10.3.1 The heating effect of an electric current
The heating effect
Heating effect of a current
The heating effect of a current is the rise in temperature of a conductor caused by moving electrons colliding with its ions and transferring energy to them.
- Whenever there is a current in a conductor, the conductor becomes warmer than its surroundings. This is the heating effect of an electric current.
- The effect occurs in every metal conductor: in connecting leads, in a resistor, in the element of a kettle and in the filament of a lamp. Even a thick copper lead carrying a small current warms up by a small amount.
- The energy that appears in the thermal store of the conductor is transferred from the chemical store of the cell, or from the mains supply, along the electrical working pathway.
- The heating effect is not a fault and not a special case. It is the unavoidable consequence of driving charge through a material that has resistance.
- The effect is easy to observe. A resistor in a circuit feels warm after a minute or two, the filament of a lamp reaches about 2500 ∘C2500\ ^\circ\text{C}2500 ∘C and glows white hot, and the wire inside a fuse can become hot enough to melt.
- Two quantities travel through the wire and they must be kept apart. The charge passes through the conductor and out the other side, while the energy carried by that charge is left behind in the metal.
Inside a metal conductor
Lattice
A lattice is the regular, repeating arrangement of positive ions that makes up a metal.
- A metal consists of positive ions arranged in a regular repeating pattern called a lattice. Each ion has a fixed average position in that pattern.
- Each metal atom has released one or more of its outer electrons. Those electrons are no longer attached to any one atom, so they are described as delocalised or free electrons.
- The delocalised electrons are free to move throughout the whole piece of metal. This sea of mobile electrons is what makes a metal a good conductor of electricity.
- The lattice ions are not free to move through the metal. They vibrate about their fixed positions, and the higher the temperature of the metal, the larger the amplitude of that vibration.
- With no potential difference across the metal the free electrons still move, and move fast, but their directions are random. As much charge moves one way as the other, so there is no net flow of charge and no current.
- The random speed of the electrons is very high, but the extra drift that a potential difference adds is slow, typically less than 1 mm s−11\ \text{mm s}^{-1}1 mm s−1. The heating effect comes from that slow drift, not from the random motion.
The collision mechanism
- A potential difference applied across the ends of the wire sets up an electric field along the wire, and that field exerts a force on every free electron inside it.
- The force gives each electron a steady drift in one direction along the wire, added on top of its fast random motion. This net drift of charge in one direction is the current.
- A drifting electron does not travel freely from one end of the wire to the other. It repeatedly collides with the vibrating positive ions of the lattice.
- In each collision the electron transfers some of its kinetic energy to the ion it strikes. Energy passes from the electrons to the lattice.
- The ion that receives the energy then vibrates with a larger amplitude about its fixed position, so the kinetic energy of the lattice has increased.
- The internal energy of the metal is the total kinetic and potential energy of all of its particles. More vigorous lattice vibration therefore means the internal energy of the metal has risen.
- A rise in the internal energy of the lattice appears as a rise in the temperature of the wire, so the wire becomes hotter than its surroundings.
- Because the wire is now hotter than its surroundings, energy is transferred out of it by heating: by conduction into whatever the wire touches, by convection in the air around it, and by radiation from its surface.
- The electrons lose energy at every collision, yet the current does not die away. Between collisions the potential difference does work on each electron again and accelerates it again, so the supply keeps replacing the energy the electrons hand to the lattice.
- The temperature of the wire stops rising when it reaches a steady state, in which the wire transfers energy to the surroundings at the same rate as the supply transfers energy to the wire.
- Every link in this chain depends on the one before it. No potential difference means no drift, no drift means no collisions, and no collisions means no rise in the internal energy of the lattice.
- Energy is transferred out of the chemical store of the cell by the electrical working pathway, into the internal energy of the metal lattice, and then to the surroundings by heating.
- The mechanism that does the transferring is drifting electrons colliding with the vibrating lattice ions, giving them energy at every collision.
