Andres Martinez
6th Period
Chemistry Honors
Dr. Abusalih
Title: Heat of Fusion Lab Report
Purpose:
(a) The purpose of the heat of fusion lab is to determine the heat of fusion for water by
monitoring the changes in temperature while a known mass of ice melts in a cup of water.
(b) The purpose of part B: melting Ice Experiment, is to analyze the rates in which ice
melts with distilled water versus salt water to find out whether ice melts faster in distilled
water, or salt water.
Introduction:
In science, there are processes known as phase changes. This is when a substance
in the form of one of the three phases solid, liquid, or gas, changes into a different form.
When a substance melts from a solid to a liquid, the reaction has a specific name, fusion.
In this lab we are finding the heat of fusion of water. The heat of fusion of anything is the
amount of heat required to convert unit mass of a solid into the liquid without a change in
temperature. In other words, it is the amount of energy that is needed to melt a unit mass
of a solid into a liquid, without a change in temperature. This amount of energy will vary
based on the mass of the material undergoing the change. To find the heat of fusion of
water, first you have to isolate the water from any other reactions by placing it in a
calorimeter.
Materials:
ice, 400 ml beaker, Bunsen Burner, thermometer, Styrofoam cup and lid, distilled
water.
Method:
Procedure:
(a) Be careful to make accurate measurements! First a 400 mL beaker was filled to about
half full of distilled water and heated on the Bunsen burner to about
40-45°C. Next a thermometer was placed into a bucket of ice and the temperature reading
on the thermometer was allowed to stabilize. The temperature was then recorded in the
column of Tice in row 6. Then the Styrofoam cup calorimeter was assembled with the
cover and the mass of both the calorimeter and the cover was measured and recorded in
row 1. The inner cup of the calorimeter was filled to about half full with warm, distilled
water and the cover was replaced. The mass of the calorimeter and warm water was
recorded in row 2. The thermometer was then placed into the calorimeter and stirred for
one minute. The temperature of the water was then recorded in row seven; Tw. Next two
ice cubes were obtained and dried with a paper towel, being careful not to touch the ice
with bare hands to not risk melting, and placed into the calorimeter. The calorimeter was
then covered and stirred with the thermometer until all of the ice was melted. The
temperature should have then been 8-12 °C. If it was not in that temperature range more
ice was added. Then the temperature of the water with melted ice was recorded in Tfinal in
the data table. The thermometer was removed from the calorimeter and the mass of the
calorimeter, cover, and cool water was recorded in row 4. The process was then repeated
in case errors occurred.
(b) First the calorimeter was placed on a scale to record the mass on a data
table. Then 150 mL of water was added to the calorimeter and it was placed on
the scale to record the mass of the Calorimeter and the water. Then the two
values were subtracted from each other to find the value of the mass of the water
itself. A thermometer was then placed in the water and the temperature was
recorded. Another thermometer was placed in the bucket of ice provided and that
temperature was also recorded. Then 5-6 ice cubes were added to the
calorimeter careful not to spill it with the bouncing of ice. The temperature is then
recorded when the ice cubes have melted completely. The process may then be
repeated in case of errors. The average of the three trials was recorded on the
data table. The averages were then used to determine the heat of fusion of the
ice.
Data Table:
1. Mass of
Calorimeter and
Cover (g)
2. Mass of
Calorimeter and
cover and warm
water (g)
3. Mass of warm
water
4. Mass of
Calorimeter and
cover and cool
water (g)
5. Mass of Ice mice (g)
6. Tice (°C)
7. Tw (°C)
8. Tf (°C)
9. Heat lost Warm
Water
10. Heat Gained, Cold
Water
11. Heat Gained, Ice
12. Hfusion
13. Average Hfusion
14. Percent Error
Trial 1
1.749g
Trial 2
1.744g
225.378g
229.65g
223.636g
225.38g
246.20g
250.34g
20.822g
0 °C
44°C
14 °C
6709.08 cal/g°C
20.7g
0 °C
42 °C
16 °C
5859.88 cal/g°C
291.508 cal/g°C
331.2 cal/g°C
6417.572 cal/g°C
308.211 cal/g°C
287.648 cal/g°C
5528.68 cal/g°C
267.085 cal/g°C
Calculations:
Make sure that you keep track of your units
1. Subtract the mass of the calorimeter (row 1) from the mass of the calorimeter + warm
water (row 2), enter this difference as the mass of warm water (mw in row 3).
