In this article, we shall study the concept of energy, types of mechanical energies, and the law of conservation of energy

Energy:

Different types of energy are mechanical energy, sound energy, heat energy, light energy, chemical energy, electrical energy, atomic energy, nuclear energy. Mechanical energy is further classified into kinetic energy and potential energy.

Kinetic Energy:

The energy possessed by the body on account of its motion is called kinetic energy. e.g Energy possessed by flowing water and wind, moving bicycle

Consider a body of mass ‘m’ lying on the smooth horizontal surface, is acted upon by a constant force of magnitude ‘F’ which displaces it through a distance ‘s’ in its own direction. Then the work done by the force is given by

W  =  F .  s     ……….. (1)

By Newton’s second law of motion

F  =  m . a     ……….. (2)

Where ‘a’ is the magnitude of the acceleration in the body.

From equations  (1) and (2)

∴  W  =  m a s  …………… (3)

By equation of motion we have

v² = u²   +  2as

Where u  = magnitude of the initial velocity. In this case u = 0

v =  magnitude of final velocity after covering the distance ‘s’

∴  v² =  2 a s

∴ as =  v²/2

Substituting in equation (3) we get

∴  W  =  mv²/2

∴  W  =  ½mv²

This work is stored as kinetic energy in the body. Thus the kinetic energy of the body is given by

K.E. = ½mv²

This is an expression of the kinetic energy of a body.

Potential Energy:

The energy possessed by a body or a system on account of its position and configuration is called potential energy. e.g. energy possessed by water stored in a dam, in wound spring of a watch

Suppose that body of mass ‘m’ be raised to some height say ‘h’ against the gravitational force which is equal to the weight of the body ‘mg’. Where ‘g’ is an acceleration due to gravity.

As the applied force and the displacement of the body are in the same direction.

Work = Force × Displacement

W = mg × h

∴  W = mgh

This work is stored as the potential energy in the body.

∴  P.E. = mgh

This is an expression for the gravitational potential energy of a body, raised to some height above the earth’s surface.

Units and Dimensions of Energy and that of Work are the Same:

The capacity of a body to do work is called energy. Hence energy is measured in terms of work. Therefore, the units and dimensions of energy and that of work are the same.

kilowatt-hour:

a kilowatt-hour is a unit of measuring energy. This unit is a general unit of energy consumption bills (Electricity bills)

Now Work = Power x time

Hence, 1 kilowatt hour= 1 kilowatt × 1 hour

If the power of 1 kilowatt is used for 1 hour, the work done or energy consumed is said to be 1 kilowatt hour.

1 kWh   = 1kW x 1 hour

= 1000 W x 60 x 60 sec

= 1000 J/s x 3600 s

= 3600000 J

= 3.6 x 10⁶ J

Kinetic Energy is Always Positive:

The kinetic energy of a body is given by the expression. K.E. = ½mv²

The right-hand side contains the term mass ‘m’ which is always positive and a term square of velocity which is also positive. Thus the right-hand side of the expression is always positive. Thus kinetic energy is always positive.

Principle of Conservation of Energy:

The energy cannot be created nor it can be destroyed but can be converted from one form to another. Thus the total energy of the isolated system remains the same.

Energy can be converted from one form to another Examples

• In an electrical bulb, electrical energy is converted into light energy and heat energy.
• When the hammer strikes the nail mechanical energy gets converted into sound energy and heat energy.

Working of Hydroelectric Power Station :

The principle of conservation of energy can be explained by the example of a hydroelectric power station.

Water is stored in the artificial reservoirs created in the mountains by constructing a dam across the river. Thus the kinetic energy of flowing water is converted into potential energy of stored water. This stored water is brought downhill i.e. at the foot of the mountain through pipes. This water is then directed on blades of the wheel of the turbine. Thus the kinetic energy of water is used to rotate the coil in the turbine. Due to rotation of the coil in the magnetic field the kinetic energy gets converted into electrical energy. This energy can further be converted into different forms of energy like sound, heat, light, magnetism, etc.

Principle of Conservation of Mass:

The mass cannot be created nor it can be destroyed but can be converted from one form to another. Thus the total mass of isolated system remains the same.

Einstein’s Mass-Energy Relation:

According to Albert Einstein, the mass and energy are interconvertible and the equivalence between them is given by the relation

E  =  m c²

Where   E = amount of energy

M = Mass

c = speed of light in vacuum

This relation is known as Einstein’s mass-energy relation.