- The charge is conserved and passes through the wire, while the energy it carries is deposited in the metal.
Resistance and dissipation
Dissipation
Dissipation is the transfer of energy to the thermal store of the surroundings, where it becomes spread out and can no longer be used usefully.
- The collisions that heat the wire are the same collisions that obstruct the flow of charge through it. Resistance and the heating effect are two views of one mechanism, not two separate pieces of physics.
- Energy transferred out of the electrical pathway into the internal energy of the lattice, and then spread into the surroundings, is said to be dissipated.
- Dissipated energy is not destroyed. It becomes shared between an enormous number of particles in the surroundings, spread so thinly that it can no longer be gathered up and used for the same job again.
- A conductor of higher resistance takes more energy from each coulomb of charge that passes through it, because that charge meets greater obstruction on its way through.
- The more strongly the lattice ions vibrate, the larger the volume they sweep out and the more often the drifting electrons run into them. This is why the resistance of a metal wire is greater when the wire is hot than when it is cold.
- A perfect conductor, one of resistance 0 Ω0\ \Omega0 Ω, would dissipate no energy at all. No ordinary metal at everyday temperatures behaves like that, so every real conductor warms up at least a little.
What makes the effect larger
- For a given wire, a larger current means more charge passes every second, so there are more electron collisions with the lattice every second and the wire heats up faster.
- For a given current, a larger resistance means each electron gives up more energy to the lattice on its journey through, so more energy is dissipated every second.
- Current has the stronger influence of the two. Doubling the current in a fixed resistance raises the rate at which energy is dissipated by considerably more than double.
- The time for which the current flows matters as well. The longer the current continues, the more energy in total is dissipated in the conductor.
- A conductor of very low resistance, such as a short thick copper lead, still dissipates energy, but so little that the temperature rise is hard to detect by touch.
- A conductor of high resistance carrying a large current can reach a very high temperature, which is why the thin filament of a lamp glows white hot while the thick leads feeding it stay cool.
- Do not say that the electrons get hot. Temperature belongs to the metal as a whole, and it is the vibration of the lattice ions that increases.
- Do not write that the electrons slow down and stop, and that this is why the wire heats up. The potential difference re-accelerates them between collisions, so a steady current continues.
- Do not say that energy is lost. Say it is dissipated to the surroundings, because the total energy is always conserved.
- Do not say that the charge is used up in the wire. The charge passes through and is conserved, and only the energy it carries is transferred to the lattice.
- Do not claim that only high-resistance components heat up. Every conductor carrying a current dissipates some energy.
- Do not describe the collisions vaguely as electrons bumping into atoms or into one another. The transfer is from the electrons to the vibrating ions of the lattice.
Explaining the heating effect
- Answer a question about why a wire gets hot as a chain of linked steps: potential difference, electron drift, collisions with the vibrating lattice ions, larger ion vibration, rise in internal energy and temperature, then transfer to the surroundings by heating.
- Use the wording the electrons collide with the vibrating ions of the lattice, because a loose statement about electrons hitting things will not gain the mark.
- Name both the store and the pathway when a question asks about the energy: the electrical working pathway fills the thermal store of the wire, and then the thermal store of the surroundings.
- Write dissipated rather than lost when describing where the energy goes.
- Always pair a change with its effect, for example a larger current gives more collisions each second and therefore a greater rate of heating.
- Keep the roles of the two moving quantities straight in an explanation. The electrons carry the charge through, and the lattice keeps the energy.
- Name the two kinds of particle in a metal that take part in the heating mechanism, and state which of the two moves through the wire.
- Describe, one step at a time, how a potential difference across a wire ends up raising the temperature of the wire.
- Explain why a steady current continues even though the electrons lose energy at every collision.
- Explain why the temperature of a current-carrying wire stops rising after a time.
- State what dissipation means and explain why dissipated energy is not described as lost.
10.3.2 Reducing energy transfer in wires
Why cables warm up
- Every connecting lead and every mains cable is made of metal, and every metal has some resistance, so charge passing along a cable always leaves a little energy behind in the cable itself.
