Energy is often moved around during chemical and physical changes, and this energy usually takes the form of heat.  Calorimetry is the science of measuring such heat transfers.  The word derives from the term calorie, an old unit used to measure heat energy.  The conversion between calories and the SI unit Joules is 1 cal = 4.184 J.  The dietetic Calorie (with a capital “C”) is actually equal to a scientific kilocalorie.  

Heat is a type of energy and thus not directly measured.  Instead, it is calculated from properties like temperature that can be measured.  Temperature and heat are not the same thing, but they are related.  Temperature is a measure of the average motion of the molecules in a sample; heat is the total of all the energy of all the molecules in the sample.  We'll see below how heat can be calculated from temperature measurements.

Water freezing into ice, a chemical reaction that gets hot, and the dissolving of sodium acetate in water are all processes that release heat.  Other processes absorb heat instead.  Chemists wish to measure the amounts of heat absorbed or released in chemical and physical changes.

Measuring the energy absorbed or released during a particular process can be a bit tricky because the heat energy has a disturbing tendency to escape the beaker and disperse into the environment while we are trying to measure it.  To avoid this, chemists use an insulated container called a calorimeter.  A calorimeter can be a fancy and expensive device or it can be as simple as a thermos or a Styrofoam coffee cup.  Given our budget, we’re going with the Styrofoam cup!  Now we can trap the heat of a process inside the cup and measure it before it escapes.

Part A: Heat of Fusion of Ice

Imagine taking a very cold sample of ice and gradually adding heat energy to it.  The temperature will rise in a linear fashion, implying that the heat energy (Q) is directly proportional to the temperature (T).  In fact, the relationship is

Q = msDT

where Q = heat energy, m = mass of sample, and DT = the change in temperature.  The term s is a constant for the material itself called the specific heat.  For water, s = 4.184 J/goC, meaning that it takes 4.184 J of heat energy to raise 1 g of water by 1oC.  If we continue heating the ice, we will eventually reach the melting point, 0oC.  At this temperature the ice melts, but it remains at the same 0oC temperature until every last bit of ice is melted.  In other words, the temperature remains constant during a phase change even though we continue to add heat energy.  So where does the energy go?  It goes into the phase change itself, used to overcome the forces holding the particles in their crystal lattice.  The amount of energy needed to do this is called the heat of fusion, and it is a constant for that material.  It is measured in J/g, which is just the Joules of energy needed to melt each gram of the material.  Once we pass the melting point the water again rises in temperature in a linear fashion, at least until we hit the heat of vaporization at the boiling point.  We're going to use calorimetry to find the heat of fusion of ice.

Methodology

1.	Heat a sample of water to roughly 50oC.  Measure out 100.0 g of the hot water and add it to the calorimeter.  Record the exact temperature of the water in the calorimeter as T1.  Record the mass of water as m1.

2.	Immediately add a few ice cubes to the water in the calorimeter.  Stir the ice water mixture continually.  When the last of the ice is about to melt, add more ice.  Continue with this procedure, making sure that there is always some ice in the water.

3.	Continue adding ice until the temperature no longer drops.  Record this temperature as T2.  Carefully remove any unmelted ice, making sure not to lose any water from the calorimeter.

4.	Measure the new mass of water in the calorimeter and record it as m2.

5.	If time permits, repeat the experiment and average your results.

Analysis

The heat of fusion is simply the amount of heat energy lost to the ice divided by the mass of ice that absorbed the energy.  We can use the equation

Hf = DQ / mice

where DQ is the heat lost by the original water and mice is the mass of the ice added to the water.  The mass of ice can be easily found by subtracting the original and final masses of water

mice = m2 – m1

To find the heat lost by the original water, we use the equation Q = msDT.  Plugging in specific values from the lab, we get

DQ = (m1) (sH2O) (T1 – T2)

where sH2O = 4.184 J/goC.  Once we have DQ and mice, we can divide them to get the Hf value for ice.  Look up the accepted value and find the relative error for your result.

Part B: Heat Capacity of the Calorimeter

Before we move on, we should address a potential source of error in part A.  We have been assuming that the calorimeter itself is a perfect insulator, but that’s not entirely true.  Some of the heat energy inside is sure to transfer into the walls of the cup itself.  The amount of heat that a calorimeter will absorb in this way is a constant for that particular calorimeter and is called the heat capacity of the calorimeter.  It can be found by mixing known samples in the calorimeter and then directly measuring how much energy was lost to the calorimeter itself.  This lost energy must be accounted for when calculating the energy transfers for the chemical reaction.

An additional difficulty is that heat transfers are not instantaneous.  When we mix substances in the calorimeter, we must wait until they thoroughly mix and reach equilibrium, but the longer we wait the more error we get due to the inevitable escape of heat into the environment.  Our solution to these conflicting issues is to take several temperature measurements over a period of time after mixing the substances, then extrapolating back to what the temperature must have been at time zero (the time when the substances were first mixed).