Thus mass and energy are not two different physical quantity or the mass is a form energy.

Examples of Mass-Energy Interconversion:

Phenomenon of pair production :

In the phenomenon of pair production, the energy of gamma rays photons is converted under proper conditions, into a positron-electron pair. Thus here energy gets converted into mass.

Phenomenon of pair annihilation:

In the phenomenon of pair annihilation, a positron and electron under proper conditions combine to form the gamma-ray photon. Thus the particles (mass) are converted into energy.

Note:

Positrons and electrons both are similar particles having the same mass only difference is their charges. Positrons are positively charged while electrons are negatively charged.

Modified Law of Conservation of Mass and Energy:

The total amount of mass and energy in the universe is always constant.

Einstein’s Formula for the Variation of Mass with Velocity:

When the velocity of light is comparable with that of light, then, the mass of the particle in motion is given by

Where m_o = mass of a body at rest.

m = mass of a body when moving with a velocity ‘ v ’

c = velocity of light in vacuum.

This relation is known as Einstein’s formula for the variation of mass with velocity. This relation shows that the mass of a body increases with the increase in its velocity.

Science > Physics > Work, Power, and Energy > Conservation of Energy

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In this article, we shall study the composition of two SHM. Sometimes particle is acted upon by two or more linear SHMs. In such a case, the resultant motion of the body depends on the periods, paths and the relative phase angles of the different SHMs to which it is subjected.

Consider two SHMs having same period and parallel to each other, where a1 and a2 are amplitudes of two SHMs respectively. a1 anda2 are initial phase angle of two SHMs respectively, whose displacements are given by

x₁ = a₁ Sin (ωt + α₁)   and x₂ = a₂ Sin (ωt + α₂)

Resultant displacement of the particle subjected to above SHMs is given by

x = x₁ + x₂

∴ x  = a₁ Sin (ωt + α₁)  +  a₂ Sin (ωt + α₂)

∴   x = a₁ [Sinωt . Cosα₁ + Cosωt . Sinα₁] + a₂ [Sinωt . Cosα₂ + Cosωt . Sinα₂]

∴   x = a₁ Sinωt . Cosα₁ + a₁ Cosωt . Sinα₁ + a₂ Sinωt . Cosα₂ + a₂ Cosωt . Sinα₂

∴   x = a₁ Sinωt . Cosα₁ + a₂ Sinωt . Cosα₂ + a₁ Cosωt . Sinα₁ + a₂ Cosωt . Sinα₂

∴   x = Sinωt .(a₁  Cosα₁  + a₂ Cosα₂) + Cosωt . (a₁ Sinα₁ + a₂  Sinα₂) ………….. (1)

Let, (a₁ Cosα₁  + a₂ Cosα₂)   = R Cos δ … (2)

(a₁ Sinα₁ + a₂ Sinα₂) = R Sin δ    ……(3)

From Equations (1), (2) and (3)

x = Sin ωt (R Cos δ)   + Cos ωt (R Sin δ)

∴   x = R (Sin ωt Cos δ   + Cos ωt Sin δ)

∴ x = R Sin (ωt + δ)  ………..(4)

Equation (4) indicates that resultant motion is also a S.H.M. along the same straight line as two parent SHMs and of the same period and initial phase δ .

Squaring equations (2) and (3) and adding them

(R Cos δ)²+    (R Sin δ)² =   (a₁  Cosα₁  + a₂ Cosα₂)² +   ( a₁ Sinα₁ + a₂  Sinα₂ )²

∴   R² Cos² δ+    R² Sin² δ =    a₁² Cos² α₁ + a₂² Cos² α₂ +2 a₁ a₂ Cos α₁ Cos α₂

+ a₁² Sin² α₁ + a₂²Sin²α₂ + 2 a₁ a₂ Sin α₁ Sin α₂

∴   R² (Cos² δ + Sin² δ) =    a₁² (Cos² α₁ + Sin² α₁) + a₂² (Cos² α₂ + Sin²α₂)

+2 a₁ a₂ (Cos α₁ Cos α₂ +Sin α₁ Sin α₂)

∴   R² (1) =    a₁² (1) + a₂² (1) +2 a₁ a₂ Cos (α₁ – α₂)

∴   R² =    a₁² + a₂² +2 a₁ a₂ Cos (α₁ – α₂)

Dividing equation (3) by (2)

From Equations (6) and (7) we can find the resultant and initial phase of resultant S.H.M.

