Định luật jun len xơ cho biết điều gì? Hệ thức jun len xơ & bài tập vận dụng (vật lý 9)

In the 9th Physics program in particular and the general physics program, Jun Lenz's law is an extremely important part, requiring students to clearly understand and deeply understand all knowledge. So what is Junlen's law? Which is the most correct statement about Junlen's law? What is the application of Junlen's law? All of the above questions will be answered in detail by timhieulichsuquancaugiay.edu.vn in the following article.

Where electrical energy is transformed into heat energy

To help you better understand Jun Len X's law, timhieulichsuquancaugiay.edu.vn will provide some basic information about electricity.

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Electricity, in other words, is the energy of electric current. You need to note that electric current always carries a certain amount of energy. The reason is because electricity has the ability to perform work and change the thermal energy of objects.

It is possible to observe a number of devices that use electricity in daily life. Through observation, you will notice the heating of electrical devices when used for a long time such as computers, televisions, light bulbs, fans, etc. This is the property of electricity that can cause objects to change temperature. .

Energy is not created or destroyed, it only changes from one form to another. Electricity is similar – there will be a conversion of energy forms if too much electricity is generated at the same time. Electricity will be converted into useful or useless forms of energy.

The generated electricity is not only used entirely to operate electrical equipment, but a part of the generated electricity can also be converted into another form of energy, most typically electricity converted into heat energy. .

There are 2 cases:

Electrical energy partially converted into thermal energy: In this case, only a portion of the generated electrical energy is converted into thermal energy.

The remainder will be converted into light or mechanical energy. Manufacturers often apply this property to produce electrical devices used in daily life such as:

• Electricity is converted into heat and mechanical energy: drills, electric fans, water pumps,…

• Electricity is converted into heat and light: LED lights, fluorescent lights, filament bulbs, …

Electrical energy is completely transformed into heat energy: Unlike the case where a part of the electrical energy is converted into heat energy, when all the generated electricity is converted into heat energy, the phenomenon of electrical energy conversion will no longer occur. into forms of energy such as light or mechanical energy.

For example: Rice cooker, water heater, iron, electric water heater, etc. These items have one common characteristic: the ability to radiate a large amount of heat.

State the content of the joule len fiber law

Below is some necessary information when learning about Jun Lenz's law

The birth of Jun len fiber law

In 1841, while still an amateur researcher, Jun began the process of researching the development of electric current based on his own research results.

Besides, Jun also relied on suggestions from Faraday and many other physicists. After many years of exploration and research, physicist Jun discovered the relationship between the intensity of electric current and the heat emitted from the wire. From there, Jun's statement was born.

By 1844, physicist Lenz also began conducting a series of research experiments and formulated Lenz's law, similar to Jun's previous law. Therefore, the last law is called Jun-lenz's law.

State the law

Junlen's law is stated as follows: The amount of heat released in a wire when a current flows through it is proportional to the square of the current intensity, the resistance of the wire and the time the current flows through.

What does the joule len fiber law say?

Junlen's Law also states that electrical energy is converted into heat energy. This is one of the important phenomena in physics and is widely applied to produce necessary equipment in life.

Jun Lenz's law relation

Through the statement of Jun Lenz's law, the formula of the law is as follows:

In there:

• R: is the resistance of the conductor. The unit of R is Ohm (Ω).

• I: is the current intensity flowing through the conductor. The unit of electric current is Ampere (A).

• t: is the time that the current flows through the conductor. The unit of time is seconds (s).

• Q: is the heat emitted from the conductor. The unit of Q is the joule (J).

Note: In addition to the units mentioned above, when applying to solving calculation exercises, you also need to pay attention to the relationship between the unit Joule (J) and the unit calorie (cal). The conversion rate is as follows:

1 J = 0.24 cal ; 1 cal = 4.18 J

Application of the joule len fiber law

Joule's law is used to determine the relationship between heat output, current intensity and conductor resistance. This is the basis for producing electrical equipment that meets safety standards and prevents excessive heat causing dangerous fires and explosions during use.

In addition, based on the relation of Jun Len's law, the amount of heat energy emitted by an object can be calculated. From there, people will choose the right materials for each specific type of equipment.

Typically, devices where electrical energy is converted into thermal energy will use conductors made from Constantan or Nickel.

See more: What is the work of electric current? Calculation formulas and exercises with textbook solutions

Exercises to apply Jun Len's law

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To help students understand and remember theoretical knowledge in the best way, timhieulichsuquancaugiay.edu.vn will provide exercises that apply Jun Len's law commonly found in the 9th grade Physics curriculum.

Lesson 1: Which of the following statements is not true?

Heat is released from the wire when current flows through it:

A. Proportional to the current intensity, to the resistance of the wire and the time the current flows through.

B. Proportional to the square of the current intensity, to the resistance of the wire and to the time the current flows through.

C. Proportional to the square of the potential difference between the two ends of the wire, to the time the current flows through and inversely proportional to the wire resistance.

D. Proportional to the voltage difference between the two ends of the wire, to the current intensity and to the time the current flows through.

Answer: A. (Based on the theory and relations of Jun Lenz's law)

Lesson 2: Why does the filament of a light bulb heat up to a high temperature with the same current flowing through it, while the wire connected to the light bulb barely heats up?

Reply:

• Because the bulb filament and the connecting wire are connected in series, the current flowing through both has the same intensity. According to Jun Len's law, the heat radiated in the filament and in the connecting wire is proportional to the resistance of each wire segment.

• In addition, the filament has a large resistance, so a lot of heat is emitted, so the filament heats up to a high temperature and glows. The wire has a small resistance, so little heat is emitted and most of it is transmitted to the surrounding environment. Therefore, the extension cord will hardly heat up.

