ELECTRICAL EQUIVALENT OF HEAT Presented by: Group 2 Wednesday (10th Jan 2024) PHY 636
1. 2. To determine the Electrical Equivalent Of Heat (J) from the experiment To determine the efficiency of an Incandescent Lamp OBJECTIVES
In the context of this experiment, the phenomenon under investigation is commonly referred to as Joule Heating. The electrical equivalent of heat can be determined by the application of the principle of conservation of energy. The electrical equivalent of heat refers to the quantity of electrical energy, measured in Joules, that is equivalent to the amount of thermal energy contained in one calorie. Joule heating refers to the phenomenon in which the flow of electric current through a conductor result in the generation of thermal energy. The manifestation of thermal energy can be observed as an increase in the temperature of the conductor material, hence the word "heating" . Joule heating can be observed as a process in which electrical energy is converted into thermal energy, in accordance with the principle of energy conservation. THEORY
PROCEDURE EXPERIMENT 1
Step 1 EXPERIMENT 1 Step 2 Step 3 The room temperature is measured and recorded as Tr. The EEH jar is weighed together with the lid and its mass is recorded as Mj. The lid of the EEH jar is removed the the jar is filled with cold water until it reach 16°C. The Electrical Equivalent of Heat 0.0839 kg
The power supply is attached to the terminals of the EEH jar along with the voltmeter and ammeter. Step 4 EXPERIMENT 1 Step 5 Step 6 10 drops of India ink are added into the water. The EEH jar is placed in the styrofoam calorimeters. The Electrical Equivalent of Heat
The current and voltage is recorded. The ammeter and voltmeter are monitored throughout the experiment to make sure the values do not shift significantly. Step 7 EXPERIMENT 1 Step 8 Step 9 The thermometer is inserted through the hole at the top of the jar. The water is gently stirred. Once the temperature warms about 20°C (Ti), the power supply is turned on. Once the temperature is reached 32°C and above, the power supply is turned off and the time (tf) is recorded. The water is continued to stirred. The peak temperature (Tf) is recorded. The Electrical Equivalent of Heat
The EEH jar that filled with water is weighed and the value (Miw) is recorded. EXPERIMENT 1 Step 10 The Electrical Equivalent of Heat 0.2814 kg
RESULTS AND ANALYSIS FOR EXPERIMENT 1
EXPERIMENT 1 Temperature,T (±0.2)(°C) TIme,t (±0.01)(min) Voltage, V (±0.1)(V) Current, I (±0.01)(A) 20.0 (Ti) 0.00 (ti) 21.4 0.76 21.4 2.00 21.4 0.76 23.6 4.00 21.4 0.76 25.8 6.00 21.5 0.76 27.8 8.00 21.5 0.76 29.4 10.00 21.5 0.76 31.6 12.00 21.5 0.76 34.0 14.00 (tf) 21.5 0.76 34.8 16.00 21.5 0.76 34.8 18.00 21.5 0.76 34.8 (Tf) 20.00 21.5 0.76
EXPERIMENT 1 Tr = (26.0 ± 0.2) °C Mj = (0.0839 ± 0.0001) kg Mjw = (0.2814 ± 0.0001) kg Vave = (21.48 ± 0.01) V Iave = (0.76 ± 0.01) A ti = (0.00 ± 0.01) min tf = (14.00 ± 0.01) min Ti = (20.0 ± 0.2) °C Tf = (34.8 ± 0.2) °C DATA
EXPERIMENT 1 ∆t = tf - ti ∆t = 14.00 min - 0.00 min ∆t = 14.00 min @ 840 s E = Vave • Iave •∆t E = (21.48 V) • (0.76 A) • (840 s) E = 13.71 kJ E, THE ELECTRICAL ENERGY DELIVERED TO THE LAMP
EXPERIMENT 1 Mw = Mjw - Mj Mw = 0.2814 kg - 0.0839 kg Mw = 0.1975 kg @ 197.5 g Me = 23.0 g H = (Mw + Me)(1 cal/g °C)(Tf - Ti) H = (197.5 + 23.0)(1 cal/g °C)(34.8 °C - 20.0 °C) H = (220.5 g)(1 cal/g °C)(14.8 °C) H = 3.26 kcal H, THE ELECTRICAL HEAT TRANSFERRED TO THE WATER
EXPERIMENT 1 J = E / H J = 13.71 kJ / 3.26 kcal J = 4.21 J/cal J, THE ELECTRICAL EQUIVALENT OF HEAT
EXPERIMENT 1 (Jtheo = 4.186 J/cal) %error = [(|Experiment - Theory|) / Theory] × 100% %error = [(|4.21-4.186|) / 4.186] × 100% %error = 0.57% PERCENTAGE ERROR CALCULATION
PROCEDURE EXPERIMENT 2
The procedures in Experiment 1 is repeated for this experiment but without the use of India ink. EXPERIMENT 2 Step 1 The Electrical Equivalent of Heat 0.2840 kg
RESULTS AND ANALYSIS FOR EXPERIMENT 2