Special Cases:

Case 1: When the two SHMs are in the same phase then (α₁ – α₂) =  0

If the two SHMs have the same amplitude then, a₁ = a₂ = a

∴ R   =  a + a   =  2a

Case 2: When the two SHMs are in opposite phase then, (α₁ – α₂) = π

If the two SHMs have the same amplitudes then, a₁ = a₂ = a

R   = a – a = 0

Case 3: When the phase difference is (α₁ – α₂)   = π / 2

If the two SHMs have the same amplitude then, a₁ = a₂ = a

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Next Topic: Theory of Simple Pendulum

Science > Physics > Oscillations: Simple Harmonic Motion > Composition of Two SHM

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In this article, we shall study to solve numerical problems to calculate potential energy, kinetic energy, and total energy of particle performing S.H.M.

Example – 01:

A particle of mass 10 g performs S.H. M. of amplitude 10 cm and period 2π s. Determine its kinetic and potential energies when it is at a distance of 8 cm from its equilibrium position.

Given: Mass = m = 10 g, amplitude = a = 10 cm, Period = T = 2π s, displacement = x = 8 cm

To Find: Kinetic energy =? and Potential energy = ?

Solution:

Angular velocity ω = 2π/T = 2π/2π = 1 rad/s

Kinetic energy = 1/2 mω² (a² – x²) =1/2 x 10 x 1²(10² – 8²)

∴ Kinetic energy = 5 x (36) = 180 erg = 1.8 x 10^-5 J

Potential energy = 1/2 mω²x²

∴ Potential energy = 1/2 x 10 x 1² x 8² = 320 erg = 3.2 x 10^-5 J

Ans: Kinetic energy = 1.8 x 10^-5 J and potential energy = 3.2 x 10^-5 J

Example – 02:

A particle of mass 10 g executes linear S.H.M. of amplitude 5 cm with a period of 2 s. Find its PE and KE, 1/6 s after it has crossed the mean position.

Given: Mass = m = 10 g, amplitude = a = 5 cm, Period = T = 2 s, time elapsed = 1/6 s, particle passes through mean position, α = 0.

To Find: Kinetic energy =? and Potential energy =?

Solution:

Angular velocity ω = 2π/T = 2π/2 = π rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴ x = 5 sin (π x 1/6 + 0)

∴ x = x = 5 sin (π/6) = 5 x 1/2 = 2.5 cm

Kinetic energy = 1/2 mω² (a² – x²) =1/2 x 10 x π² (5² – 2.5²)

∴ Kinetic energy = 5 x 3.142²(25- 6.25) = 5 x 3.142²(18.75)

∴ Kinetic energy = 925.5 erg = 9.26 x 10^-5 J

Potential energy = 1/2 mω²x²

∴ Potential energy  = 1/2 x 10 x π² x 2.5² = 5 x 3.142² x 2.5²

= 308.5 erg = 3.09 x 10^-5 J

Ans: Kinetic energy = 9.26 x 10^-5 J and potential energy = 3.09 x 10^-5 J

Example – 03:

The total energy of a particle of mass 0.5 kg performing S.H.M. is 25 J. What is its speed when crossing the centre of its path?

Given: Mass = m = 0.5 kg, Total energy T.E. = 25 J

To Find: Maximum speed = v_max =?

Solution:

The speed when crossing mean position is a maximum speed

Total energy = 1/2 mω²a²

∴ 25 = 1/2 x 0.5 x ω²a²

∴ ω²a² = 25 x 2/ 0.5 = 100

∴ ωa = 10 m/s

But ωa = v_max = 10 m/s

Ans: The speed when crossing mean position is 10m/s

Example – 04:

A particle performs a linear S.H.M. of amplitude 10 cm. Find at what distance from the mean position its PE is equal to its KE.

Given: P.E. = K.E.

To Find: Distance = x=?

Solution:

P.E. = K.E.

∴ 1/2 mω²x² = 1/2 mω² (a² – x²)

∴ x² =  a² – x²

∴ 2x² = a²

∴ x = ± a/√2 = ±10/√2 = ±5√2 cm

Ans:  At a distance of 5√2 cm from either side of the mean position K.E. = P.E.