- That energy is transferred by the drifting electrons colliding with the vibrating ions of the metal, and it ends up in the thermal store of the cable and then of the surroundings.
- Energy dissipated in a cable is wasted twice over. It never reaches the appliance, and it raises the temperature of the cable, which can damage the cable's covering.
- Cable design therefore aims to make the energy dissipated in the cable as small as is reasonably possible, and there are four separate levers available: the metal chosen, the thickness of the wire, the length of the wire, and the size of the current it carries.
- The first three levers all work by lowering the resistance of the wire. The fourth works by reducing how much charge is pushed through that resistance every second.
- None of the four levers can remove the heating completely. A real cable always dissipates some energy, and the target is to make that amount small enough to ignore.
Choosing the metal
Electrical conductor
An electrical conductor is a material containing charges that are free to move, so it allows a current to flow through it.
- A good electrical conductor holds a very large number of delocalised electrons that are free to move, so a small potential difference is enough to drive a large current through it.
- Copper is the standard choice for connecting leads and mains cables. It has a very low resistance for its size, it can be drawn into thin flexible strands, and it does not corrode quickly.
- Silver is a slightly better conductor than copper, but it is far too expensive to use for ordinary wiring, so it appears only in specialised contacts.
- Iron and steel have a much higher resistance than copper of the same length and thickness, so an iron lead dissipates far more energy for the same current in it.
- Nichrome has a deliberately high resistance and is used for heating elements, which makes it precisely the wrong metal for a connecting lead.
- Aluminium is a poorer conductor than copper but much less dense, so it is chosen where the mass of the cable matters more than the last part of its resistance.
- Changing the metal lowers the resistance without changing the size or shape of the cable at all, which is why it is usually the first design decision made.
Comparing two leads
- A copper lead of length 2.0 m2.0\ \text{m}2.0 m has a resistance of 0.030 Ω0.030\ \Omega0.030 Ω, and an iron lead of the same length and thickness has a resistance of 0.18 Ω0.18\ \Omega0.18 Ω.
- Each lead in turn carries the same current of 5.0 A5.0\ \text{A}5.0 A to the same appliance.
- At a fixed current, the energy dissipated in a lead each second is proportional to the resistance of the lead.
- The ratio of the two resistances is 0.180.030=6.0\dfrac{0.18}{0.030}=6.00.0300.18=6.0.
- The iron lead dissipates six times as much energy each second as the copper lead, which is why copper is used.
Thickness and length
- The cross-sectional area of a wire is the area of the circular face that would be exposed if the wire were cut straight across.
- A wire of larger cross-sectional area has a lower resistance, because the drifting electrons have a wider channel and more routes past the lattice ions, so a smaller fraction of them is obstructed.
- For the same metal and the same length, resistance is inversely proportional to cross-sectional area, so doubling the area halves the resistance.
- A longer wire has a higher resistance, because each electron must pass a greater number of vibrating lattice ions on its way through and so makes more collisions.
- For the same metal and the same thickness, resistance is directly proportional to length, so a 4.0 m4.0\ \text{m}4.0 m lead has twice the resistance of a 2.0 m2.0\ \text{m}2.0 m lead cut from the same reel.
- Halving the resistance of a cable halves the energy dissipated in it each second, provided the current through it is unchanged.
- This is why the flex on a kettle is short and made of thick multi-strand copper, while the hair-thin tracks inside a piece of electronics are acceptable only because the currents in them are tiny.
- Thickness and length cannot be pushed as far as a designer might like. Thick cable costs more, uses more copper and bends less easily, and a cable must be long enough to reach the socket.
Doubling the thickness
- A copper wire of length 3.0 m3.0\ \text{m}3.0 m and cross-sectional area 1.0 mm21.0\ \text{mm}^{2}1.0 mm2 has a resistance of 0.052 Ω0.052\ \Omega0.052 Ω.
