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Joule's law of Heating is the mathematical expression of the rate at which resistance in a circuit is converted from electrical energy into heat energy. As we all know, Electric current has a heating effect. Heat is generated by the collision of electrons in the current-carrying conductor. The heat that is generated due to the flow of electric current is measured in terms of joules. This relationship between the resistance in the circuit and the heat energy generated is given by Joule’s first law which is expressed as:
| Q=I2RT Where, Q → Amount of heat; I → Electric current; R → The amount of electric resistance in the conductor; T → Time |
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Key Terms: Joule’s Law, Joule’s First Law, Electrical Resistance, Electrical Energy, Heat Energy, Electrical Circuit
What is Joule’s Law of Heating?
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Joule's law is a mathematical explanation of how quickly resistance in a circuit turns electric energy into heat energy. James Prescott Joule, an English physicist, established in 1840 that the quantity of heat generated per second in a wire carrying a current is proportional to the wire's electrical resistance and the square of the current. He calculated that the heat emitted every second is equal to the absorbed electric power or the power loss.
Joule’s Law Experiment
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Joule’s First Law of Heating
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The relationship between heat created by passing an electric current through a conductor is given by the joule's first law.
Q=I2RT
Where,
- Q → Amount of heat
- I → Electric current
- R → The amount of electric resistance in the conductor
- T → Time
- When the current in the circuit and the flow of current remain constant, the quantity of heat generated is proportional to the wire's electrical resistance.
- When the electrical resistance and current supply are constant, the quantity of heat generated in a conductor carrying current is proportional to the square of the current flow through the circuit.
- When the resistance and current flow are held constant, the amount of heat created by the current flow is proportional to the time of flow.
Joule’s First Law
Also Read: Unit of Electric Charge
The mechanical equivalent of heat is the number of units of work that must be performed on a system to produce a unit quantity of heat. The mechanical equivalent of heat is determined by the amount of work performed on the system and the amount of heat produced by it.
| J = 4.2 joules/cal (and 1 joule = 107 ergs) = 1400 ft. lbs./CHU = 778 ft. lbs/B Th U |
These numbers provide precise results that are very close to the true values during heat calculations trials.
Joule’s Law Graph
Solved Example on Joule’s First Law of HeatingExample: Find the heat energy produced in a resistance of 10 Ω when 5 A current flows through it for 5 minutes. Solution: R = 10 Ω, I = 5 A, t = 5 minutes = 5 × 60 s H = I2 R t = 52 × 10 × 5 × 60 =25 × 10 × 300 =25 × 3000 =75000 J (or) 75 kJ |
Joule’s Second Law of Heating
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The internal energy of an ideal gas is independent of its volume and pressure and only depends on its temperature, according to Joule's Second Law.
Internal Energy for Ideal Gas
αT = 1 for an ideal gas characterized by proper microscopic postulates, implying that the temperature change of such an ideal gas during a Joule–Thomson expansion is zero. This theoretical result suggests that for such an ideal gas:
A perfect gas's internal energy is solely determined by its temperature (not pressure or volume).
Joule discovered this rule for real gases in his experiments, and it is known as Joule's second law.
Solved Example on Joule’s Second Law of HeatingExample: 3000 J of heat produced in 5sec in 8 Ω resistance. Calculate the potential difference? Solution: Given: H = 3000 J R = 8 Ω t = 5 sec V =? I = √H/Rt = √[3000/(8×5)] = √3000/(40) I = √75 = 8.66A From Ohm’s law V = IR = 8.66 × 8 = 69.28V V = 69.28 V Thus, the potential difference is 36.28 V. |
Also Read: Heat Engine
Applications of Joule’s Law of Heating
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The following are the applications of Joule’s law.
Electric Heating Device
The heating effect of electric current is used in some electrical equipment such as the electric iron, electric toaster, and electric heater. In many electrical devices, Nichrome (an alloy of nickel and chromium) is employed as a heating element. This is due to the following factors:
- Nichrome possesses a high level of specific resistance.
- The melting point of nichrome is extremely high.
- Nichrome is resistant to oxidation.
Electrical Heating device
Fuse Wire
Fuse wire is a lead-tin alloy with a 37 percent lead content and a 63 percent tin content. In electric circuits, the fuse wire is always connected in series. Because of its high resistance and low melting point, the fuse wire melts when a considerable amount of electric current travels through it, opening the circuit and preventing any damage to the electrical devices.
Fuse Wire
Light Bulb
Because the filament of an electric bulb has a high resistance to the flow of electric current, a lot of heat is generated. When heated to incandescence, this filament generates light. Tungsten is the most often used filament, with a melting point of 3380°C. An electric bulb's filament is encased in a glass that holds a low-pressure inert gas.
Electric Bulb
The heating effect of electrical current is also used in electric arc and electric welding.
When it comes to systems like transformers and dynamos, Joule's law heating impact of
Electric current is useless. These are devices that aid in reducing energy loss owing to the heating impact of electric current.
Joule’s law of heating is also seen in the working of water heaters, induction stoves, thermistors, etc.
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Things to Remember
- In electricity, Joule's law is a mathematical explanation of how quickly resistance in a circuit turns electric energy into heat energy.
- James Prescott Joule calculated that the heat emitted every second is equal to the absorbed electric power, or the power loss.
- The relationship between heat created by passing an electric current through a conductor is shown by the joule's first law Q=I2RT
- The mechanical equivalent of heat is the number of units of work that must be performed on a system to produce a unit quantity of heat.
- The mechanical equivalent of heat is determined by the amount of work performed on the system and the amount of heat produced by it.
