LAB 4: HEAT OF FUSION
Learning Objectives
1. Measure the Heat of Fusion of water.
2. Compare the experimentally measured Heat of Fusion with the accepted standard and compute the
percentage difference.
3. Learn the principles of design and operation of a simple calorimeter.
4. Learn to read a laboratory thermometer to 0.1oC by estimating between the lines of the thermometer
graduations, or by directly reading a digital electronic thermometer.
5. Increase the accuracy of a laboratory measurement by calculating the average of three different
measurements.
6. Learn the Specific Heat relationship between amount of heat absorbed during fusion of ice and the
mass of the ice.
7. Utilize the Specific Heat of water by using the relationship between the amount of heat gained by by
a given amount of water and the temperature increase of the water.
8. Utiilize the principle of heat transfer in an insulated calorimeter that the heat absorbed by the ice
during fusion will be equal to the heat lost by the water.
Background Scientific Principles
Heat of Fusion
Changes of state (phase changes) involve the conversion or transition of matter from one of
the common states (solid, liquid or gas) to another. Examples include fusion or melting (change from
the solid to the liquid state), freezing (change from the liquid to the solid state), evaporation or
vaporization (change from the liquid to the gaseous state), condensation (change from the gaseous
state to the liquid state), sublimation (change from the solid directly to the gaseous state), and
deposition (change from the gaseous state directly to the solid state). Heat of Fusion is the amount of
heat released as 1 kilogram of liquid freezes (solidifies or crystallizes), or the amount of heat absorbed
as a solid melts (fuses or liquefies). See your textbook, p. 115-116. The exact value of the Heat of
Fusion is different for different substances. For water, the Heat of Fusion is 335 kiloJoules per
kilogram, or 335 Joules per gram, also expressed as 80 calories per gram.
Although the Joule is the official SI unit for measurement of energy, it will be simpler to
calculate in this experiment if we use the calorie as an energy unit. 1 calorie = 4.184 Joules. The
calorie is simple to use in problems involving water, because 1 calorie of heat energy will raise the
temperature of 1 gram of water by 1oC.
Typically, changes of state of a pure substance, such as melting, freezing, evaporation,
condensation, etc., occur at a constant temperature. Temperature change only occurs when all of the
phase changes are completed. This means that during the phase change, energy is being absorbed or
emitted with no change in temperature of the material being studied. This energy can be absorbed or
supplied by some other substance in the environment, and absorbing or supplying that energy can alter
the temperature of that material.
Specific Heat
When a substance changes temperature, but does not change its state, its temperature goes up
by an amount proportional to the amount of heat energy absorbed or lost. The specific amount of
energy that must be lost or gained to change the temperature of a standard mass
(1 kilogram for physics) of a substance by 1o Celsius is called the Specific Heat. The exact value of
the specific heat varies, depending on the particular substance. For water, the Specific
Heat is 4.2 kiloJoules per kilogram per o Celsius, or 4.2 joules per gram per o Celsius, also expressed as
1.0 calories per gram per oC.
Application of Heat of Fusion and Specific Heat in this experiment
In our study today, we are going to study the Heat of Fusion of water. We are going to
place some solid water (ice) at its melting temperature of ~0oC into some warm liquid water that has
been heated to a warmer temperature than the lab room. By doing this procedure in an insulated
container called a calorimeter, we will limit transfer of heat to between the ice and the water. Because
of the insulation, very little heat will be transferred between the water or ice and the environment.
The calorimeter simply consists of a small styrofoam cup within a larger styrofoam cup. An
insulating cardboard or styrofoam cover is pierced by a hole to allow insertion of a thermometer,
which also serves as a stirring device. The insulation (styrofoam,cardboard and air spaces) of the
calorimeter will prevent heat transfer to any other substance or material in the environment. By
assuming that the heat absorbed by the styrofoam calorimeter and air is approximately zero, we can
approximate the heat absorbed by the ice as it melts as being equal to the heat lost by the water as it
cools off.
Heat gained = Heat lost
equation 1
The heat gained by the ice as it melts is calculated by multiplying the mass of the ice times the
heat of fusion of ice. (See Text, p. 115.)
Heat gained (ice) = mice  Hf
equation 2
Once the ice melts, the cold water from the melted ice (which will be at the same temperatue as
the ice) will gain some heat from the warm water in the calorimeter. The heat gained by the cold water
is calculated by multiplying the mass of the ice times the specific heat of water times the rise in
temperature of the cold water. (See Text, p. 97.)
Heat gained (ice water) = mice  Hsp (Tice – Tfinal)
equation 3
The total heat gained is equal to the sum of equations 2 and 3.
Total Heat gained = mice  Hf + mice  Hsp (Tice – Tfinal)
equation 4
The heat lost by the warm water in the calorimeter will be calculated by multiplying the mass
of the warm water times the specific heat of water times the drop in temperature.
