Physics 222 Lab Report Joule Equivalent Of Electrical Energy

Joule Equivalent of Electrical Energy Emily A. Gatlin P a rt ne r: W h i t ne y He a s t o n D a t e P e r fo rm e d : 1 8 F e b r u a ry 20 0 9 2 5 F e b r u a ry 2 0 0 9 J o h n C a ru t h

OBJ ECTI VE The primary goal of this experiment is to show how the concept of heat energy relates to electrical energy. It furthers the understanding of calorimetry through measuring the electrical energy and calculating the Joule equivalent of electrical energy.

INTRO DUCTI ON The theory of heat energy measured in quantities separately defined from the laws of mechanics and electricity and magnetism. Sir James Joule studies of these separate phenomena lead him to the discovery of the proportionality 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 P V  t t

Power is the rate of performing work and electrical power is the amount of electrical energy expanded over time. Since in an electrical circuit, the energy Electrical and

P V I

mechanical energy are measure in units of joules, but heat energy uses the measurement units of kilocalories.

(1.2)

W  P t

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 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 heat energy will increase its container and its constituents’ temperature. Thus, the change in heat energy of the container and water will be the sum of the heat energies of each as shown in the equation below. 1

Emily Gatlin

Joule Equivalent of Electrical Energy

Q  mc cc T  mw cw T where mw  mass of water; mc  mass of container (1.4)

PROC EDURE

VI t  J mc cc T  mw cw T  VI 1 J mc cc  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, electrical leads, digital multimeter, and Pasco® 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 the temperatures at specified time interval of 5-seconds. After the computer is completely set-up, the rest of the apparatus is assembled. Water is added to the calorimeter until it is about 2-inches away from being completely full. In order to lower the temperature of the water, a few ice cubes are added. Once the ice is completely melted, the calorimeter is carefully placed into the apparatus to ensure that they 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 to the 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 moved up and down to stir the water as it is heated. This ensures that all the water and its container will come into thermal equilibrium with each other. The data acquisition stops automatically after ten-minutes. The results are graphed in the plot, temperature

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Emily Gatlin Joule Equivalent of Electrical Energy vs. time. The DataStudio™ software also allows the direct plotting of the temperature vs. time graph and the calculation of the slope in the form y  mx  b

J

m

VI

mc cc  mw cw  I t

T t

(1.5)

The entire experiment is repeated using 8-volts from the power source instead of 6-volts.

DATA & CA LCU LATI ON S DATA SET #1 Mass of container and Water

mcw  278.0grams

Mass of Container

mc  42.00 grams m w  236.0 grams

Mass of Water Specific Heat of Water

cw  1.0

Specific Heat of Aluminum

cc  0.21

kcal kg oC

oC

V  6.1 volts

Voltage Drop Across Heater Coil Current Flowing in Heater Coil Slope of Temperature versus Time

kcal kg

I  4.82 C

m

T  0.025  6.5105 t

J  4792.37602853

Joule Equivalent of Heat

Joules

kcal

J  4817.36134569 kcal %Error  14.4858105236% %Error  15.0826886214% Joules

% Error U s i n g e q u a t i o n ( 1. 5 ) J

J

J J

VI

mc cc  mw cw 

T t

6.1 volts  4.82 C

   kcal kcal  5 0.042 kg 0.21 kg  0.236 kg 1.0 kg 0.025  6.510     C C Theoretical Value Measured Value 29.402 v  C % Error = ×100

Theoretical Value

0.00882 kcal o C  0.236 kcal  C0.024935 o C seconds

% Error =

29.408 v  C

0.24482 kcal  oC0.024935

C

second



29.408 v  C J  4817.36134569 0.0061045867 kcal second J  4817.36134569

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Joules

kcal

4186

Joules

kcal  4792.37602852 4186 Joules kcal

%Error  14.4858105236% Joules

kcal

Joules

kcal

100

Emily Gatlin

J

Joule Equivalent of Electrical Energy

VI

mc cc  mw cw 

T t

6.1 volts  4.82 C

J

   kcal kcal  0.025  6.5105  0.042 kg  0.21  0.236 kg  1.0  kg kg   C C Theoretical Value Measured Value 29.402 v  C % Error = ×100 J oC Theoretical Value o  0.00882 kcal  C  0.236 kcal  C 0.025065 seconds 4186 Joules kcal  4817.36134569 Joules kcal % Error = 100 29.408 v  C Joules J 4186 kcal 0.24482 kcal  oC 0.025065  C second %Error  15.0826886214% 29.408 v  C J  4792.37602853 Joules kcal 0.0061364133 kcal second









DATA SET #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

cw  1.0

kcal kg o C

Specific Heat of Aluminum

cc  0.21

kcal kg o C

Voltage Drop Across Heater Coil Current Flowing In Heater Coil

I  6.38 Amps

Slope of Temper

T  0.0416  1.0 104 t

Joule equivalent of Heat % Error

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V  8.0 volts

J  4757.69224409

Joules

J  4759.98014611 %Error  13.6572% %Error  13.7119%

Joules

kcal

kcal

Emily Gatlin VI J T mc cc  mw cw  t

J

J J

Joule Equivalent of Electrical Energy

8.0 volts  6.38 Amp

   kcal kcal  5  0.042 kg  0.21  0.249 kg  1.0  0.0416  110  kg kg  C C  51.04 v  A

0.00882 kcal o C  0.249 kcal  C0.04161 o C seconds 51.04 v  A

% Error =

0.25782 kcal  oC0.04161  C second 

51.04 v  C J  4757.69224409 0.107278902 kcal second

J

VI

mc cc  mw cw 

T t

J

Joules

% Error = kcal

Theoretical Value Measured Value ×100 Theoretical Value 4186

kcal  4757.69 4186 Joules kcal

Joules

Joules

kcal

100

%Error  13.6572%

8.0 volts  6.38 Amp

   kcal kcal  5  0.042 kg  0.21  0.249 kg  1.0  0.0416 110  kg kg  C C  Theoretical Value Measured Value % Error = ×100 51.04 v  A Theoretical Value J 0.00882 kcal o C  0.249 kcal  C 0.04159 o C seconds 4186 Joules kcal  4759.98 Joules kcal % Error = 100 51.04 v  A 4186 Joules kcal J %Error  13.7119% 0.25782 kcal  oC 0.04159  C second





J





51.04 v  C  4759.98014611 0.0109887338 kcal second

J  4759.98014611

Joules

Joules

kcal

kcal

The error within the experiment could stem from multiple sources. For example, the minute presence of ice not completely melted might have been present when the experiment begins. This would cause a much cooler temperature than expected at the specified voltages and current readings. Additionally, the mass of the water and container might be skewed due to the presence of ice. Ice is less dense than liquid water and this would cause disparity in the data used to calculate the joule equivalent of electrical energy using equation(1.5). Additionally, the water within the calorimeter might still not possess a uniform temperature due to the stirring, which would also produce error within the data. 5

Emily Gatlin

Joule Equivalent of Electrical Energy

CONC LUSIO N This experiment demonstrated the relationship between the equivalence of electrical energy and heat energy using calorimetry to 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 electrical energy. Equation(1.3) shows the direct correlation of these fundamental principles to each 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 versus time curve shows the direct linear relationship associated with the joule electrical equivalent of energy. This correlation is shown in 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 as seen in Equation(1.4). The error present in the experiment still demonstrated the concepts effectively and allowed for a calculation of the joule equivalent of electrical energy.

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