89
Specific Heat Capacity
PRE-LAB ASSIGNMENTS:
To be assigned by your lab instructor.
STUDENT LEARNING OUTCOMES:
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The First Law of Thermodynamics
Thermal energy as a function of temperature
The units of energy
Sign conventions in heat flow
Heat capacity
Specific heat capacity
Factors in experimental error
EXPERIMENTAL GOALS:
You will measure the specific heat capacity of a known metal. Then you will measure the
specific heat capacity of an unknown to identify the metal.
INTRODUCTION:
Heat Capacity
A large part of chemistry is studying and understanding the physical and chemical
changes that matter undergoes. The driving force for most chemical and physical changes is
energy, often in the form of thermal energy (heat). Temperature and heat are not the same
quantity. The temperature of an object is an indicator of the amount of thermal energy it
“possesses”, but heat is a measure of the transfer of energy to or from an object. We know that
the more a substance is heated, the more its temperature will rise. This relationship is a direct
proportionality that may be expressed as q  T , where q is the heat absorbed or released by the
system and ΔT is its temperature change (ΔT =Tfinal – Tinitial). This proportionality may be also
be written as an equality,
q
 constant
T
or
q  constant  T
[1]
The constant of proportionality is the heat capacity, C, and is defined as the quantity of heat
required to change an object’s temperature by one unit of temperature (usually 1 oC or 1 K).
Hence we can rewrite Equation 1 as
q  C  T
[2]
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It follows that heat capacity has units related to
energy
temperature
and is commonly expressed as
J
C
o
(also written as J/°C or J °C-1).
The heat capacity of a system is dependent on a few factors. The heat capacity is an
extensive property and thus dependent on the amount of a sample present as shown in Figure 1.
A larger amount of sample requires more heat than a smaller amount of the same substance to
raise the substance’s temperature by one unit. In addition to the amount of substance present, the
kind of substance has a direct bearing on its heat capacity. The same amounts of different
substances usually have different heat capacities, and heat capacity is often used for samples of a
fixed size. It would be more convenient if the heat capacity depended only on the type of
material, and not the amount of material. If heat capacity were an intensive property,
independent of the amount of substance, it could be used as an analytical tool to identify a
substance, much like density.
100 g
C=q
200 g
C' = 2q
o
Figure 1. Heat required for 1 C change in temperature showing dependence
of heat capacity, C, on amount of substance.
Specific Heat Capacity
In order to overcome the fact that heap capacity varies with the amount of a substance,
chemists often use the specific heat capacity, Cs, also known as the specific heat. The specific
heat capacity of a substance is defined as the amount of heat required to raise the temperature of
1 gram of the substance by 1oC (see Figure 2).
23.0C
1 g H2O
22.0C
4.184 J
Figure 2. Specific heat capacity.
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The relationship between the thermal energy and specific heat capacity may be expressed as
q  m  C s  T
[3]
Equation 3 is similar to Equation 2 with the additional parameter of mass, m. This extensive
property of a given sample, heat capacity (as in Figure 1), may be converted to its specific heat
capacity by dividing the mass of the sample into C to give
Cs 
C
C'

100 g 200 g
By accounting for the amount of substance, the specific heat depends only on the kind of
substance and is therefore an intensive property. This quantity has units related to
energy
. The units of specific heat are commonly expressed as J (also written as
mass  temperature
-1
g o C
-1
J/g·°C or J g °C ).
In this laboratory, a calorimeter is used to measure the specific heat of various known
metals. A calorimeter is an apparatus used to measure the heat released or absorbed by a
physical or chemical process. The calorimeter used here is composed of a pair of stacked
Styrofoam coffee-cups with a lid (see Figure 3 in the Procedure section). The lid has two holes,
one for a thermometer and the other for a stirring wire, and helps minimize heat loss to the
atmosphere. In addition to measuring the specific heat of a known metal, an unknown metal will
be identified based on the determination of its specific heat capacity.