Heat lost (warm water) = mw  Hsp (Tw – Tfinal)
equation 5
The total heat lost is equal to equation 4. Since heat lost equals heat gained (equation 1), then
equation 5 equals equation 4.
mice  Hf + mice  Hsp(Tice – Tfinal) = mw  Hsp (Tw – Tfinal)
equation 6
Equation 6 can be solved for the Heat of Fusion.
mw  Hsp (Tw – Tfinal) – mice  Hsp (Tice – Tfinal)
Hf = ___________________________________________
mice
equation 7
Value and use of the average of several measurements
Scientists have long known by experience, and statisticians have proved mathematically, that
the average of several measurements is more likely to be closely representative of the actual value
than is any one of the individual measurements. During the process of making several measurements,
we can also evaluate if any of our measurements deviate seriously from the others. Such deviant
values may reflect serious experimental errors in either our measurements, calculations or procedures.
Deviant measurements, called outliers, are usually discarded, and the measurements are repeated until
the required number of reasonable values have been obtained. For our purposes, we will use the the
standard that any value of Hf that deviates by more than 20% from the others is an outlier, and will be
discarded*.
Materials
Calorimeter:
larger and smaller styrofoam cups
cardboard or styrofoam cover
Thermometer
400 mL Beaker
Laboratory balance
Ice
Distilled water
Hot plate
Paper towels
Please note, if the Physical Science thermometers measure temperature only in degrees
Fahrenheit (oF), the temperatures measured must be converted to oC, by using the formula
C = (oF + 32) ÷ 1.8
o
_____________________________________________________________________________
*The Science of Statistics, as well as the Art of Accounting have often been characterized by
the aphorism “Figures lie and liars figure.”
Experimental Procedure
1. Determine the room temperature by setting the thermometer on the table for 5 minutes, and then
reading it. Record the temperature in oF, as it is displayed, then calculate and record the
temperature in oC to the same number of decimal places as the thermometer reading.
2. Fill the 400 ml beaker about half full of distilled water from the special supply bottle.
3. Place the beaker of water on the hot plate and heat to a temperature about 15oC (27oF) above
room temperature. If the water becomes too hot, you can cool it down by adding room
temperature distilled water from the supply bottle.
4. Assemble the inner and outer calorimeter cups and put on the cardboard cover.
5. Weigh the assembled calorimeter, and record the weight.
6. Fill the inner calorimeter cup about half full of warm water from the beaker on the hot plate.
7. Re-cover the calorimeter and weigh the combined cup and the water, and record the mass.
8. Insert the thermometer into the ice container.
9. Allow the temperature to stabilize, a few minutes, and then read the thermometer, recording the
temperature as Tice.
10. Insert the thermometer into the calorimeter, and stir briefly.
11. When the temperature stabilizes, read the thermometer, estimating between the lines to the
nearest 0.1oC, recording the temperature of the warm water as Tw.
12. Acquire a tablespoon or so of ice, dry the pieces of ice with paper towels, and dump the dried
ice into the calorimeter. Be careful not to touch the ice with your bare fingers or hands, as this
will cause melting and create errors in your results.
13. Quickly cover the calorimeter and stir with the thermometer, until all the ice has melted. If
necessary, add more dried ice until the temperature reaches at least 10oC below room
temperature.
14. When all the ice has melted, and the temperature reaches a low point, around 5-10oC (40-50oF),
record the temperature, as Tfinal.
15. Remove the thermometer and weigh the covered calorimeter and cool water. Record the mass.
16. Subtract the mass of the calorimeter from the mass of the calorimeter and warm water. Record
as the mass of the warm water, mw.
17. Subtract the mass of the calorimeter and warm water from the mass of the calorimeter and cold
water. Record as the mass of the ice, mice.
18. Use equation 7 to calculate the Heat of Fusion of water.
mw  Hsp (Tw – Tfinal) – mice  Hsp (Tice – Tfinal)
Hf = _____________________________________________
mice
equation 7
Use the value of 1.0 calories per gram per oC for the Hsp of water.
19. Repeat steps 6, 7, and 11 through 18 until you obtain 3 non-deviant values for Hf.
20. Add the three values of Hf to obtain a total.
21. Divide the total obtained in step 20 by three to obtain the average.
22. Subtract the average experimental value of Hf for water, obtained in step 21, from the
accepted standard value of 80.0 calories per gram, divide by the standard and multiply times
100 to obtain the per cent difference between your value and the standard.
Standard – Experimental
80.0 – Average
Per Cent Difference = ______________________  100% = _______________  100%
Standard
80.0
Be sure and mention this per cent difference in the Results and Discussion sections of your lab
report.
23. In the Discussion section of your lab report, state your conclusion by answering the following
question:
Did your experimental determination of the Heat of Fusion of water agree with, or come close to the
accepted standard value? Explain why or why not.