Example – 05:

Find the relation between amplitude and displacement at the instant when the K.E. of a particle performing S.H. M. is three times its P.E.

Given: K.E. = 3 x P.E.

To Find: Distance = x=?

Solution:

K.E. = 3 x P.E.

∴ 1/2 mω² (a² – x²)  = 3 x 1/2 mω²x² 

∴ a² – x² = 3x² 

∴ 4x² = a²

∴ x = ± a/2, where a = amplitude

Ans:  At a distance of a/2 cm from either side of the mean position K.E. = 3 x P.E.

Example – 06:

When is the displacement in S.H.M. one-third of the amplitude, what fraction of total energy is kinetic and what fraction is potential? At what displacement is the energy half kinetic and half potential?

Part – I:

Given: x = a/3

To Find: K.E/T.E. =? and P.E./T.E. =?

Solution:

Part – II

Given: P.E. = K.E.

To Find: Distance = x =?

Solution:

P.E. = K.E.

∴ 1/2 mω²x² = 1/2 mω²(a² – x²)

∴ x²  =  a² – x²

∴ 2x² = a²

∴ x = ± a/√2

Ans:  The fraction of K.E = 8/9, fraction of P.E. = 1/9, required displacement = ± a/√2 unit

Example – 07:

An object of mass 0.2 kg executes S.H.M. along the X-axis with a frequency of 25 Hz. At the position x = 0.04 m, the object has a K.E. of 0.5 J and P.E. of 0.4 J. Find the amplitude of its oscillations.

Given: Mass = m = 0.2 kg, frequency = n = 25 Hz, displacement = x = 0.04 m = 4 cm, K.E. = 0.5 J, P.E. = 0.4 J

To Find: Amplitude = a =?

Solution:

Angular speed ω = 2πn = 2 x π x 25 = 50π rad/s

∴   5x² = 4a² – 4x²

∴   9x² = 4a²

∴ 4a² = 9x 4² = 144

∴ a² = 36

∴ a = 6 cm

Ans: The amplitude = 6 cm

Example – 08:

The amplitude of a particle in S.H. M. is 2 cm and the total energy of its oscillation is 3 x 10^-7 J. At what distance from the mean position will the particle be acted upon by a force of 2.25 x 10^-5 N when vibrating?

Given: amplitude = a = 2 cm, Total energy = T.E. = 3 x10^-7 J = 3 x10^-7 x 10⁷ = 3 erg, Force = 2.25 x 10^-5 N = 2.25 x 10^-5 x 10⁵ = 2.25 dyne

To Find: Distance = x =?

Solution:

T.E =1/2 mω²a²

∴ 3 =1/2 mω²(2)²

∴ mω² =3/2 ………… (1)

Now Force F = mf = mω²x

∴ 2.25 = (3/2)x

∴ x = 2.25 x 2 /3 = 1.5 cm

Ans: At a distance of 1.5 cm from the mean position will the particle be acted upon by a force of 2.25 x 10^-5 N

Example – 09:

A body of mass 100 g performs S.H.M. along a path of length 20 cm and with a period of 4 s. Find the restoring force acting upon it at a displacement of 3 cm from the mean position? Find also the total energy of the body.

Given: mass = m = 20 g, Path length = 20 cm, amplitude = a = 20/2 = 10 cm, Period = T = 4s,

To Find: Restoring force = F =? Total energy = T.E. = ?

Solution:

Angular speed ω = 2π/T = 2π/4 = π/2 rad/s

Restoring force F = mf = mω²x

F = 100 x (π/2)² x 3 = 740.4 dyne = 740.4 x 10^-5 N = 7.404 x 10^-3 N

T.E. = 1/2 x 100 x (π/2)²x 10² = 1.234 x 10⁴ erg

T.E. = 1.234 x 10⁴ x 10^-7 J = 1.234 x 10^-3 J

Ans: Restoring force = 7.404 x 10^-3 N; total energy = 1.234 x 10^-3 J

Example – 10:

A particle of mass 200 g performs S.H.M. of amplitude 0.1m and period 3.14 second. Find its K.E. and P.E. when it is at a distance of 0.03 m from the mean position.

