Download Intro to Physics Lab

Bellevue College
PHY115 Physics Lab:
SPECIFIC HEAT OF A METAL/HEAT OF FUSION OF WATER
Purpose



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To review the relationships associated to flow of heat from a hotter (metal) to cooler
object (water at room temperature).
To understand and apply the principle of conservation of thermal energy – Heat
energy gained by cold water is equal to heat lost by the hot metal.
To determine the specific heat capacity of two unknown samples of metal in the
temperature range between room temperature (water) and water boiling temperature
(metal).
To compare the accepted values of specific heat for 2 different metals
Theory
Heating is a result of energy transfer from object with higher temperature to an object with
lower temperature. Two objects in contact that start at different temperatures will reach the
same temperature – thermal equilibrium. If the two objects are enclosed in the environment
that no heat is exchanged with surrounding, the amount of heat which flows from the hot object
must equal the amount of heat which enters the cold object – the low of conservation of energy.
As a result the thermal energy (heat) of the hotter object will decrease and that of the cooler
object will increase.
The metric unit for heat energy is Joule (J). Besides Joule, the unit of calorie “cal” is also used in
metric system. A calorie is the amount of energy or heat, required to increase the
temperature of one gram of water for one degree Celsius. The relationship between these
two units is:
1 cal = 4.19 J
or
1 J = 0.239 cal
The amount of heat energy that is required to raise the temperature of one gram of a substance
by one degree Celsius is called the specific heat capacity, or simply the specific heat of that
substance.
J
or Joules per kilogram · Kelvin
kg × K
Water, for instance, has a specific heat of: 4190
Some examples of specific heats in
Al:
900
Lead: 128
Cu:
Iron:
J
kg × K
J
are:
kg × K
385
449
Other specific heats are listed on p. 381
The amount of heat energy involved in changing the temperature of a sample of a particular
substance depends on three parameters
 the specific heat of the substance (cv),
 the mass of the sample (m),
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
and the magnitude of the temperature change. The Greek letter delta ( ) is used to indicate
a change.
T = temperature
final
- temperature
initial
The amount of heat energy that is transferred in the process of producing a temperature change
can be calculated from this information, according to the following equation:
Change in heat energy  mass X specific heat X change in temperatur e
Q  m  cv  T
In this experiment, you will determine the specific heat of a metal. A heated sample of this
metal will be poured into cool water contained in a plastic foam cup. Shortly after mixing, the
water and the metal will have come to the same temperature. Because plastic foam is a good
insulator, heat cannot easily escape to the surroundings. Therefore, the heat lost by the metal
can be said, for the purpose of this experiment, to be equal to the heat gained by the water. The
amount of heat energy gained by the water will be calculated in the following formula:
Heat gained by water  mass of weater X specific heat of water X change in temperatur e of water
Qwater  mwater  c water  Twater
The amount of heat energy lost by the metal will be calculated in the following formula:
Heat lost by the metal  mass of the metal X specific heat of the metal X change in temperatur e of the metal
Qmetal  mmetal  cmetal  Tmetal
Because the heat gained must equal the heat lost, a third equation can be written.
Qmetal  Qwater
mmetal  cmetal  Tmetal  mwater  c water  Twater
The specific heat of water is known. The temperature changes of the water, and of the metal,
can be measured, as can the mass of the water and the mass of the metal. Using this data, the
specific heat of the metal can be calculated using equation:
cmetal 
mwater  c water  Twater
mmetal  Tmetal
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PROCEDURE:
1. Fill a 250-mL beaker about 3/4 full of water. Place the beaker on electric plate (see
drawing). Use a electric plate to bring the water to a slow boil. While the water is
heating, proceed to step 2.
.
2. Obtain a sample of the metal being used that will fill the test tube about 1/4 full. Find the
mass of the metal to the nearest 0.01 g and record on the table.