- It is replaced by a wire of the same metal and the same length but of cross-sectional area 2.0 mm22.0\ \text{mm}^{2}2.0 mm2.
- Resistance is inversely proportional to cross-sectional area, so the area has doubled and the resistance is halved.
- The new resistance is 0.0522=0.026 Ω\dfrac{0.052}{2}=0.026\ \Omega20.052=0.026 Ω.
- At the same current, halving the resistance halves the energy dissipated in the wire each second.
Keeping the current small
- The energy dissipated in a cable rises steeply as the current in it rises, because a larger current means more charge passing every second and therefore more collisions with the lattice every second.
- Doubling the current in a given cable raises the rate of heating in it by considerably more than double, which makes current the most powerful of the four levers.
- A cable feeding several appliances at once carries the sum of their currents, so plugging a fan heater and a kettle into the same lead can make that lead noticeably warm.
- Thin extension leads sold for lamps and radios are built for small currents only. Running a 13 A13\ \text{A}13 A appliance through such a lead forces far more charge through its thin wires each second than they were designed for, and the lead gets hot.
- Cables are therefore marked with a current rating in amperes, and that rating must be at least as large as the current the appliance draws through them.
- Two leads of the same thickness behave quite differently in use if one carries 0.50 A0.50\ \text{A}0.50 A to a phone charger and the other carries 10 A10\ \text{A}10 A to an electric heater. The heater lead needs the thicker copper.
Insulation and cable ratings
Electrical insulator
An electrical insulator is a material with almost no free charges, so it does not allow a current to flow through it.
- An electrical insulator has almost no free charges to move, so charge cannot flow through it. PVC, rubber and dry air are good insulators.
- A cable pairs the two kinds of material deliberately: a conductor core of copper to carry the charge with as little dissipation as possible, wrapped in an insulator of coloured plastic to keep that charge inside the cable.
- The insulation does nothing to reduce the heating. The energy is dissipated in the copper, and the plastic simply surrounds it and slows the escape of that energy to the air.
- Insulation is damaged by heat. A cable that runs too hot softens, then melts its plastic, after which the conductors inside can touch each other or be touched by a person.
- An undersized cable is therefore a fire risk. The energy dissipated in its thin conductors raises the temperature of the insulation and of anything packed around the cable.
- An undersized cable has a second and quieter drawback. A larger share of the supply potential difference is used up along the cable, so less is left across the appliance and the appliance runs below its intended output.
- A well-designed cable is a compromise between four demands: low dissipation, low cost, enough flexibility to be used, and enough length to reach.
- Do not say a thicker wire has more resistance because there is more metal in it. A larger cross-sectional area gives the charge more room, so the resistance is lower.
- Do not write that copper has no resistance. Its resistance is small, not zero, and a long copper cable carrying a large current still gets warm.
- Do not answer with a vague improvement such as use a better wire. Name the change: a metal of lower resistance, a larger cross-sectional area, a shorter length, or a smaller current.
- Do not confuse the insulation with the conductor. The plastic sheath carries no current, so changing the sheath does not change the resistance of the cable.
- Do not suggest thicker insulation as a way of reducing the heating. Extra insulation traps the dissipated energy inside the cable and makes it hotter.
- Do not explain a long lead by saying the electrons get tired. The extra length means extra resistance, and so more energy dissipated in the lead itself.
Answering on cable design
- State the change and then its effect, for example use copper rather than iron, because copper has a lower resistance, so less energy is dissipated in the lead.
- Name the metal when a question asks which material to choose. Copper earns the mark where a general phrase such as a good conductor may not.
- Say larger cross-sectional area rather than bigger wire, since area is the quantity resistance depends on.
- Give genuinely different levers when two improvements are asked for. A shorter lead and a thicker lead count as two; thicker and wider count as one.
- Use directly proportional only where it is earned: resistance against length for a wire of fixed thickness qualifies, resistance against area does not.
- Finish an explanation at the energy, not at the resistance. Lower resistance is only worth marks once it is linked to less energy dissipated in the cable.