Given: mass = m = 200 g, amplitude = a = 0.1 m = 10 cm, period = T = 3.14 s, Distance = x = 0.03 m = 3 cm,

To Find: K.E. =? and P.E. = ?

Solution:

Angular speed ω = 2π/T = 2π/3.14 = 2 rad/s

Kinetic energy = 1/2 mω²(a² – x²) =1/2 x 200 x 2²(10² – 3²)

∴ Kinetic energy = 100 x 4 x (100 -9) = 3.64 x 10⁴ erg

∴ Kinetic energy = 3.64 x 10⁴ x 10^-7 J = 3.64 x 10^-3 J

Potential energy = 1/2 mω²x²

∴ Potential energy = 1/2 x 200 x 2² x 3² = 3.6 x 10³ J

∴ Potential energy = 3.6 x 10³ x 10^-7 = 3.6 x 10^-4 J

Ans: K.E. = 3.64 x 10^-3 J; P.E. = 3.6 x 10^-4 J

Previous Topic: Energy of Particle Performing S.H.M.

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Science > Physics > Oscillations: Simple Harmonic Motion > Numerical Problems on Energy of Particle Performing S.H.M.

The post Numerical Problems on Energy of Particle Performing S.H.M. appeared first on The Fact Factor.

In this article, we shall study the concept and expression of the total energy of a particle performing S.H.M. and its constituents.

Kintetic Energy of Particle Performing Linear S.H.M.:

Consider a particle of mass ‘m’ which is performing linear S.H.M. of amplitude ‘a’ along straight line AB, with the centre O.  Let the position of the particle at some instant be at C, at a distance x from O.

This is an expression for the kinetic energy of particle S.H.M.

Thus the kinetic energy of the particle performing linear S.H.M. and at a distance of x₁ from the mean position is given by

Special cases:

Case 1: Mean Position:

The kinetic energy of particle performing S.H.M. is given by

For mean position x₁ = 0

Case 2: Extreme position:

The kinetic energy of particle performing S.H.M. is given by

For mean position x₁ = a

Potential Energy of Particle Performing Linear S.H.M.:

Consider a particle of mass ‘m’ which is performing linear S.H.M. of amplitude ‘a’ along straight line AB, with the centre O.  Let the position of the particle at some instant be at C, at a distance x from O.

Particle at C is acted upon by restoring force which is given by F = – mω²x

The negative sign indicates that force is restoring force.

Let. External force F’ which is equal in magnitude and opposite to restoring force acts on the particle due to which the particle moves away from the mean position by small distance ‘dx’ as shown. Then

F’ = mω²x

Then the work done by force F’ is given by

dW =  F’ . dx

dW = mω²x dx

The work done in moving the particle from position ‘O’ to ‘C’ can be calculated by integrating the above equation

This work will be stored in the particle as potential energy

This is an expression for the potential energy of particle performing S.H.M.

Special cases:

Case 1: Mean Position:

The potential energy of particle performing S.H.M. is given by

For mean position x₁ = 0

∴ E_P = 0

Case 2: Extreme position:

The potential energy of particle performing S.H.M. is given by

For mean position x₁ = a

Total Energy of Particle Performing Linear S.H.M.:

The Kinetic energy of particle performing S.H.M. at a displacement of x₁ from the mean position is given by

The potential energy of particle performing S.H.M. at a displacement of x₁ from mean position is given by

The total energy of particle performing S.H.M. at a displacement of x₁ from the mean position is given by

Since for a given S.H.M., the mass of body m, angular speed ω and amplitude a are constant, Hence the total energy of a particle performing S.H.M. at C is constant i.e. the total energy of a linear harmonic oscillator is conserved. It is the same at all positions. The total energy of a linear harmonic oscillator is directly proportional to the square of its amplitude.

Variation of Kinetic Energy and Potential Energy in S.H.M Graphically:

Relation Between the Total Energy of particle and Frequency of S.H.M.: 

The quantities in the bracket are constant. Therefore, the total energy of a linear harmonic oscillator is directly proportional to the square of its frequency.

Relation Between the Total Energy and Period of S.H.M.: 

The quantities in the bracket are constant. Therefore, the total energy of a linear harmonic oscillator is inversely proportional to the square of its period.