The specific heat can be determined by immersing a known mass of warm metal of a
known temperature into a known volume of water, at a known, lower temperature. Heat energy
will flow from the hot metal to the water until, after a period of time, the water and metal achieve
the same temperature. From the First Law of Thermodynamics, the sum of the heat lost by the
metal and the heat gained by the water in a perfect experiment is qmetal + qwater = 0, which we can
rearrange as
qmetal = -qwater
[4]
The negative sign in Equation 4 comes about because of the ΔT term (see Equation 3). The final
temperature of the metal is lower than the initial temperature, and the resulting value of qmetal
will be negative. By convention, the flow of heat out of a body is a negative quantity.
An ideal calorimeter would provide perfect insulation so that its contents, the system,
could undergo its temperature change without any loss or gain of heat to the surroundings or to
the calorimeter itself. Such a calorimeter does not exist. Instead, we must build an apparatus
that gains (or loses) a minimum amount of heat energy itself while also minimizing the flow of
heat into or out of the Universe. Experimentally, the total heat gained by the water that we
measure will not be the actual heat lost by the metal since the apparatus will gain some of the
heat as well as permit some leakage to the atmosphere. In this experiment, the heat lost by
leakage to the atmosphere will be ignored, since the Styrofoam calorimeter is a reasonably good
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insulator. However, the heat lost by the system to the calorimeter is small, but not negligible,
and it must be considered. With this correction, Equation 4 becomes
qmetal = -(qwater + qcal)
[5]
To determine calorimeter, the heat capacity of the calorimeter, Ccal, must first be determined. With
Ccal, the heat absorbed by the calorimeter may be accounted for in the determination of the
specific heat of a metal. The determination of Ccal is called the calibration of the calorimeter.
To calibrate the calorimeter, warm water of a known mass and temperature is mixed with
a known mass and temperature of cooler water in the apparatus. In an ideal apparatus, the heat
gained by the cooler water would exactly equal the heat lost by the warmer water. However, in
the real world, the heat lost by the warm water will be equal to the heat gained by the cooler
water plus heat lost to the calorimeter,
qlost by hot = -(qgained by cold + qcal)
[6]
qcal = -(qlost by hot + qgained by cold)
[7]
Rearrangement yields
which can be used to determine the heat capacity of the calorimeter.
In this experiment, the actual calibration of the calorimeter will not be performed.
Instead the Ccal is given as 8.36 J/oC. With this value of Ccal for the coffee-cup calorimeter,
Equations 3 and 5 may be used to determine the specific heat of a metal. Table 1 lists the
specific heat capacities of various metals.
Table 1. Specific heat capacities of various metals.
Specific Heat
Capacity, Cs
Metal
(J/g∙oC)
Aluminum
0.897
Copper
0.385
Iron
0.449
Lead
0.129
Nickel
0.444
Tin
0.228
Zinc
0.388
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SAMPLE CALCULATION OF THE CALIBRATION OF A CALORIMETER:
Water with a mass of 50.0 g at 80.1 oC was added to 50.0 g of water at 27.2 oC. Upon mixing,
the final temperature attained was 51.9 oC. What is the heat capacity of the calorimeter, Ccal?
(Note: This example calculates the heat capacity of the calorimeter, whereas in the laboratory
you will need to calculate the specific heat of the metal. The calculations are similar, but the
distinction needs to be recognized.)
Heat lost by warm water:
qlost = (50.0 g)∙(4.18 J/g∙oC)∙(51.9 oC – 80.1 oC) = -5894 J
Heat gained by cool water:
qgained = (50.0 g)∙(4.18 J/g∙oC)∙(51.9 oC – 27.2 oC) = 5162 J
Heat absorbed by calorimeter:
qcal = -(-5894 J + 5162 J) = 732 J
Heat capacity of calorimeter:
C cal 
732
 29.6 J/oC
51.9  27.2
PROCEDURE:
A. Measure the Cs of a Known Sample.
1. Obtain two nested coffee-cups. Zero the balance and weigh the empty, dry cups to the
nearest milligram. Add approximately 50 mL of water to the cups and determine the mass of
water by difference. Assemble the coffee-cup calorimeter as shown in Figure 3. First, insert
the thermometer into the split stopper, and slide the stopper toward the top of the
thermometer. Next, slide the thermometer into the center hole of the calorimeter lid. Place
the ring stand clamp on the stopper, and adjust the clamp on the ring stand to where it will
hold the thermometer bulb about 1 cm from the bottom of the cup. Finally, insert the stirrer
in the second hole of the lid with the loop encircling the bulb of the thermometer. Wait
approximately 5 minutes before recording the temperature of the water to the nearest 0.1 oC.