3. Transfer the metal into the inner container of the calorimeter. Place the metal on the “ring”.
Position the test tube so that the metal is below the level of water in the beaker (be sure the
bottom of the test tube does not touch the bottom of the beaker). Adjust the knob on the
electric plate so the water is just boiling gently. Allow the test tube to remain in the boiling
water bath for at least 10 minutes. Proceed to step 4 while the metal is heating.
4. Carefully measure out 100 mL water in a graduated cylinder, and pour the water into a plastic
foam cup. Place the cup in a 400-mL beaker for support. Place a thermometer and stirring rod in
the cup. Record on the table the mass of the 100.0 mL water sample. REMEMBER that 1 milliliter
of water has a mass of 1 gram.
5. After the metal has been heating for at least 10 minutes, using the thermometer, measure
the temperature of the hot water. The temperature of the metal is the same as the water.
Record this temperature to the nearest 0.5 oC as the initial temperature of the metal sample in
the table. Read the temperature of the water in the calorimeter to the nearest 0.5 oC and record
in the table as the initial temperature of the water.
6. Remove the metal from the boiling water, using the clamp as a holder. Carefully, but quickly,
place the metal into the water in the second cup (use a paper towel to keep any hot water on
the tube from dropping). Note the temperature frequently. As the temperature begins to change
more slowly, watch the thermometer continuously so as not to miss the maximum temperature
reached. Record this maximum temperature on the table to the nearest 0.5 oC, as the final
temperature of water and metal.
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DATA TABLE:
Mass of
Mass of
the metal
water in sample
Metal calorimeter
mmetal
Sample
mwater
kg
Initial
temperature
Initial
Final
of metal
temperature temperature
sample
of water in of water and
calorimeter
metal
kg
K
K
K
1
2
3
4
CALCULATIONS TABLE:
Change in
Change in
temperature temperature Heat energy
Metal
of water
of the metal gained by specific heat
Sample
Twater
Tmetal
the water of the metal
K
J
K
J
kg × K
Error
%
1
2
3
4
7. Calculate the changes in temperature of the water ( Twater) and of the metal shot ( Tmetal).
These are just simple subtractions. Remember, the water and metal had the same final
temperature, but different initial temperatures. Record the date on table.
T = temperature
final
- temperature
initial
8. Calculate the heat energy gained by the water. Use formula for Qwater from the introduction for
this calculation.
9. Remember that the heat gained by the water is equal to the heat lost by the metal, calculate
the specific heat of the metal using the formula for cv: .
cmetal 
mwater  c water  Twater
mmetal  Tmetal
The specific heat of water is again, in
J
kg × K
Look up the specific heat for your metal on a textbook, and compare your experimental answer
the questions.
10. Compare calculated values for the specific heats to those from textbook for the
corresponding metal by calculating the % error values.
Repeat the whole procedure for the second and third different metal sample.
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Questions and applications
1. If you double the mass of metal, keeping everything else the same, what will happen to the
ΔT of the water in a beaker?
2. If you double the mass of water, keeping everything else the same, what will happen to the
ΔT of the water in a beaker?
3. Calculate the amount of heat required to raise the temperature of 78.2g of water from 10C to
35C.
4. What is the specific heat of a 75 gram sample that requires 1200 cal to change the
temperature from 25C to 85F?
5. Calculate the amount of heat, in joules, need to raise 34g of ice from 55C to 67C.
6. Calculate the specific heat for a 102g sample that requires 1430J to raise the temperature
from 8.7C to 12.5C.
7. What will the final temperature be if, at room temperature (25C), 1300 cal are added to a
76g sample of iron?
8. What would the change in temperature (T) be if an 89g sample of copper required 678
calories of heat?
9. Convert the specific heat capacity values to units of
J
J
Note: The units
are the
kg × K
kg × K
same as J/kg. oC, since a temperature change in oC is the same as a temperature change in K
units
10. Calculate the % difference of the experimental values from the accepted values.
Part II
Heat of Fusion of Water
Devise an experiment that will experimentally obtain
the Heat of Fusion of Water (Solid/Liquid phase
change). Be sure to include appropriate data tables,
graphs, calculations and data analysis.
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