Methodology

1.	Add 50.0 g of cold water to a Styrofoam calorimeter.  Allow the calorimeter and water to come to a constant temperature and record that temperature as T1.

2.	Heat a sample of water to roughly 50oC.  Measure out 50.0 g of the hot water, find its exact temperature, and immediately pour it into the calorimeter with the cold water.  Record the hot water temperature as T2.

3.	Stir the water in the calorimeter to mix it thoroughly.  Record the temperature of the water exactly 30 s after mixing.  Record the temperature again after 60 s, and once again after 90 s.  In the Analysis section, you will use this data to extrapolate to time t0 and find the heat capacity of the calorimeter.

Analysis

To find the heat capacity of the calorimeter, you must first extrapolate the time and temperature data.  Draw a graph with time on the x-axis and temperature on the y-axis.  Draw the best possible straight line through the graph and extend it back to a time of 0 s.  Record the temperature here (where the graph crosses the y-axis) as T0.

Now we must find the heat lost by the hot water and the heat gained by the cold water using Q = mcDT.  Plugging the hot water data into the equation, we get

Qh = mh s (T2 – T0)

as the heat lost by the hot water. Using the cold water data, we get

Qc = mc s (T0 – T1)

as the heat gained by the cold water.  Now subtract the two (Q = Qh – Qc) to get the heat gained by the calorimeter.  The heat capacity of the calorimeter is then just the heat gained divided by the change in temperature,

C = Q / (T0 – T1)

Save that answer; we'll need to use it in part C.

Part C: Heat of Reaction

Many chemical reactions absorb or release heat.  Those that absorb heat are called endothermic and those that release heat are called exothermic.  The process of calorimetry can be used to measure the exact amount of energy absorbed or released by a chemical reaction.  In this experiment, we will measure the heat released by the exothermic reaction between hydrochloric acid and potassium hydroxide.

When an acid and a base are mixed they neutralize each other and produce a salt, water, and heat energy.  We can mix them in a calorimeter and measure the amount of heat released.  We must be careful to mix equal quantities of each reactant, though, or there will be some of the excess reagent left over after the reaction is complete.  Of course, the amount of energy released will depend on the total amount of reactants we use.  For uniformity, chemists often work with the molar heat of reaction, which is the heat released when one mole of product are formed.  After we find the heat of reaction, we can calculate the molar heat of reaction.

Methodology

1.	Determine the number of grams of KOH needed to make 100 mL of a 2.00 M solution.  Obtain the correct amount of KOH and dissolve it with distilled water in a 100 mL volumetric flask.

2.	Pour 50.0 mL of the KOH solution into a beaker and mass it.

3.	Obtain 50.0 mL of 2.00 M HCl and determine its mass.  Add the masses of the two solutions together to get the total mass of reactants.

4.	Record the temperature of the two solutions.  If they are not the same temperature, record the average temperature.

5.	Carefully pour the two solutions into a Styrofoam cup calorimeter and continually stir with a thermometer.  Record the temperature after 30 s, 60 s, and 90 s.  Once again, you will use this data to extrapolate the temperature at t0.

6.	If time permits, repeat the experiment and average your results.

Analysis

Now that we have the heat capacity of the calorimeter, we can move on to part B, finding the heat of reaction.  Once again, we’ll start by drawing a temperature-time graph for the data from part B, step 5.  Draw a linear regression and extrapolate the line to the y-axis, recording the temperature there as T0.

Now we simply use Q = msDT to find the heat gained by the solution.  Substituting the terms from the experiment, the equation becomes

Qsoln = (mtot) (sH2O) (T0 – T1)

Note that we are using the specific heat of water because the solution is mostly made of water.  We have to account for the heat absorbed by the calorimeter, which is given by the equation Q = CDT.  Using our symbols, we get

Qcal = C (T0 – T1)

The total heat released by the reaction is the sum of the heat we measured plus the heat absorbed by the calorimeter:

Qtot = Qsoln + Qcal

This is the heat of reaction, but it only applies to the specific amounts that we used.  To make it more universal, we divide by the number of moles of water molecules produced in the reaction

Hrxn = Qtot / n

where n is the number of moles.  You can get that from the stoichiometry of the reaction.  By the way, your value should be given a negative sign to show that the reaction is exothermic, i.e., that the energy was lost from the reactants.  The unit on the value is J/mol, but since it will be a very large number, we often express it in kJ/mol.  The accepted value for the heat of reaction for this neutralization reaction is -59.85 kJ/mol, so you can do a relative error on your result.  

Questions

1.	How could we have improved the accuracy of our heat of fusion result in part A?

2.	After the ice water in part A reaches its minimum temperature, a student waits a while before removing the ice, allowing more to melt.  How will this affect his results?

3.	Would the results of this experiment have been different if we had measured temperatures in Kelvin?  Explain.

4.	If we had used a better insulated calorimeter, how would the graph in parts B and C have changed?

5.	Why do you need to know the Molarity of the acid and base solutions in part C?