Expressions for Potential Energy, Kinetic Energy and Total Energy of a Particle Performing S.H.M. in Terms of Force Constant:

Potential energy: 

The potential energy of particle performing S.H.M. is given by

This is an expression for the potential energy of particle performing S.H.M. in terms of force constant.

Kinetic energy: 

The kinetic energy of particle performing S.H.M. is given by

This is an expression for Kinetic energy of particle performing S.H.M. in terms of force constant.

Total energy: 

The total energy of particle performing S.H.M. is given by

This is an expression for the total energy of particle performing S.H.M. in terms of force constant.

Previous Topic: Graphical Representation of S.H.M.

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Science > Physics > Oscillations: Simple Harmonic Motion > The Energy of Particle Performing S.H.M.

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The internal energy can be defined as the sum of all possible kinds of energies that a system possesses. Every chemical reaction is associated with heat change. Heat is either evolved or absorbed.  This is possible only when every substance involved in chemical reaction possess certain fixed amount of energy which is called internal energy. Every substance is composed of molecules, atoms and subatomic particles.  The position and motions of the molecules, atoms and subatomic particles is the origin of internal energy.

Constituents of Internal Energy:

Kinetic or Thermal Energy:

The energy that a body possesses due to its movements is called as kinetic energy. Since kinetic energy depends upon the temperature it is called thermal energy.  Thermal energy is directly proportional to the temperature. K.E. is of three types

Translational Energy (E_trans):

The energy associated with the molecules due to the continuous, rapid, random movements along straight line path is called translational energy. Molecules of gases or liquids are in a state of constant random movement. Hence the molecules of gases and liquids have translational energy.

Vibrational Energy (E_vibr):

The energy associated with the molecules due to atomic vibrations is called vibrational energy.

There is a repulsive force between nuclei of two atoms, and electrons of two atoms.  There is an attractive force between the nucleus of one atom and electrons of other atoms and vice versa. As a result of these attractions and repulsions, atoms are in a state of to and fro movement.  Vibrations are of two types viz. stretching and bending vibration.

Rotational Energy  (E_rot):

The energy associated with the molecules by virtue of their rotations about their axes is called as rotational energy. All diatomic and polyatomic molecules rotate about an axis perpendicular to the axis of molecule.

Thus   the total kinetic energy of the system is given by

K.E.  = E_trans + E_vibr +  E_rot

Potential or Bonding Energy:

The energy associated with the body by virtue of its position is called is potential energy. Potential energy is independent of temperature.  It arises due to the bonding between atoms in a molecule. It is classified into two types.

Intermolecular Energy (E_intermole):

The amount of energy required to separate molecules from each other is called as intermolecular energy. In solids and liquids, molecules are held together by means of weak physical forces of attraction called Vander Waals forces. These forces are strong in solids hence intermolecular energy is more in solids than in liquids.

Intramolecular Energy (E_intramole):

The energy required to break the molecule into its constituent atoms is called as intramolecular energy. Atoms are held together in the molecule by certain electrostatics force of attraction called the chemical bond.  So intramolecular energy is nothing but the energy required to break chemical bonds so as to isolate constituent atoms from each other.

Thus the total potential energy of the system is given by

P.E. =   E_intermole + E_intramole.

Total Internal Energy of a System:

Internal energy is the sum of K.E. and P.E. The absolute value of it cannot be determined but the change in it can be determined.

Significance of Internal energy:

• It depends on the quantity of a substance, hence it is extensive property.
• Change in it represents the heat evolved or absorbed in a reaction at constant temperature and constant volume.
• For isothermal process change in it is zero.
• For the process involving the evolution of energy change in it is negative and for the process involving absorption of energy change in it is positive.

Characteristics of Internal energy:

• The internal energy of a system is extensive property.
• It is a state property. The change in internal energy is independent of the path followed.
• Change in it of a cyclic process is zero.

Notes:

• In practice, the absolute value of internal energy (U) is not known and cannot be measured because it is very difficult to determine accurately the most of the quantities that contribute to the internal energy of a system. But in thermodynamics, the quantity internal energy (U) is not important while the change in it (ΔU) is important.
• The quantity change in internal energy (ΔU) is associated with many other thermodynamic quantities by simple mathematical relations. Using those relations the change in internal energy (ΔU) can be determined. Such relations are ΔH = ΔU   +  PΔV  and ΔH = ΔU   + ΔnRT Hence  the change in internal energy (ΔU) can be determined.

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