Lesson 3: An electric stove when operating normally has a resistance R = 80Ω and the current through the stove is I = 2.5A.

a) Calculate the amount of heat the stove radiates in 1 second.

b) Use an electric stove to boil 1.5 liters of water with an initial temperature of 25 degrees Celsius, the boiling time is 20 minutes. Considering that the heat provided to boil water is useful, calculate the efficiency of the stove. Given that the specific heat capacity of water is c = 4200J/kg.K.

c) Use this electric stove for 3 hours every day. Calculate the electricity bill to be paid for using that electric stove for 30 days, if the price of 1kW.h is 700 VND.

Solution:

a) The heat that the stove radiates in 1 second is:

(The heat capacity of the stove can be determined as P = 500W).

b) The heat that the stove emits in 20 minutes is:

Qtp = Q.20.60 = 500.20.60 = 600000 (J).

The amount of heat required to boil the given amount of water is:

Qi = cm(t2 – t1) = 4200.1.5.(100-25) = 472500 (J)

The efficiency of the stove is: H = Qi/Qtp = 472500/600000 = 78.75%.

c) The amount of electricity consumed by the stove in 30 days (in kWh) is:

A = Pt = 500.30.3 = 45000 (Wh) = 45 (kW.h)

The electricity bill payable for using an electric stove is:

T = 45,700 = 315,000 (VND).

Lesson 4: Two resistors R1 = R2 = 100 Ω. People connect those two resistors in two ways: in parallel and in series, then connect them to a circuit with a voltage of 100V.

a) Calculate the current through the resistors in each case.

b) Determine the heat released on each resistor in two cases within 30 minutes. Any comments on the results found?

Solution:

a) In case two resistors are connected in series: Rnt = R1 + R2 = 200 (Ω).

• The current intensity running through the set is: Int = U/Rnt = 100/200 = 0.5(A)
• In case two resistors are connected in parallel: Rss = (R1.R2) / (R1+R2) = 50 (Ω).
• The current intensity through the set is: Iss = U/Rss = 100/50 = 2 (A).

b) When two resistors are connected in series, the heat released is:

Q = I^2.Rnt.t = 0.52.200.30.60 = 90000 (J)

When two resistors are connected in parallel, the heat released is:

Q = I^2.Rss.t = 22.50.30.60 = 360000 (J)

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From the above calculations, we can see that when connecting resistors in parallel, the heat generated is 4 times greater than when connecting resistors in series.

Lesson 5: A line connecting from the city's electrical network to a family's electrical network is a copper conductor with a total length of 60m, a cross-section of 0.6mm2, and a resistivity of 1.7.10-8Ω.m. Knowing that the total capacity of the family's electrical appliances is 176W. The average daily electricity usage time is about 4 hours. Calculate:

a) Resistance of the entire line connecting the common network to that family.

b) Current intensity flowing in the wire when using the above given capacity.

c) Heat is released on this wire for 10 days.

Solution:

a) The resistance of the entire line connecting the common network to the family is:

R = ρ.l/S = (1.7.10-8.60)/(0.6.10-6) = 1.7 (Ω).

b) The current flowing through the wire when using the above conduction power is:

I = P/U=176/200 = 0.8 (A)

c) The heat released on this wire in 10 days is:

Q =I^2.Rt = 1,7.0,82.10.4.3600 = 156672 (J)

Lesson 6: Let the length of a twisted wire in an electric stove be 7m and the cross-section be 0.01mm2. Knowing that 1.1.10-6 Ω.m is the resistivity tune, ask:

a) What is the magnitude of the resistance in the twisted wire?

b) In case the stove is connected to a voltage of 220V, how much heat will the stove emit in a period of 25 minutes?

c) Assuming heat loss is ignored, in a period of 25 minutes, how much water can the stove make from a state of 250C to a state of boiling (1000C)? (Know that water has a specific heat capacity of 4200J/kgK).

Solution:

a) The resistance running in the wire has a magnitude of:

R = ρ.l/S = (1,1.10-6.7) / (0,1.10-6) = 77 (Ω).

b) In case the stove is connected to a voltage of 220V, the twisted wire has a current flowing through it with an intensity of:

I = U/R = 220/77 = 2.86 (A)

During a period of 25 minutes, the twisted wire emits an amount of heat of:

Qradiation = I^2.Rt = 2,862.77.25.60 = 944643.8 (J)

c) In order for water to go from a state of 250C to a state of 1000C, an amount of heat needs to be absorbed:

Qthu = mcΔt

According to the heat balance equation, we get the amount of heat released equal to the amount of heat absorbed

944643.8 = mcΔt = m.4200.(100 – 25)

m = 3 (kg)

=> The amount of water boiled from a temperature of 250C is 3 liters (because the density of water is 1kg/liter)

Lesson 7: People use an electric stove to boil 2 liters of water from a temperature t = 2000C. To boil that amount of water in 20 minutes, what capacity of electric stove must be used? Knowing the specific heat capacity of water c = 4.18.103 J/kg.degree, the efficiency of the stove H = 80%.

Solution:

Heat required to boil water: Q1 = mc(t2 – t1)

Useful heat provided by the stove during time t: Q2 = HPt

Where P is the capacity of the stove, H is the efficiency

We have the heat balance equation: Q1 = Q2

⇒ mc(t2 – t1) = HPt

So we must use an electric stove with a capacity of 697W.

Conclusion

Above is all the knowledge about Junlen's Law that is necessary for students in the process of studying and understanding the 9th grade Physics program. Hopefully, with the useful information that timhieulichsuquancaugiay.edu.vn has provided, you will can self-train and improve math solving skills as well as skills to analyze related phenomena in everyday life. Children can learn other lessons through the basic knowledge section

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