / Procedure:

The following items were used:

- Calorimeter - the exchange between the metal cylinder and the water occurred inside the calorimeter to minimize the heat loss to the environment.  

- thermometer - was inserted into the Calorimeter

- boiling water bath - the metal cylinder was originally submerged here, an initial temp of @ 100ºC was achieved 

- metal cylinder of unknown composition

- laboratory balance that measures to 1/1000 of a gram to measure the mass of the cylinder and the mass of the water

Procedure:

Step 1:             In order to determine the specific heat of the metal cylinder an observable temperature change was needed, so the cylinder was started in a boiling water bath, a thermometer measured the temp of the bath to a tenth of a degree Celsius.

Step 2:             Room temperature water was added to the calorimeter and the Mass of the Water was found by taking the difference between the ‘mass of the calorimeter with the water’ and the ‘mass of the empty calorimeter’.

Step 3:             The temperature of the calorimeter and water was allowed to come to equilibrium the temperature was read and recorded, this was used as the Initial Temp of the Water.

Step 4:             The temperature of the boiling water bath was recorded before removing the cylinder, this was used as the Initial Temp of the Cylinder.

Step 5:             The hot metal cylinder was placed quickly into the calorimeter and the temperature of the water was monitored with the digital thermometer every 10 seconds for 3 minutes gently swirling the contents every 20 seconds to avoid areas of differently heated water.

Step 6:             The highest temperature that the water reached was recorded, this temperature was used as the Final temp of the Water. ( to avoid using spikes in temp readings, we graphed  the results to obtain our reading from there )

Step 7:             The Metal Cylinder was removed from the Calorimeter, dried and its mass was measured.

             

Results:

       Since the metal was fully submerged in the water when it lost its heat we can say that the heat flowed from the cylinder to only the water.  So it can also be assumed that:

Assumption – “The heat lost by the metal is equal to the heat gained by the water. ”

 

The heat gained by the water was calculated first, this was needed to determine the heat lost by the metal  With the value of the Heat Lost by the Metal  the Specific Heat of the Metal can be found using the same equation.

 

With the data that was collected the Heat the Water Gained was found using the equation

    H = (m)(Spec. Ht.)(∆T).  

             Where:

            H = Heat Gained by the Water

            m = Mass of the Water  = 149.6grams

            Specific Heat of Water = 1 ^cal/_{g x C}º

 

The graph was used to determine the Final Temperature of the water. (see Fig 1)

            ∆T = Change in Temp of the Water =

( Final Temp of the Water – Initial Temp of the Water ) = 

25.2ºC – 23.2ºC = 2.0Cº

 

Heat Water Gained =  ( Mass of Water ) ( Specific Heat of Water ) ( Change in Temp of water )

 Heat Water Gained       =     ( 149.6g )    ( 1 ^cal/_{g x Cº} )       ( 2.0Cº )        

            Heat Water Gained      =  299.2cal

 

Since the metal cylinder was submerged in the water as they reached that final temp we assumed they were at the same temperature and thus used the Final temp of the Water as the Final Temp of the Cylinder.  

              

As was stated the Heat the Water Gained will be the same as the Heat the Metal Lost.

The same formula was used to solve for the Specific Heat of the Metal. 

H = (m)(Sp. Ht.)(∆T)

H = Heat Lost by the Metal = 299.2cal

m = Mass of the Metal  = 42.57grams

            Specific Heat of Metal = ?? ^cal/_{g x Cº}

            ∆T = Change in Temp of the Metal = ( Final Temp of the Metal – Initial Temp

of the Metal ) =  100.2ºC – 25.2ºC = 75.0Cº

 

Heat Metal Lost  =  ( Mass of Metal ) ( Specific Heat of Metal ) ( Change in Temp of Metal )

      299.2cal    =            ( 42.57g )        ( Sp.Ht. )                           ( 75.0Cº )

            Specific Heat of Metal =  0.094 ^cal/_{g x Cº}

 

   Discussion:

 

     The value of Final Temp of the Water/Metal was determined from the graph.  The spike at time = 30 seconds was ignored as being caused by the thermometer having too close a proximity to the hot cylinder.  The temperature reading determined by the cooling line was used.

     The measured value for Spec. Heat was then compared to the list of known Specific Heats(1).

Laboratory Value =  0.094 ^cal/_{g x Cº}    Some possible substances (metals) with Known Values from List had specific heat values equal to 0.095 ^cal/_{g x Cº}, a difference of only 0.001 ^cal/_{g x Cº} seemed within an acceptable range to consider. 

     From the list of known Specific Heats the 3 possibilities were: Brass, Bronze and Copper.  Based on this information and by looking at the cylinder’s other properties such as color and factors like the expense and likelihood of Bronze it was determined that the cylinder’s composition was Copper. 

            The value determined in lab was very close to the value in the Table of Specific Heat.

% ERROR:

            %Error = |(Lab Value) – (Known Value)|   x100   =

                                        Known Value

 

( 0.094 ^cal/_{g x Cº} ) – ( 0.095 ^cal/_{g x Cº} )  x100     =        0.1 ^cal/_{g x Cº}          =  1.05% Error

_________________________________               ___________

                      0.095 ^cal/_{g x Cº}                                                   0.095 ^cal/_{g x Cº}

 

A % Error of 1.05% is acceptable given the equipment used.  Some possible sources of error could have been:

            - thermometers not accurate or not read properly

            - heat lost to the cup and air not factored in

Overall this exercise should be considered successful.

 

Conclusion:  

        It was shown that finding the Specific Heat of an unknown substance can be used to help identify the object’s composition.

 

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Bibliography

Lawliss, M,.  Lab Handout- Identifying Materials with Specific Heat,  Applied Chemistry-CHE 101 Spring 05

(1) - CRC Handbook of Chemistry and Physics, 81st Edition (c)1999 CRC publishing .....etc.

(2) - Chemistry An Introduction to General, Organic and Biological Chemistry, Timberlake 8th Edition ... etc

 

 

FIG 1 

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