- List the four changes that reduce the energy dissipated in a connecting cable.
- Explain why a wire of larger cross-sectional area has a lower resistance.
- State why copper rather than iron is used for a mains cable.
- Explain why a thin extension lead becomes warm when a heater is plugged into it.
- Give two problems caused by fitting a cable that is too thin for the current it carries.
10.3.3 Advantages and disadvantages of the heating effect
Heating as the purpose
- Some appliances are built so that the heating effect is the entire point of the device. The working part is a heating element, a length of wire chosen to have a high resistance.
- Elements are usually made of nichrome, an alloy of nickel and chromium, because it has a high resistance, keeps its strength when red hot and does not oxidise away quickly.
- The element is fed by thick copper conductors of very low resistance, so nearly all the energy is dissipated in the element and almost none in the leads that reach it.
- The element is often wound into a tight coil, so that a great length of resistance wire fits into a small space and a large resistance sits exactly where the heat is wanted.
- In every device of this kind the energy arrives along the electrical working pathway and fills a thermal store: the water, the air, the food, or the metal of the tool itself.
- The heating effect is an advantage when the thermal store it fills is the store the user wanted filled.
- It is a disadvantage when it fills a thermal store nobody wanted, because that energy is then wasted and it also raises the temperature of components that work better cold.
- The physics is identical in the two cases. Only the purpose of the device decides which side of the balance sheet it falls on.
Devices that heat water
- An electric kettle carries a coiled element rated at about 3 kW3\ \text{kW}3 kW, either sitting in the water or hidden in the base plate below it. The element dissipates energy and the thermal store of the water fills until the water reaches 100 ∘C100\ ^\circ\text{C}100 ∘C.
- Because the element touches the water directly, energy passes into the water by conduction and is then spread through the whole kettle by convection currents, so the water heats evenly rather than only in the layer against the element.
- An immersion heater is a long sealed element lowered into a hot water tank. It fills the thermal store of the stored water slowly, so that hot water is ready hours later.
- An electric shower heats the water as it flows past the element, so the energy must be dissipated very quickly. This is why a shower is one of the highest-current appliances in a house, drawing tens of amperes and needing its own thick cable and its own circuit.
- A dishwasher and a washing machine each hide an element that fills the thermal store of the water in the drum, because grease is removed and detergents work far better hot than cold.
- In all of these the heating effect is the advantage, because the useful output of the appliance is hot water and nothing else.
Heating food and air
- A toaster holds flat nichrome ribbons behind a wire mesh. They dissipate energy until they glow dull red, and the energy reaches the bread mainly as infrared radiation, filling the thermal store of the bread.
- An electric oven ring holds a heavy element under a metal or ceramic surface. Energy passes by conduction into the pan and then into the food, so the thermal store of the food fills.
- A hair dryer sets a nichrome element in front of a small fan. The element fills the thermal store of the air, and the fan drives that hot air over wet hair so the water evaporates faster.
- A soldering iron dissipates energy in an element inside its shaft, filling the thermal store of the metal tip until the tip is hot enough to melt solder onto a joint.
- A car rear-window demister is a set of thin conducting strips baked into the glass. A current in the strips fills the thermal store of the glass, and the condensation or ice on the surface then evaporates or melts.
- An electric blanket, a heated towel rail and a fan heater all place resistance wire where the warmth is wanted, and a fan heater is the one case where the thermal store of the surroundings is the intended destination.
- A heating appliance can transfer nearly all of its input usefully, because the store the user wants filled is the very store the dissipation fills.
- Even a kettle wastes a little, since the case, the base and the escaping steam all carry energy away from the water.
Light from a filament
Filament lamp
A filament lamp is a lamp that emits light because the current heats a thin coiled metal wire until it glows.
- A filament lamp contains a very thin coiled tungsten wire. Its resistance is high and its surface small, so the energy dissipated in it raises its temperature to roughly 2500 ∘C2500\ ^\circ\text{C}2500 ∘C.