 Trail 1: 225.378g-1.749g= 223.636g Trail 2: 229.65g-1.744g= 225.38g
2. Subtract the mass of the calorimeter and warm water (row 2) from the mass of the
calorimeter + cool water (row 4), enter this difference as the mass of the ice (row 5).
 Trail 1: 246.20g- 225.378g= 20.822g Trail 2: 250.34- 229.64= 20.7g
3. The warm water in the cup loses heat to melt the ice and then to warm up the new cold
water formed when the ice cube melts. The heat lost by the warm water can be
calculated using the equation below: (Hsp = 1 cal/g °C)
Heat lost (warm water) = mw x Hsp x (Tw -Tfinal) (enter in row 9)
 Trail 1: Heat lost= 223.636g x 1cal/g°C X ( 44°C -14°C)=223.636 cal/g°C X
(30° C) = 6709.08 cal/g°C
 Trail 2: Heat lost= 225.38g x 1cal/°C x (42-16) = 225.38 cal/°C x (26) =
5859.88 cal/g°C
4. In our experiment, the ice cube melts and turns into cold water (which has the same
mass as the ice cubes). This cold water gains heat from the warm water, and the heat
gained can be calculated from the equation below: (Hsp = 1 cal/g ° C)
Heat gained (cold water) = mice x Hsp x (Tfinal -Tice) (enter in row 10)
 Trail 1:Heat gained (cold water)= (20.822g)x1 cal/g°C x(14) = 291.508 cal/g°C
 Trail 2:Heat gained (cold water)= (20.7g) x x 1 cal/g°C x (16) = 331.2 cal/g°C
5. Obviously, the ice cubes must gain heat from the warm water in order to melt. The law
of conservation of energy says that the heat gained by the ice and the cold water must be
equal to the heat lost by the cold water.
Heat gained (ice) + Heat gained (cold water) = Heat lost (warm water)
We need to know the heat gained (ice), so the above equation can be rearranged and
solved:
Heat gained (ice) = Heat lost (warm water) - Heat gained (cold water) (enter in row
11)
 Trail 1: Heat gained (ice)=6709.08 cal/g°C-291.508 cal/g°C= 6417.572 cal/g°C
 Trail 2: Heat gained(ice)= 5859.88 cal/g°C-331.2 cal/g°C= 5528.68 cal/g°C
6. We are trying to determine the heat of fusion for ice and compare it to the accepted
value. The heat gained by the ice depends on the mass of ice multiplied by Hfusion. To
find Hfusion, the equation can be rearranged and solved:
Heat gained (ice) = mice x Hfusion
Hfusion = Heat gained (ice) = row 11
mice row
5(enter in row 12)
 Trail 1: Hfusion= heat gained (ice)/ mice = 6417.572 cal/g°C / 20.822g = 308.211
cal/g°C
 Trail 2: Hfusion=heat gained (ice)/ mice= 5528.68 cal/g°C / 20.7g = 267.085
cal/g°C
7. Compute the average Hfusion for your three trials.
 (308.211 cal/g°C +267.085 cal/g°C)/2 = 287.648 cal/g°C
8. Compare your average Hfusion with the accepted value of 80.0 calories/gram. The most
convenient way to express their relationship is by calculating the % difference.
% difference = 100 x (80.0 - Hfusion)/80.0
 100 (80-287.648)/80= -269.56
Method (heat formula and percent error) two points
Calculations( must show steps 4-6, and 8) 4 points
Analysis of parts a and b (must explain major findings and explain it) 2 points
Conclusion1 point
21/20