2. Measure the mass of a weighing boat, add approximately 10 g of a known sample of metal,
and obtain the mass of the metal by difference. (The metals used should be large pieces of
metal; metal shot and foil don’t work very well.) Transfer the sample into an appropriately
sized test tube, ensuring that no sample is lost.
3. It is recommended that a trial run be performed in conjunction with your partner before data
is actually recorded for steps 4 and 5. You actions need to be swift and accurate to ensure
quality data (and results!), but safety cannot be compromised. Take the time to familiarize
yourself the procedure and determine the best way to execute it.
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split stopper
clamp
thermometer
stirrer
lid
2 nested Styrofoam cups
Figure 3. The coffee-cup calorimeter setup
4. To a 250-mL beaker, add approximately 150 mL of water and bring to a rolling (but not
violent!) boil. Once the water is boiling, immerse the test tube containing the metal into the
water using a clamp to support the test tube. After ~10 minutes in the boiling water, the
sample inside the test tube and the boiling water should be at the same temperature. Record
the temperature of the water to the nearest 0.1 oC.
5. Quickly transfer the hot metal in the test tube to the calorimeter containing the water. Take
care to not splash any water out of the calorimeter! Immediately replace the lid,
thermometer and stir wire, and record the temperature at time zero. Gently stir the water
with the wire and record the temperature in 15 second intervals until the temperature
decreases for at least five measurements. This may take a few minutes. It is extremely
important that the water be constantly stirred throughout the experiment. Finally, record
when the temperature is at a maximum, which will be immediately before the temperature
begins to descend. Return your metal sample to the waste container provided.
6. Using your data, compute the specific heat capacity, Cs, of the known sample. Also report
the percent error from the value listed in Table 1.
B. Measure the Cs of an Unknown Sample.
1. Use the same procedure for determining the specific heat of a known sample to determine the
specific heat of an unknown sample. Be sure to indicate which unknown you have.
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LAB REPORT
Specific Heat Capacity
Name ________________________________
Date _________
Partner ________________________________
Section _________
Report Grade ______
A. Specific Heat of a Known Metal.
1. Mass of the coffee-cups
____________
Mass of the cups + water
____________
Mass of the water
____________
Temperature of cool water
____________
2. Identity of metal sample
____________
Mass of metal sample
____________
Temperature of boiling water
____________
3. Time vs. temperature data:
Time Temperature Time Temperature Time Temperature
(sec)
(°C)
(sec)
(°C)
(sec)
(°C)
0
105
210
15
120
225
30
135
240
45
150
255
60
165
270
75
180
285
90
195
300
Maximum System Temperature ____________
4. Calculate the specific heat of the known and percent error . Show work!
Cs = __________ J/g∙oC
% error = ____________
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B. Specific Heat of an Unknown Metal.
1. Unknown Number
____________
Mass of the coffee-cups
____________
Mass of the cups + water
____________
Mass of the cool water
____________
Temperature of water
____________
2. Mass of unknown
Temperature of boiling water
____________
____________
3. Time vs. temperature data:
Time Temperature Time Temperature Time Temperature
(sec)
(°C)
(sec)
(°C)
(sec)
(°C)
0
105
210
15
120
225
30
135
240
45
150
255
60
165
270
75
180
285
90
195
300
Maximum System Temperature ____________
4. Calculate the specific heat of the unknown and identify it. Show work!
Cs = __________ J/g∙oC
Identity = ____________