# Law of Conservation of Energy – Example

## Law of Conservation of Energy

The law of conservation of energy is one of the basic laws of physics along with the conservation of mass and the conservation of momentum. The law of conservation of energy states that energy can change from one form into another, but it cannot be created or destroyed. Or the general definition is:

The total energy of an isolated system remains constant over time.

Energy can be defined as the capacity for doing work. It may exist in a variety of forms and may be transformed from one type of energy to another in hundreds of ways.

For example, burning gasoline to power cars is an energy conversion process we rely on. The chemical energy in gasoline is converted to thermal energy, which is then converted to mechanical energy that makes the car move. The mechanical energy has been converted to kinetic energy. When we use the brakes to stop a car, that kinetic energy is converted by friction back to heat, or thermal energy.

A consequence of the law of conservation of energy is that a perpetual motion machine of the first kind, which produces work without the input of energy, cannot exist.

## Example: Conservation of Mechanical Energy

Example of Conservation of Mechanical Energy - Pendulum
Assume a pendulum (ball of mass m suspended on a string of length L that we have pulled up so that the ball is a height H < L above its lowest point on the arc of its stretched string motion. The pendulum is subjected to the conservative gravitational force where frictional forces like air drag and friction at the pivot are negligible.

We release it from rest. How fast is it going at the bottom?

The pendulum reaches greatest kinetic energy and least potential energy when in the vertical position, because it will have the greatest speed and be nearest the Earth at this point. On the other hand, it will have its least kinetic energy and greatest potential energy at the extreme positions of its swing, because it has zero speed and is farthest from Earth at these points.

If the amplitude is limited to small swings, the period T of a simple pendulum, the time taken for a complete cycle, is:

where L is the length of the pendulum and g is the local acceleration of gravity. For small swings the period of swing is approximately the same for different size swings. That is, the period is independent of amplitude.

## Example: Elastic Collisions

Example: Elastic Nuclear Collision
A neutron (n) of mass 1.01 u traveling with a speed of 3.60 x 104m/s interacts with a carbon (C) nucleus (mC = 12.00 u) initially at rest in an elastic head-on collision.

What are the velocities of the neutron and carbon nucleus after the collision?

Solution:

This is an elastic head-on collision of two objects with unequal masses. We have to use the conservation laws of momentum and of kinetic energy, and apply them to our system of two particles.

We can solve this system of equation or we can use the equation derived in previous section. This equation stated that the relative speed of the two objects after the collision has the same magnitude (but opposite direction) as before the collision, no matter what the masses are.

The minus sign for v’ tells us that the neutron scatters back of the carbon nucleus, because the carbon nucleus is significantly heavier. On the other hand its speed is less than its initial speed. This process is known as the neutron moderation and it significantly depends on the mass of moderator nuclei.

## Example: Inelastic Collisions

Example: Ballistic Pendulum

A ballistic pendulum is a device for measuring the velocity of a projectile, such as a bullet. The ballistic pendulum is a kind of “transformer,” exchanging the high speed of a light object (the bullet) for the low speed of a massive object (the block). When a bullet is fired into the block, its momentum is transferred to the block. The bullet’s momentum can be determined from the amplitude of the pendulum swing.

When the bullet is embedding itself in the block, it occurs so quickly that the block does not move appreciably. The supporting strings remain nearly vertical, so negligible external horizontal force acts on the bullet–block system, and the horizontal component of momentum is conserved. Mechanical energy is not conserved during this stage, however, because a nonconservative force does work (the force of friction between bullet and block).

In the second stage, the bullet and block move together. The only forces acting on this system are gravity (a conservative force) and the string tensions (which do no work). Thus, as the block swings, mechanical energy is conserved. Momentum is not conserved during this stage, however, because there is a net external force (the forces of gravity and string tension don’t cancel when the strings are inclined).

## Equations governing the ballistic pendulum

In the first stage momentum is conserved and therefore:

where v is the initial velocity of the projectile of mass mP. v’ is the velocity of the block and embedded projectile (both of mass mP + mB) just after the collision, before they have moved significantly.

In the second stage mechanical energy is conserved. We choose y = 0 when the pendulum hangs vertically, and then y = h when the block and embedded projectile system reaches its maximum height. The system swings up and comes to rest for an instant at a height y, where its kinetic energy is zero and the potential energy is (mP + mB)gh. Thus we write the law of conservation of energy:

which is the initial velocity of the projectile and our final result.

When we use some realistic numbers:

• mP = 5 g
• mB = 2 kg
• h = 3 cm
• v = ?

then we have:

## Example: Combustion of Hydrogen

Consider the combustion of hydrogen in air. In a flame of pure hydrogen gas, burning in air, the hydrogen (H2) reacts with oxygen (O2) to form water (H2O) and releases energy.

Energetically, the process can be considered to require the energy to dissociate the H2 and O2, but then the bonding of the H2O returns the system to a bound state with negative potential. It is actually more negative than the bound states of the reactants, and the formation of the two water molecules is therefore an exothermic reaction, which releases 5.7 eV of energy.

2H2(g) + O2(g) → 2H2O(g)

The balance of energy before and after the reaction can be illustrated schematically with the state in which all atoms are free taken as the reference for energy.

