CHEMISTRY LAB #15-1
CONSERVATION OF ENERGY
INTRODUCTION: The first law of thermodynamics states that the total energy in the universe
is constant. On a grand scale, the universe may be considered as a large closed system. A closed
system does now allow energy in or out of the system and the surroundings are not affected. In
practice, scientists often study the behavior of matter by enclosing a small system and assuming
that the heat energy can be held constant without loss to the surroundings. Experimentally, this
situation can be approached rather closely, or better yet, the heat transferred in or out of the
system can be calculated as a constant value and used to collect the experimental data.
A calorimeter is a device for measuring heat energy changes. In this series of experiments a
plastic cup will be used as a calorimeter. Although an inherent degree of experimental error
exists, we will assume that the calorimeter cup is acting as a small enclosed system with no heat
loss. If there is no heat loss from the calorimeter, then the heat energy lost by a “high energy”
substance in the calorimeter will equal the heat energy gained by a “low energy” substance in the
calorimeter. Expressed by equation:
qlost (Heat loss) + qgain (Heat gained) = 0
Part I: Hot Water/Cold Water
Here is a problem to solve. The temperature gauge on a hot water heater is broken making it
difficult to know the temperature inside the heater. The hot water heater cannot be opened.
Your task is to determine the temperature of water in the water heater by pouring the hot water
into water of known mass and temperature.
Materials: Calorimeter, 100 ml Graduated Cylinder, Hot water, Cold Water
Procedure:
 Measure 100 mL of cold water from a graduated cylinder into calorimeter.
 Measure the temperature and immediately add hot water to calorimeter.
 Stir and find the final temperature of mixed water.
 Pour water from calorimeter into graduated cylinder and determine volume of hot water
added.
Data:
Volume of water in cup
Before mixing
After mixing
100. mL
175 mL
Temperature of water in cup 12.0 ºC
27.0 ºC
Calculations:
Determine the temperature of water in hot water heater. (Show your work!)
After whole class review: List your procedural errors and their effect on your final calculation.
(Did it increase/decrease your calculated temperature?
Part II: Two Different Substances Mixed Together
The heat change for a substance is a function of three factors: the mass of substance (m), the
temperature change (  t), and the specific heat (c ). The specific heat is a constant for a given
substance. Specific heat is defined as the heat necessary to raise one gram of substance one
degree on the Celsius scale. For example, the heat lost by the metal when placed in the
calorimeter cup can be expressed as:
Heat loss (metal) = (cm) (mm) ( Tm)
*T = T(final) - T(initial)
Similarly, the heat gained by the water can be expressed as:
Heat gained (water) = (c H2O) (m H2O) (  T H2O)
*T = T(final) - T(initial)
Since the calorimeter cup is assumed to be a closed system, the heat loss equals the heat gain.
qlost + qgain = 0
OR
(cm) (mm) ( Tm) + (c H2O) (m H2O) (  T H2O) = 0
The purpose of this experiment is to find the specific heat of a metal. This is easily calculated
from the foregoing equation after obtaining the necessary data. The specific heat of water is
exactly one calorie per gram per degree Celsius (1cal/gᵒC). Also: Since 4.18 Joules = 1 calorie,
1cal/gᵒC = 4.18 J/ gᵒC.
PROCEDURE:
DATA
(+/-0.01g)
Mass of metal used
45.09
Mass of water used
(Equals volume of water)
Temperature of boiling
water and shot
Temperature of water in
calorimeter
Temperature of water and
shot after mixing
100. mL
97.7
23.2
24.4
Gr.# Sp. Heat
(J/g°C)
1
2
0.131
3
0.132
4
0.115
5
0.147
6
0.101
7
0.133
8
0.122
Ave
Calculations
Determine the specific heat of your metal sample. (Show your work!)
Deviation
Part III. Energy to Melt Ice
INTRODUCTION: The amount of energy required to convert a solid to a liquid, at constant
pressure and temperature, is called the heat of fusion of the substance. It is often convenient to
speak of the molar heat of fusion. The molar heat of fusion is the amount of energy required to
completely change one mole of a solid, at its melting point, into a liquid. In this experiment, the
molar heat of ice will be determined.
The ice will be melted by placing it in a known volume of hot water contained in a calorimeter.
The heat lost by the water will be absorbed by the melting ice. The volume of ice that melts can
be determined by measuring the volume of the water in your calorimeter before the ice is added
and after the ice has melted.
(2)
Volume(melted ice) = Volume of Water(final) + Volume of Water(initial)
Using the density of water as 1.0 g/ml, the molar mass of ice, and dimensional analysis, we can
then calculate the molar heat of ice.
PURPOSE: To determine the molar heat of fusion of ice in units of kJ/mol.
PROCEDURE:
1. Measure out 100.0 ml of warm water (approximately 50 C) into a calorimeter. Measure
the temperature of the water with a thermometer and record in the data table.
2. Obtain several ice cubes. (Use approximately 40 g worth.) Shake the excess water from
the cubes. Place the ice in the warm water calorimeter and stir carefully use your
thermometer until the temperature of the ice-water mixture reaches a MINIMUM
temperature. This should be near, not necessarily at O C.
3. Remove the excess ice from the mixture using tongs. Measure the volume of water
remaining in the calorimeter to the nearest 0.1 ml and record in the data table
Volume & Temperature of warm water in calorimeter.
(Before adding ice.)
Volume & Temperature of water in calorimeter.
(After adding the ice allowing it to melt.)
Extra volume of water = Mass of ice
Change in temperature of water
Change in temperature of ice (Assume initial temp. of 0 C.)
Calculations
Determine the molar heat of fusion of ice. (Show your work!)
Mass
(g)
100
Temp.
(C)
42.6
138
9.5
Questions:
Part I.
1. What is the final temperature of water in a calorimeter if 25.0 g of 55℃ water is added to
45.0 g of 40 ℃ water?
2. What is the final temperature of ethanol in a calorimeter if 25.0 g of 55℃ ethanol is
added to 45.0 g of 40 ℃ ethanol?
Part II
3. A 15.0 g piece of metal at 100℃ is dropped into 65.0g of 20.0 ℃ water. The final
temperature of the mixture in the calorimeter is 22.5 ℃.
a. What is the total heat gain by the hot water in calories?
In joules?
b. What is the total heat loss by the metal in in calories?
In joules?
c. What is the specific heat of the metal in cal/g℃?
In J/g℃?
Part III
4. The heat of fusion of water is 333.9J/g. What is this value in J/mole?
5. How much heat is required to melt 10.0 grams of ice?
6. How much heat is required to melt 10.0 moles of ice?
7. What will the final temperature of a calorimeter system be if 10.0 grams of ice at 0.0 ℃
are added to 222 grams of 60.0 ℃ water?