O B JEC T I V E
The primary goal of this experiment is to show how the concept of heat energy relates to electrical e lectrical energy. It furthers the understanding understanding of calorimetry calorimetry through measuring the the electrical energy and and calculating the Joule equivalent of electrical energy. I N T RO D U C T I O N
The theory of heat energy measured in quantities separately defined from the laws of mechanics and electricity and magnetism. Sir James Joule studies studies of these separate phenomena lead him to the discovery of the proportionality constant constant known as the Joule equivalent of heat, denoted by J . The Joule equivalent of heat is the amount of mechanical or electrical energy within a unit of heat energy.
(1.1)
W Q V W Q V P t t
Power is the rate of performing work and electrical power is the amount of ele ctrical energy expanded over time. Since in an electrical circuit, the energy Electrical and and
P V
I
mechanical energy are measure in units of joules, but heat energy uses the measurement units of
kilocalories.
W P t (1.2) W V I t
The change in the heat energy of a material Q is directly proportional to the change in temperature of the material Q T , which also depend on the material and its its specific heat. The transfer of electrical energy to heat energy equals
W J Q if
J Joule equivalent of heat or the machanical energy equivalent of heat J = 4186 Joule/kilocalorie
(1.3)
If a constant current flows through a resistive heating element, producing a constant maintained potential drop V across the element. This energy expanded into into heat energy will increase increase its container container and its constituents’ constituents’ temperature. Thus, the change in in heat energy of the container and water will be the sum of the heat energies of each as shown s hown in the equation below. 1
Emily Gatlin
Joule Equivalent of Electrical Energy
Q mcccT mw cw T where mw mass of wat wat er; mc mass of conta cont ainer (1.4)
VI t J mcccT mw cw T J
P R O C ED U R E
VI 1 mccc mw cw T t
APPARATUS The apparatus contains a resistive heating-coil, stirrer, and electrical connector posts, double-wall aluminum calorimeter, low voltage, high current power supply, digital voltmeter and ammeter, ele ctrical leads, digital multimeter, and Pasco® P asco® 750 Science Workshop data acquisition system with temperature sensor.
DATA ACQUISITION In order to obtain the data, the DataStudio™ software is set up to use the temperature sensor on the apparatus to collect systematically systematically the temperatures at specified time interval interval of 5-seconds. After the computer is completely set-up, the the rest of the apparatus is assembled. Water is added to the the calorimeter until it is about 2-inches away from being completely completely full. In order to lower the temperature temperature of the water, a few ice cubes are added. Once the ice ice is completely melted, the calorimeter calorimeter is carefully placed into the apparatus to ensure ensure that they heating-coil heating-coil and temperature probe do not touch. The voltage is set to a constant amount of approximately 6-volts. The voltmeter is wired directly directly to the heating heating coil assembly and is used to gain an accurate measurement of voltage between the two ends of the heatingcoil. The computer is now ready to collect the data. While the data is recorded into the computer directly from the temperature probe, the stirrer rod is constantly constantly moved up and down to stir the water as it is heated. This ensures that all the water water and its container container will come into thermal equilibrium with each other. The data acquisition stops automatically after ten-minutes. ten-minutes. The results are graphed in the the plot, temperature
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Emily Gatlin
Joule Equivalent of Electrical Energy
vs. time. The DataStudio™ software also allows the direct dire ct plotting of the temperature vs. time graph and the calculation of the slope in the form y mx b
VI
J
mccc mw cw m
T t
(1.5)
I t
The entire experiment is repeated using 8-volts from the power source instead of 6-volts. D AT A & C ALC U LAT I O N S
DATA S ET #1 Mass of container and Water
mcw 278.0grams
Mass of Container
mc 42.00 grams m w 236.0 grams
Mass of Water
cw
Specific Heat of Water
Slope of Temperature versus Time Joule Equivalent of Heat % Error
mccc mw cw
cc 0.21
J
kg oC
volts s V 6.1 volt I 4.82 C T m 0.025 6.5105 t oul es J 4792.37602853 Joule kcal oul es J 4817.36134569 Joule kcal %Erro %Errorr 14.48 .4858105236% %Erro %Errorr 15.08 .0826886214%
6.1 6.1 volts 4.8 4.82 C
kcal kcal 0..21 kg 0.236 kg 1. 1.0 kg 0.025 6.5 10 5 0.042 kg 0 C C
J
kcal
T t
J
kg
VI
J
J
kcal oC
Specific Heat of Aluminum Voltage Drop Across Heater Coil Current Flowing in Heater Coil