## Example: Law of Conservation of Mass-Energy – Mass-Energy Equivalence

Example: Mass defect of a 63Cu
Calculate the mass defect of a 63Cu nucleus if the actual mass of 63Cu in its nuclear ground state is 62.91367 u.

63Cu nucleus has 29 protons and also has (63 – 29) 34 neutrons.

The mass of a proton is 1.00728 u and a neutron is 1.00867 u.

The combined mass is: 29 protons x (1.00728 u/proton) + 34 neutrons x (1.00867 u/neutron) = 63.50590 u

The mass defect is Δm = 63.50590 u – 62.91367 u =  0.59223 u

Convert the mass defect into energy (nuclear binding energy).

(0.59223 u/nucleus) x (1.6606 x 10-27 kg/u) = 9.8346 x 10-28 kg/nucleus

ΔE = Δmc2

ΔE = (9.8346 x 10-28 kg/nucleus) x (2.9979 x 108 m/s)2 = 8.8387 x 10-11 J/nucleus

The energy calculated in the previous example is the nuclear binding energy.  However, the nuclear binding energy may be expressed as kJ/mol (for better understanding).

Calculate the nuclear binding energy of 1 mole of 63Cu:

(8.8387 x 10-11 J/nucleus) x (1 kJ/1000 J) x (6.022 x 1023 nuclei/mol) = 5.3227 x 1010 kJ/mol of nuclei.

One mole of 63Cu (~63 grams) is bound by the nuclear binding energy (5.3227 x 1010 kJ/mol) which is equivalent to:

• 14.8 million kilowatt-hours (≈ 15 GW·h)
• 336,100 US gallons of automotive gasoline
Example: Mass defect of the reactor core
Calculate the mass defect of the 3000MWth reactor core after one year of operation.

It is known the average recoverable energy per fission is about 200 MeV, being the total energy minus the energy of the energy of antineutrinos that are radiated away.

The reaction rate per entire 3000MWth reactor core is about  9.33×1019 fissions / second.

The overall energy release in the units of joules is:

200×106 (eV) x 1.602×10-19 (J/eV) x 9.33×1019 (s-1) x 31.5×106 (seconds in year) = 9.4×1016 J/year

The mass defect is calculated as:

Δm = ΔE/c2

Δm = 9.4×1016 / (2.9979 x 108)2 = 1.046 kg

That means in a typical 3000MWth reactor core about 1 kilogram of matter is converted into pure energy.

Note that, a typical annual uranium load for a 3000MWth reactor core is about 20 tonnes of enriched uranium (i.e. about 22.7 tonnes of UO2). Entire reactor core may contain about 80 tonnes of enriched uranium.

## Mass defect directly from E=mc2

The mass defect can be calculated directly from the Einstein relationship (E = mc2) as:

Δm = ΔE/c2

Δm = 3000×106 (W = J/s) x 31.5×106 (seconds in year) / (2.9979 x 108)= 1,051 kg

## Example: Conservation of Energy in Nuclear Reactions

Example: Exothermic Reaction - DT Fusion

The DT fusion reaction of deuterium and tritium is particularly interesting because of its potential of providing energy for the future. Calculate the reaction Q-value.

3T (d, n) 4He

The atom masses of the reactants and products are:

m(3T) = 3.0160 amu

m(2D) = 2.0141 amu

m(1n) = 1.0087 amu

m(4He) = 4.0026 amu

Using the mass-energy equivalence, we get the Q-value of this reaction as:

Q = {(3.0160+2.0141) [amu] – (1.0087+4.0026) [amu]} x 931.481 [MeV/amu]

= 0.0188 x 931.481 = 17.5 MeV

Example: Endothermic Reaction - Photoneutrons
In nuclear reactors the gamma radiation plays a significant role also in reactor kinetics and in a subcriticality control. Especially in nuclear reactors with D2O moderator (CANDU reactors) or with Be reflectors (some experimental reactors). Neutrons can be produced also in (γ, n) reactions and therefore they are usually referred to as photoneutrons.

A high energy photon (gamma ray) can under certain conditions eject a neutron from a nucleus. It occurs when its energy exceeds the binding energy of the neutron in the nucleus. Most nuclei have binding energies in excess of 6 MeV, which is above the energy of most gamma rays from fission. On the other hand there are few nuclei with sufficiently low binding energy to be of practical interest. These are: 2D, 9Be6Li, 7Li and 13C. As can be seen from the table the lowest threshold have 9Be with 1.666 MeV and 2D with 2.226 MeV.

In case of deuterium, neutrons can be produced by the interaction of gamma rays (with a minimum energy of 2.22 MeV) with deuterium:

The reaction Q-value is calculated below:

The atom masses of the reactant and products are:

m(2D) = 2.01363 amu

m(1n) = 1.00866 amu

m(1H) = 1.00728 amu

Using the mass-energy equivalence, we get the Q-value of this reaction as:

Q = {2.01363 [amu] – (1.00866+1.00728) [amu]} x 931.481 [MeV/amu]

= -0.00231 x 931.481 = -2.15 MeV

References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987, ISBN: 978-0471805533
7. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
9. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.