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Physics Factsheet
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Number 65
Storing Energy
The purpose of this Factsheet is to look at the various devices which can
store energy and investigate similarities and differences between them.
This is the sort of comparison which you might be called upon to make on
the synoptic paper for A2.
At any point in between, the total energy is a mix of G.P.E and K.E.
Assuming no energy losses, the total energy would remain the same, and
Total energy
Before studying this Factsheet you should be familiar with the following
ideas from GCSE studies, AS and other units of A2:
• Energy is defined as the capacity to do work;
• Types of energy include: thermal, light, sound, kinetic, gravitational
potential , chemical , electrical , magnetic , nuclear , elastic potential;
• Energy is transformed from one form to another by various devices
and total energy is conserved, but it is not always in a form useful to us
- efficiency is defined as the % useful energy output /total energy
input;
• Most waste energy ends up as heat, which is dissipated into the
atmosphere;
• Energy can be stored in various devices;
• A simple pendulum performs Simple Harmonic Motion, (Factsheets
20 and 54) and changes K.E, into G.P.E and back again continuously;
• A mass on a spring also performs S.H.M. and exchanges K.E. with
elastic potential energy (Factsheet 53);
• A capacitor stores charge and thus electrical energy (Factsheet 29);
• Nuclei store energy, which can be released as kinetic energy and/or
thermal energy;
• A heat engine stores heat energy, which can be converted into mechanical
work (Factsheet 44 );
• An electric field stores energy, which can do work to accelerate a
particle (Factsheet 38);
G.P.E
K.E.
the graph of energy stored would be as below:
Total energy stored = K.E + G.P.E = mg∆h + ½ mv2
Of course, in practice, some energy is dissipated as heat doing work against
air resistance, some energy is transferred through the support and a minute
amount also does work in lifting the string, so the total energy does not
remain constant but declines (a free oscillation), and in order to maintain
the oscillation, energy would need to be put in (a forced oscillation –
Factsheet 48).
Example:
A pendulum with a bob of 100g swings through a vertical height of
4cm.
a) Assuming no external energy transfers, what is the total energy
stored in the pendulum?
b) In practice, energy is transferred to the surroundings give two
mechanisms by which this transfer occurs.
The Simple Pendulum
At the bottom of swing, the speed is a maximum and the height a
minimum, so if we take the bottom of the swing as the zero of G.P.E, then
all the pendulum’s energy is K.E., which = ½ mv2, where m is the mass of
the bob and v is the maximum speed.
a) Total energy = mg∆h = 0.1 × 9.81 × 0.04 = 0.039J
b) Energy is transferred to the surrounding by doing work against air
resistance, and by being transmitted by movement of the support.
all G.P.E
h
Mass/Spring system
When a mass is placed on the end of a spring, the spring extends to an
equilibrium position, where the tension in the spring just balances the
weight of the mass. If the mass is then displaced a small distance downwards
and released, the system will oscillate, exchanging K.E. for Elastic Potential
Energy. As with the pendulum, if external energy transfers are ignored,
then the total energy of the system will remain constant and it will merely
continue to exchange K.E. for Elastic Potential Energy.
all K.E.
K.E. = ½ mv2
m = mass of bob
v = velocity
At the high points of the swing, the speed is zero, so all the energy is
G.P.E., which = mg∆h, where ∆h is the difference in height between the
lowest point of the swing and the highest points at the sides.
The Elastic Potential Energy stored in a spring = ½ kx2
where: k is the spring constant
x the displacement from the midpoint.
Thus the total energy will be ½ ka2,
where a is the initial displacement, the amplitude of the oscillation.
This also equals mg∆h.
G.P.E. = mg∆h
Again, in practice, energy will be transferred to the surroundings.
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Physics Factsheet
65-Storing energy
Capacitor
Energy stored in a heat engine
A capacitor is a device which stores charge when a potential difference is
applied across its plates. The charge transferred is proportional to the
P.D. and given by:
A heat engine is a device in which thermal energy transferred from a hot
source to a colder sink is used to do work. There is a fundamental theoretical
limit to the % of the available energy which can be transferred to work.
(The theory of this is covered in Factsheet 44).
