Experiment 10

Specific Heat Measurement


The objective is to measure the specific heat of a selected material or substance.


A calorimeter, a piece of metal (aluminum, brass, copper, zinc, or iron), a mass scale, two thermometers, a water heater, and a calculator


Units of heat commonly used are calorie (cal), kilocalorie (kcal), and Btu.

1cal is the amount of heat that can raise the temperature of 1gr of pure water by 1˚C.

{On the average, the best range is from (58.5 to 59.5)˚C}.

1 kcal is the heat that can raise the temperature of 1 kg of pure water by 1˚C.

British Thermal Unit (Btu)

1 Btu is the heat that can raise the temperature of 1 lbm of pure water by 1˚F.

Definition of Specific Heat:

The specific heat ( c )of a substance is the amount of heat that one unit of mass of that substance absorbs for its temperature to rise by one unit of temperature.

cwater =


 because of the way 1 kcal is defined.

cwater =

 because of the way 1 cal is defined.

cwater =

 because of the way 1 Btu is defined.

The specific heat of other materials are measured and may be found in physics texts.

In this experiment, since mass is to be expressed in grams, expressing specific heats in the unit shown on the right is more convenient.

On this basis, the specific heat of aluminum turns out to be  cAl =

 Note that in general, metals such as aluminum, take less heat for their temperature to increase by a certain amount compared to water; therefore, metals have lower specific heats than water.


1.      Use the mass scale to measure the mass of the inner aluminum container of the calorimeter after making sure it is clean and dry.  Record the mass of aluminum.

2.      Fill it to 1/4 with tap water and measure its mass again.  The difference is the mass of water.

3.      Select a piece of metal for which the specific heat is known.  For example a 200-gram piece of copper, zinc, or brass.  You may obtain its specific heat from your text or a handbook.  Such value is the accepted value for this experiment.  Measure the mass of the sample you select.

4.      Connect the sample to a string and place it in the already filled water-heater and let the water come to boil.  With the sample in boiling water, the initial temperature of the sample will be very close to 100.0˚C.   Have a thermometer in the boiling water to measure the initial temperature of the sample.   As you know the boiling point of a pure substance (such as water) is a function of its pressure as well.  The atmospheric pressure is 1atm. at ocean level with no winds. 

5.      A thermometer must be placed in the calorimeter through its cap’s hole to measure the initial temperature of the calorimeter.  This is the initial temperature of water and aluminum.

6.      When, after a few minutes, the initial temperature of the calorimeter is stable and at the same time the initial temperature of the sample (while in boiling water) is also stable, in a quick and careful motion, the sample must be withdrawn from the boiling water and placed in the calorimeter.  All safety measures must be observed when working with any electric heater and boiling water to avoid injury.

7.      The temperature of the calorimeter (water plus aluminum container) keeps increasing for a while.

8.      The calorimeter could be given a mild shake (while holding its outer container in hand and lifting it slightly above the table surface).  This helps a better mixing of water in the calorimeter and a faster thermal equilibrium reach.  The thermometer must be constantly checked and the highest temperature reached must be recorded.  Record this as equilibrium temperature Teq .

9.      The gathered data must then be inserted in the thermal equilibrium equation in order to solve for the unknown.  Let the unknown be the specific heat of the metal piece (sample) used.  The value you calculate for this unknown from the equation will be the measured value in this experiment.



Given: Measured:
Substance Specific Heat

 [cal /(gr ˚C)]

Water 1.000 Mw =       Tiw =        ˚C
Aluminum 0.215 MAl =       TiAl =        ˚C
Brass 0.0924 M S  =        TiS  =        ˚C
    S = Sample TiS  =        ˚C



Equation:    The heat loss by hot object (s)  =  The heat gain by cold object (s)

or,                  - MScS (Teq – TiS =  Mwcw ( Teq - Ti +  MAlcAl ( Teq - Ti )

or,                       - MScS (Teq – TiS =  [ Mwcw  +  MAlcAl ] * ( Teq - Ti ).

Treat cS as an unknown in this equation and solve for it.  The result will be the measured value.

Comparison of the Results:

The accepted and measured values of cS may be used in the following equation to calculate a percent error.


Conclusion:   To be explained by students.

Discussion:    To be explained by students.