Temperature,T (±0.2)(°C) TIme,t (±0.01)(min) Voltage, V (±0.1)(V) Current, I (±0.01)(A) 20.0 (Ti) 0.00 (ti) 22.1 0.82 22.8 2.00 22.1 0.82 25.4 4.00 22.1 0.82 26.6 6.00 22.2 0.82 28.6 8.00 22.3 0.82 31.8 10.00 22.3 0.82 34.4 (Tf) 12.00 (tf) 22.3 0.82 34.2 14.00 22.3 0.82 33.0 16.00 22.3 0.82 32.6 18.00 22.3 0.82 32.6 20.00 22.3 0.82 EXPERIMENT 2
EXPERIMENT 2 Tr = (26.0 ± 0.2) °C Mj = (0.0848 ± 0.0001) kg Mjw = (0.2840 ± 0.0001) kg Vave = (22.24 ± 0.01) V Iave = (0.82 ± 0.01) A ti = (0.00 ± 0.01) min tf = (12.00 ± 0.01) min Ti = (20.0 ± 0.2) °C Tf = (34.4 ± 0.2) °C DATA
EXPERIMENT 2 ∆t = tf - ti ∆t = 12.00 min - 0.00 min ∆t = 12.00 min @ 720 s E = Vave • Iave •∆t E = (22.24 V) • (0.82 A) • (720 s) E = 13.13 kJ E, THE ELECTRICAL ENERGY DELIVERED TO THE LAMP
EXPERIMENT 2 Mw = Mjw - Mj Mw = 0.2840 kg - 0.0848 kg Mw = 0.1992 kg @ 199.2 g Me = 23.0 g H = (Mw + Me)(1 cal/g °C)(Tf - Ti) H = (192.2 + 23.0)(1 cal/g °C)(34.4 °C - 20.0 °C) H = (215.2 g)(1 cal/g °C)(14.4 °C) H = 3.10 kcal H, THE ELECTRICAL HEAT TRANSFERRED TO THE WATER
EXPERIMENT 2 J, the Electrical Equivalent of Heat from EXPERIMENT 1 J = 4.21 J/cal Hj = HJ Hj = (3.10 kcal)(4.21 J/cal) Hj = 13.05 kJ THE TOTAL OF ENERGY EMITTED AS VISIBLE LIGHT, HJ
EXPERIMENT 2 Efficiency: {(E - Hj) / E} × 100% Efficiency: {(13.13 kJ - 13.05 kJ) / 13.05} ×100% Efficiency: 0.61% EFFICIENCY CALCULATION (in term of percent)
DISCUSSION
EXPERIMENT 1 The amount of electrical energy, E delivered to the bulb is 13.71 kJ. The heat transferred, H to the water is 3.26 kcal. The Electrical Equivalent of Heat, J is 4.21 J/cal The percentage error of J after compared with the theoretical value of 4.186 J/cal is 0.57%
EXPERIMENT 2 The amount of electrical energy, E delivered to the bulb is 13.13 kJ. The heat transferred, H to the water is 3.10 kcal. The total of energy emitted as visible light, Hj is 13.05 kJ The percentage efficiency of the system is 0.61%
Errors Precautions Heat can be lost to surroundings, leading to underestimation of the heat produced Insulate the experimental setup as much as possible to minimize heat loss. The calorimeter used to measured the heat may not be perfectly insulated and some heat exchange with surroundings may occur. Use a well-insulated calorimeter and minimize the time between switching on the current and taking temperature readings. Errors in measuring voltage and current can lead to inaccuracies in the calculated power and subsequently the heat produced. Use calibrated instruments for measuring voltage and current. Take multiple readings and calculate average values to minimize random errors. ERRORS AND PRECAUTIONS
Designing and Optimizing Electrical System Conductors Engineers can use the information to minimize energy losses in electrical components and systems, leading to more efficient devices and reduced energy consumption. Can reveal how different materials respond to the flow of electrical current in terms of heat generation. This information is valuable in selecting appropriate materials for specific applications, considering factors like conductivity and resistance. IMPORTANCE OF FINDINGS
Calibration of Instruments Design and operation of heating devices The findings from experiments in this area contribute to the development of efficient and reliable heating technologies. IMPORTANCE OF FINDINGS The findings from such experiments are crucial for calibrating instruments that measure electrical parameters. Understanding the relationship between current, resistance, and heat production allows for accurate measurements and ensures the proper functioning of devices that rely on these principles.
CONCLUSION
CONCLUSION We are able to investigates Joule Heating process where Electrical energy is converted to thermal energy. The electrical equivalent of heat from this experiment is J = 4.21 J/cal The percentage error of the J’s results is 0.57% The percentage efficiency of the system is 0.61%
QUESTION ANSWER AND
Naqiuddin Arfan (2021859002) Muhammad Firdaus bin Mohd Rapi (2021858816) MEET THE GROUP Muhammad Amir Farhan bin Amiruddin (2021454544)
THANK YOU