1.0 1.0
% Error Error =
29.40 .402 v C
0.00882 kcal o C 0.236 kcal C 0.024935 o C seconds
% Error =
29.40 .408 v C 0.2448 482 2 kca kcal oC0.02 0.024 4935 935 0.24
C
second
29.40 .408 v C 4817.36134569 0.0061045867 kcal second
J 4817.36134569 Jouleskcal
3
T heoret ical Val Val ue M easured ured Val Value ×100 Theoretical Value 4186
Joules Joules kcal 4792.37602852 kcal Joules 4186 kcal
%Erro %Errorr 14.4 14.485 858 8105 105236% 36% Joules kcal
100
Emily Gatlin
Joule Equivalent of Electrical Energy
VI
J
mccc mwcw
T t 6.1 volts 4.82 .82 C
J
kcal kcal 1.0 kg 0.025 6.5105 0.042 kg 0.21 kg 0.236 kg1. C C T heoret ical ical Value M easured ured Val Value 29.40 .402 v C % Error Error = ×100 J oC Theore heorettical Value o 0.00882 kc kcal C 0.236 kc kcal C 0.025065 seconds 4186 Jouleskcal 4817.36134569 Jouleskcal 100 % Error = 29.40 .408 v C J 4186 Jouleskcal 0.24 0.2448 482 2 kca kcal oC 0.02 0.025 5065 C second %Erro %Errorr 15.0 5.0826 826886 886214 214% 29.40 .408 v C 4792.37602853 Jouleskcal J kcal 0.0061364133 second
DATA S ET #2
Mass of Container
mcw 291.0 grams mc 42.0 grams
Mass of Water
mw 249 grams
Mass of Container and Water
Specific Heat of Water
Specific Heat of Aluminum Voltage Drop Across Heater Coil
V 8.0 volts
Current Flowing In Heater Coil
I 6.38 Amps
Slope of Temper
T 0.0416 1.0104 t
Joule equivalent of Heat % Error
4
kcal kg o C kcal cc 0.21 kg o C cw 1.0
J 4757.69224409
Joules
kcal Joules kcal
J 4759.98014611 %Err %Erro or 13.6572% %Err %Erro or 13.7119%
Emily Gatlin VI
J
mccc mw cw J
Joule Equivalent of Electrical Energy
T t
8.0 volts volt s 6.38 Amp kcal kcal 1.0 kg 0.0416 1105 0.042 kg 0.21 kg 0.249 kg1. C C 51.04 v A
J
0.00882 kcal o C 0.249 kcal C0.04161oC seconds
J
51.04 v A
% Error Error =
0.2578 782 2 kca kcal oC0.04 0.041 161 0.25
C
second
51.04 v C 4757.69224409 J 0.107278902 kcal second
mccc mwcw
%Err %Erro or 13.65 .6572%
kcal kcal 0.21 kg 0.249 kg1. 1.0 kg 0.0416 1105 0.042 kg 0. C C 51.04 v A
0.00882 kcal o C 0.249 kcal C0.04159 o C seconds
% Error Error =
J
100
8.0 volts volt s 6.38 Amp
J
Joules Joul es kcal 4757.69 kcal Joules 4186 kcal
T t
J
4186
VI
J
J
% Error = Joules kcal
Theoret oret ical Va Value M easured ured Value Value ×100 Theoretical Value
51.04 v A 0.2578 782 2 kca kcal oC0.04 0.041 159 0.25
C
second
51.04 v C 4759.98014611 0.0109887338 kcal second
J 4759.98014611
% Error =
T heoret ical Va Val ue M easured ured Val Value ×100 T heore heoretti cal Value 4186
Joules Joules kcal 4759.98 kcal Joules 4186 kcal
100
%Err %Erro or 13.71 .7119% Joules kcal
Joules kcal
The error within the experiment could stem from from multiple sources. For example, the minute minute presence of ice not completely melted might have been present when the experiment begins. This would cause a much cooler temperature than than expected at the specified specified voltages and current readings. Additionally, Additionally, the mass of the water and container container might be skewed skewed due to the presence of ice. ice. Ice is less dense than liquid water and this would cause disparity disp arity in the data used to calculate the joule equivalent of electrical energy using equation(1.5). equation(1.5). Additionally, the the water within the the calorimeter might still not possess a uniform temperature due to the stirring, s tirring, which would also produce error within the data.
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Emily Gatlin
Joule Equivalent of Electrical Energy
C O N C L U SI O N
This experiment demonstrated the relationship between the equivalence of electrical energy and heat energy using calorimetry to show show a method to measure electrical energy. Since the formation of the concept of electrical energy revolved around the principles of mechanical energy, the correlation of electrical energy to these principles remains a crucial relationship to understanding unders tanding electrical energy. Equation(1.3) shows the direct correlation correlation of these fundamental principles to each other. other. In this experiment, the joule electrical equivalent of energy is calculated using the slope of the temperature versus time curve. This plot of the temperature temperature versus time curve shows the direct direct linear relationship associated with the joule electrical electrical equivalent of energy. This correlation is is shown in Equation(1.5). Equation(1.5). Thus, this experiment uses the key concepts behind calorimetry in order to explain its correlation to the joule electrical equivalent of energy energy as seen in Equation(1.4). Equation(1.4). The error present in the the experiment still demonstrated the concepts effectively and allowed for a calculation of the joule equivalent of electrical energy.
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