Experiment 10

Specific Heat (Heat Capacity) Measurement


To measure the specific heat (heat capacity) of a selected material or substance.


A computer with internet connection, a calculator (The built-in calculator of the computer may be used.), a ruler, a few sheets of paper, and a pencil


Units of heat commonly used are calorie (cal), kilocalorie (kcal), and Btu.  In Metric System, it is preferred to use Joules and call it Joules of heat energy.

Note that 1cal = 4.186J,  or   1kcal = 4.186 kJ.

1cal is the amount of heat that can raise the temperature of 1gr of pure water by 1˚ 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 (Heat Capacity):

The heat capacity ( 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.

   The heat capacity of water is  cwater =  

 The heat capacity of iron is  ciron =



A calorimeter is a device that is thermally insulated from its surroundings.  A thermos jug is a good calorimeter.  A double layer plastic coffee cup with a tight lid may also be used as a calorimeter.

Thermal Worth of a Calorimeter

If the mass (M) of the inner container of a calorimeter along with its heat capacity (c) are known, they can be used in heat balance calculations in different experiments to determine other unknown quantities.  If you are using a thermos jug or a double-layer coffee cup with its tight lid, both the effective mass and specific heat are unknown for it.  Knowing the product Mc is enough to perform the calculations.  For an object, the product Mc may be defined as its thermal worth.  This quantity is important to be known for a calorimeter because it is not always practical to disassemble a calorimeter and measure the mass of the effective part of its container, for example.  The method of finding  Mc is used in the following example:

Example:  A calorimeter contains 100. grams of water at 25.0C.  To measure the product Mc for the calorimeter itself, 200.grams of water at 55.0C is added to it and a final temperature of  43.0C is reached.  Find the product Mc of the calorimeter.

Solution:  In this problem, the calorimeter and the 100grams of water already in it absorb heat, but the added water loses heat because the added water is initially at a higher temperature.  The heat balance equation, may be written as:

Heat absorbed by the calorimeter + Heat absorbed by its water  =  heat loss by the added (hotter) water

[(Mc )cal  + (Mc ) water](43.0 - 25.0)C   = - [ (Mc)water added ](43.0 - 55.0)C

        [(Mc )cal  + 100. x 4.186]( 18.0)   = - [ 200. * 4.186 ](- 12.0)        ;       (Mc )cal = 139.5 J / C



Use the calorimeter in the applet to measure the heat capacity of a substance.  Suppose the objective is to measure the heat capacity of copper.   Also let the heat capacity of iron be used as known.  The experiment will have two parts.   Part 1 is to measure the Mc of the calorimeter by using the known value of the heat capacity of iron.  Once the Mc of the calorimeter is determined, it will be used in Part 2 to measure the heat capacity of copper.

Click on the following applet: http://www.chm.davidson.edu/ChemistryApplets/calorimetry/SpecificHeatCapacityOfCopper.html

Note:  In the applet the equation for qcal = Ccal ( Tf - Ti ) has a misprint in it.  Please learn the correct form of it before starting.   The correct form should read qcal = McalCcal ( Tf - Ti ).     Mcal is missing in the applet.  Remember we need to first calculate the product Mc for the calorimeter not just ccal as shown in the applet.

    Part 1: Measuring the Mc-value of the Calorimeter

1.      In Part 1 of the applet, you can choose the mass of iron as well as the mass of the water in the calorimeter.  Note that the mass of iron must be from 1 to 100 grams and that of water (in the calorimeter) is from 50 to 90 grams.  Select a mass for iron and one for water in the specified limits. 

2.      Click on the "Reset" button.  Before clicking on the "Start", you need to write down the initial temperature of the calorimeter from the thermometer.  This is also the initial temperature of the water in the calorimeter.  Note that each graduation on the thermometer is 0.2 C.

3.   Click on the "Start" button and wait for thermal equilibrium to be reached.  You need to wait about 10 seconds until the thermometer comes to a fixed temperature.  Record the final temperature.  Note that each graduation on the thermometer is 0.2 C.  Read the temperature to your best estimate.  Also, note that this final temperature is for the calorimeter, its water, as well as the iron dropped in it.  The initial temperature of iron is assumed to be 100C.  The assumption is that the piece of iron is pulled out of boiling water before being dropped in the calorimeter.  This guarantees an initial temperature of 100C or very close to it.  Record all values in Table 1 under Data Section.

4.   Now you have all data to calculate the Mc of the calorimeter.  Perform a calculation similar to the Example above.  On the left side, in the example, the mass of water is 100grams. You need to replace that with whatever mass you chose for water in your experiment (applet).  The specific heat of water is 1 cal/(gr C) or 4.186 J/(g.C).  On the right side, the added water must be replaced with iron.  Apply the mass of iron you chose, and let the specific heat of iron be 0.450J/(gr C) as given in the applet.  Perform the calculations and solve for the Mc of the calorimeter.  Record the calculated value of Mc in Table 1.

5.   Now that Mc of the calorimeter is known, it can be used in Part 2 to find the heat capacity of copper.

Part 2: The main Experiment: Measuring the Specific Heat of Copper

1.   Go to Part 2 of the applet.  Here again as in Part1 for iron, the assumption is that you are keeping the copper piece in boiling water for long enough time such that its initial temperature is 100C.  Choose a mass for copper and one for the water in the calorimeter.  Reset the applet.  Read and record the initial temperature of the calorimeter.  Start the applet.  The hot copper piece will drop into the calorimeter.  Wait long enough for the calorimeter temperature to stabilize.  Record the equilibrium temperature.

2.   Use the collected data to calculate the specific heat of copper.  On the left side you should have the calorimeter and its water as heat absorbers (colder objects).  On the right side you should have the copper piece that gives off heat (the hotter object).  Solve for ccopper.  What comes out of your calculation will be your measured value.  Then compare that with the accepted value for the specific heat of copper: 0.387J/(g.C).

3.   Calculate a %error and record all values in Table 2 of Data Section.

4.   Repeat the whole experiment (2 more times) with different choices for masses of water, iron,  and copper.  Each time you should get the same results for Mc in Part 1, and for ccopper in Part 2.


Given:  cwater = 4.186 J/(g.C),     ciron = 0.450 J/(g.C).

Trial Mass of water (grams) Mass of Iron (grams) Initial Temp. of Water (C) Initial Temp. of Calorimeter (C) Initial Temp. of Iron (C) Equilibrium Temp. (C) Mc of the Calorimeter (J/C)
1         100.    
2         100.    
3         100.    

Table 1

Measured:  ccopper =   ????

Trial Mass of water (grams) Mc of the Calorimeter (J/C) Mass of Copper (grams) Initial Temp. of Calorimeter (C) Initial Temp. of Water (C) Initial Temp. of  Copper(C) Equilibrium Temp. (C) ccopper.

 J/(gr C)

% Error
1           100.      
2           100.      
3           100.      

Table 2



Equation:      Heat gain by colder objects  =  - Heat loss by the hot object.

Complete the calculations.

Comparison of the Results:

The accepted and measured values of ccopper may be used to calculate a percent error.

Conclusion:   To be explained by students

Discussion:    To be explained by students