Q = CV
Where Q = charge stored
V = P.D. applied
C = the capacitance of the capacitor (determined
by factors such as the size of the plates and
the separation between them).
The theoretical maximum efficiency of a heat engine is given by:
T
Eff = 1 – c
Th
Where Tc and Th are the Kelvin temperatures of the cold sink and
the hot source
Unless one plate is connected to the other or any charge leaks away, the
charge will remain stored.
It can be seen from consideration of the equation above that unless the cold
sink is at absolute zero, then the efficiency can never be 100%, and of
course, in practice even this theoretical limit is seldom reached.
The energy stored for a given charge at a given P.D is given by:
E = ½ QV
Examples of heat engines include a power station and a thermocouple. In
each of these the thermal energy is converted into electrical energy.
Energy stored in a capacitor= ½ QV
= ½ CV2
Q2
= ½
C
Energy stored in an electric field
If a charged particle enters an electric field, then the field can do work on
the particle and cause it to accelerate. This is used in the Linear Accelerator
(Linac). Since potential difference is the energy per unit charge, it follows
that the energy gained by a charge q, passing through a P.D. of V , is given
by q × V
This stored energy may be converted into electrical energy by discharging
the capacitor.
Energy stored in a nucleus
According to Einstein, mass and energy can be considered as manifestations
of the same thing. Masses of nuclei and particles are more conveniently
expressed in unified atomic mass units (u = 1.66 × 10-27 kg.) The mass of
a nucleus as a whole is slightly less than the combined masses of the
individual particles which make up the nucleus. The difference (the mass
defect) may be considered to be the binding energy of the nucleus. The
binding energy per nucleon gives an indication of the stability of the
nucleus.
Energy gained by a charged particle in passing through a
P.D. is given by q × V
For the electron, q is e (1.6 × 10-19C) so the energy becomes eV. This is used
as a unit of energy - the electronvolt. 1eV = 1.6 × 10-19J.
If an electron enters an electric field which is at right angles to its velocity,
then it will experience a deflecting acceleration, whilst maintaining its forward
constant velocity. This will give it a parabolic path, like a projectile under
the influence of gravitational attraction.
With a helium nucleus, the mass of the nucleons in helium atoms is less
than the mass of the same nucleons arranged in hydrogen nuclei and thus if
hydrogen nuclei can come together with sufficient energy to fuse with each
other, the resultant nucleus has a lower mass than the original and the extra
mass is emitted as energy, according to Einstein’s famous relationship:
parallel plate
path of particle stream
E = ∆mc
2
Where
E = energy released
∆m = the difference in mass between
the two arrangements of the nucleons
c = the velocity of light.
+V
_V
parallel plate
Practice Questions
In nuclear fission, the mass of the fission products is less than the mass of
the original nuclei and so again the mass difference is emitted as energy.
1. (a) What is meant by a heat engine?
(b) State and explain two examples of heat engines.
(c) (i) What is the theoretical maximum efficiency of a heat engine?
(ii) Calculate the maximum theoretical efficiency for a power
station, whose normal source temperature is 5000C.
2. An electron enters an electric field and passes through a P.D. of 2V.
What will be its increase in speed?
3. A 6µF capacitor is charged to a P.D. of 5V.
(a) What is the energy stored in the capacitor?
It is now connected to 4µF capacitor.
(b) When the system has reached steady state, what is the energy
stored?
(c) Account for the difference in stored energy.
4. A 100g mass is placed on the end of a spring, of spring constant
400Nm-1.
(a) What will be the extension of the spring?
(b) What is the G.P.E. lost by the mass?
(c) What is the elastic potential energy gained by the spring?
(d) Account for the difference between them.
Obviously this has the potential to release huge amounts of energy from a
small mass difference.
Energy released in a nuclear fusion or fission = ∆mc2
where ∆m is the difference in mass in kgs between reactant nuclei
and product nuclei and c is the velocity of light in ms-1
Example:
The mass of a helium nucleus is 6.646782 × 10-27kg. The mass of a
proton is 1.672623 × 10-27kg and that of a neutron 1.674929 × 10-27kg.