- The internal energy of an ideal gas is independent of its volume and pressure and only depends on its temperature, according to Joule's second law. αT = 1 for an ideal gas characterized by proper microscopic postulates, implying that the temperature change of such an ideal gas during a Joule–Thomson expansion is zero.
Sample Questions
Ques. What are the differences between Joule and a Watt? (3 marks)
Ans. The joule is the SI unit of energy, and it measures the quantity of energy in the human body. Watt is the SI unit of power and represents the rate of change of energy.
Energy=Power ∗ Time
In this case, a unit of Joule equals a unit of watt for a unit of seconds.
Joules / time = Watts
A system's power is the rate at which it consumes energy.
1 Watt = 1 Joule per second
1000 Watt = 1 kW = 1 Joule every millisecond (kilowatt)
1000000 W = 1 MW = 1 Joule every microsecond (megawatt).
Ques. Calculate the amount of heat energy released by a resistance of 5Ω when a current of 3 A is sent through it for 2 minutes. (2 marks)
Ans. The formula for calculating the quantity of heat produced by a conductor is:
Q=I2RT
We get by substituting the numbers in the previous equation.
Q = 32 × 5 × 2 × 60
= 9 × 5 ×2 × 60
= 5400 J
Ques. For 30 minutes, a heater with a resistance of 300 is connected to the main supply. What is the heat produced in the heater if a current of 10 A is passed through the filament? (2 marks)
Ans. The following formula is used to compute the quantity of heat produced by the heater:
Q=I2RT
We get the solution, by substituting the values in the equation.
Q=102×300×30×60
= 100×300×30×60
= 54000000 J
= 54 MJ
Ques. Calculate the heat energy released by a resistance of 10Ω when a current of 5 A is passed through it for 5 minutes. (2 marks)
Ans. Given,
R = 10
I = 5
t = 5 minutes
= 5×60 seconds
H=I2RT
52×10×5×60
= 25×10×300
25×3000
= 75000 J or 75 KJ
Ques. How will you prove that the same current passes through every portion of the circuit including three series resistances connected to a battery using an experiment? (3 marks)
Ans. Assume R1, R2, and R3 are three resistances. As illustrated in the diagram below, we'll wire these resistances in series with an ammeter, a key, and a known-voltage battery.
Now we'll close and open the key to switch the position of the ammeter between R1 and R2, R2 and R3, and finally R3 and note the reading in each case. We'll discover that the readings in each scenario are identical.
Ques. For two minutes, 0.2 A current passes through a resistor with a resistance of 100 Ω. Calculate the amount of heat produced. (2 marks)
Ans. Given,
I = 0.2 A
R = 100 Ω
t = 2 minutes
= 2 x 60 = 120 seconds.
We know that,
H=I2RT
=0.22×100×60×20
=0.22×100×120
=480 J
Ques. For two minutes, 0.2 A current passes through a resistor with a resistance of 100 Ω. What will the heat be if the resistance is increased to 200 while I and t remain constant? (2 marks)
Ans. Given,
I = 0.2 A
R = 200 Ω
t = 2 minutes
= 2 x 60 = 120 seconds
We know that,
H=I2RT
=(0.2)2 x 200×60×20
=(0.2)2 x 200×120
=960 J
Ques. For two minutes, 0.2 A current passes through a resistor with a resistance of 100 Ω. What is the heat produced if the current is doubled but R and t remain constant? (2 marks)
Ans. Given,
I = 0.4 A
R = 100 Ω
t = 2 minutes
= 2 x 60 = 120 seconds
We know that,
H=I2RT
=(0.4)2 x 100×60×20
=(0.4)2 x 100×120
=1920 J
When the current is doubled, the heat produced will be 4 times higher.
Ques. When a 2-kW electric heater is turned on for 30 minutes, how much heat is produced? (2 marks)
Ans. H = Rating in kW x times of operation = Heat Produced
= 2 kW
1 kWh = 0.5 hour
= 36 ×105 joules
=860 kcal
Ques. What is the Gough-Joule effect, and how does it work? (3 marks)
Ans. When an elastic band is stretched first and then heated, it shrinks rather than expands. The Gough–Joule effect was initially noticed by John Gough in 1802, and it was further researched by Joule in the 1850s when it was termed the Gough–Joule effect.
"The Joule effect is a significant phenomenon that machine designers must take into account. Suspending a weight on a rubber band long enough to extend it at least 50% is the simplest technique to demonstrate this effect. When a stretched rubber band is heated with an infrared lamp, it does not extend as one might expect due to thermal expansion, but instead retracts and lifts the weight."
Ques. What exactly is the Joule–Thomson effect? (5 marks)
Ans. The Joule–Thomson effect (also known as the Joule–Kelvin effect or Kelvin–Joule effect) in thermodynamics describes the temperature change of a real gas or liquid (as opposed to an ideal gas) when it is forced through a valve or porous plug while remaining insulated so that no heat is lost to the environment. The throttling process, often known as the Joule–Thomson process, is the name given to this procedure.
All gases except hydrogen, helium, and neon cool when throttled through an orifice by the Joule–Thomson process at room temperature; these three gases have the same effect but at lower temperatures. The Joule–Thomson throttling process will warm most liquids, including hydraulic oils.
In refrigeration processes like liquefiers, the gas-cooling throttling method is often used. The warming effect of Joule–Thomson throttling can be utilized in hydraulics to locate internally leaky valves, as they will emit heat that can be identified with a thermocouple or thermal imaging camera.
Throttling is a process that cannot be reversed. Throttling caused by flow resistance in supply lines, heat exchangers, regenerators, and other (thermal) machine components is a source of losses that affects performance.
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