Calculate the binding energy per nucleon of the helium nucleus.
Mass of the individual particles = 2 × mp + 2 × mn = 6.695104 × 10-27kg
Mass of helium nucleus = 6.646782 × 10-27kg
Mass difference = 0.048322 × 10-27kg
Binding energy = 0.048322 × 10-27 × 9 × 1016 J = 4.34898 × 10-12J
Binding energy per nucleon = 1.087245 × 10-12J
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65- Stored energy
Physics Factsheet
Answers
Exam Workshop
This is a typical poor student’s answer to an exam question.The
comments explain what is wrong with the answers and how they can
be improved. The examiners’ mark scheme is given below.
1. (a) A heat engine is a device which transforms thermal energy into
mechanical work.
(b) A power station is an example of a heat engine. Thermal energy
passes from a hot source (high temperature steam) to a cold sink
(steam at 1000C) and some of the energy is extracted as work to
turn a turbine in a magnetic field, producing electrical energy.
A thermocouple is another example of a heat engine. Thermal energy
from a hot source (the hotter junction) passes to a cold sink (the
colder junction) and some of the energy is transformed into electrical
energy driving charges around the circuit.
(c)
(a) Explain why energy is released in nuclear fusion between
deuterium nuclei to produce helium.
(3)
The helium has less energy than the deuterium.
0/3
The candidate has not explained anything, merely made a statement
about relative energy. S/he has not demonstrated any understanding
of mass difference or Einstein’s relationship
(i) The maximum theoretical efficiency of a heat engine is
1 - Tc/Th
(b) Describe the difficulties involved in getting two hydrogen nuclei
to fuse to form a helium nucleus.
(3)
(ii) Assume that the temperature of the cold sink is 1000C (the
lowest possible temperature for steam under normal
pressure), then : efficiency = 1 – 373/773 = 52%
The nuclei need to be very close together.
2. In passing through the field, the electron will gain
K.E of 2 eV = 2 × 1.6 ×10-19J
so ½ mv2 = 3.2 × 10-19
6.4 × 10-19
v2 =
9.11 × 10-31
(m for electron = 9.11 × 10-31kg)
increase in v = 8.4 × 105ms-1
1/3
The candidate has stated only one difficulty and not expanded on
it and so can score only 1 mark.
Examiner’s answers
(a) Energy is released when deuterium nuclei fuse to form a helium nucleus,
because the mass of the helium is very slightly less than the total mass
of the deuterium nuclei. According to Einstein’s relationship, this mass
difference is released as energy. E = ∆mc2.
3. (a) Energy stored = ½ CV2
= ½ × 6 × 10-6× 25J
= 75µJ
(b) The charge stored on the 6µFcapacitor is given by: Q = CV
= 6 × 10-6× 5 = 3 × 10-5C
(b) In order to fuse, the nuclei need to come very close together. This is
difficult because on approaching they must overcome the coulomb
repulsion of the + charges. Very high temperatures are usually required
to give the nuclei sufficient speed to do this.
When the 2 capacitors are joined, this total charge will remain the
same and will flow until the voltage across the capacitors is the
same. Thus the capacitors can now be considered to be in parallel,
so the total capacitance of the system is C1 + C2 = 10µF
½ × 9 × 10-10
½ Q2
=
= 45 µJ
C
10 × 10-6
Thus the energy stored in the capacitor combination has decreased
by 30µJ. This energy has been dissipated in moving the charges
around the circuit, and so heating the wires.
Energy stored =
4. (a) The extension is given by: F = kx
0.1 × 9.81
x =
= 0.0024m
400
(b) Loss of G.P.E by the load = mg∆h
= 0.1 × 9.81 × 0.0024 = 0.024J
(c) Elastic strain energy gained by the spring
= ½ kx2 = ½ × 400 × 0.00242 = 0.0012J
(d) The elastic potential energy stored in the spring is 0.023J less than
the G.P.E. lost by the load. The other energy has been transformed
into other forms of energy i.e. K.E. as the mass falls, which is then
transformed into heat energy in doing work against air resistance.
Acknowledgements:
This Physics Factsheet was researched and written by Janice Jones.
The Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU
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