- At that temperature the filament emits enough visible light to be useful, so the heating effect is the means by which the lamp makes light at all.
- Tungsten is chosen because its melting point is about 3400 ∘C3400\ ^\circ\text{C}3400 ∘C, high enough for the wire to be run white hot without melting.
- The glass envelope is evacuated or filled with an unreactive gas such as argon, so the hot filament cannot react with oxygen and burn through.
- Most of the energy dissipated in the filament leaves as infrared radiation rather than visible light. A 60 W60\ \text{W}60 W filament lamp becomes too hot to touch, while an LED lamp of the same brightness draws only about 8 W8\ \text{W}8 W and stays cool.
- The filament lamp therefore appears on both sides of the balance sheet. The heating is essential to its operation and is also the reason it has been replaced by LEDs in almost every home.
A wire meant to melt
Fuse
A fuse is a safety component containing a thin wire that melts and breaks the circuit if the current rises above its stated value.
- A fuse is the one component in a circuit designed to destroy itself. Its wire is thin and made of a metal of low melting point, so it has a higher resistance than the cable it protects.
- In normal use the current is small enough that the fuse wire runs barely warm and nothing happens to it.
- If a fault drives a current larger than the fuse rating through it, the rate at which energy is dissipated in the thin wire rises steeply and its temperature climbs past its melting point.
- The wire melts and leaves a gap, so the circuit is broken, the current stops, and the appliance and its cable are saved from overheating.
- The fuse works only because it is the deliberately weakest link. It is built to reach its own melting point well before the insulation of the cable reaches the temperature that would damage it.
- This is the sharpest example of the effect being useful and destructive at the same moment, because here the destruction is the safety feature.
When heating is unwanted
Wasted energy
Wasted energy is energy transferred to a store that is not the one the device is designed to fill.
- In most electrical devices heat is not the intended output, so any energy that ends up in a thermal store is wasted energy.
- A phone charger feels warm in use. The energy dissipated in its windings and switching components never reaches the phone, so less of the energy drawn from the mains ends up in the chemical store of the battery.
- A laptop dissipates energy in its processor, its graphics chip and its power supply. It needs a fan and a metal heat sink to move that energy out, and the fan itself draws energy that does no computing.
- An electric motor in a drill, a washing machine or a fan dissipates energy in the resistance of its copper windings, so the windings run hot and the motor delivers less to the turning parts than the supply delivers to the motor.
- A transformer dissipates energy in the resistance of its coils, which is why a large one carries cooling fins or is immersed in oil, and why a small one in a plug-top adaptor feels warm to the hand.
- A long connecting cable dissipates energy along its whole length, energy that warms the cable instead of reaching the appliance plugged into its end.
- Unwanted heating brings four separate penalties: a smaller fraction of the input is transferred usefully, running costs rise, components age faster at high temperature, and there is a fire risk wherever the energy cannot escape.
- Semiconductors suffer first. A processor that overheats slows itself down or shuts down to protect itself, and a battery repeatedly run hot loses capacity permanently.
- Heating turns dangerous when the energy is trapped. A coiled extension lead is a hazard because the coiled turns insulate one another, so the energy dissipated in the copper cannot escape and the plastic can soften and melt at a current the same lead would carry safely when unwound.
- A radiator covered by clothes, an appliance pushed hard against a wall and a laptop used on a duvet all fail in the same way, because the surfaces that should be transferring energy to the air are blocked.
- Designers answer with fans, heat sinks, ventilation slots, cooling fins, thermal cut-outs and low-resistance conductors. All of them cost money and space, and none of them removes the dissipation entirely.
- Do not call an appliance inefficient merely because it gets hot. In a kettle or a toaster the heating is the useful output.
- Do not write that the energy disappears or is used up. Name the store it fills, and say that it is dissipated to the surroundings.
- Do not say a fuse breaks the circuit because the current is too big. Say the large current heats the thin fuse wire until it melts, and the melted wire leaves a gap.
- Do not claim a filament lamp gives out no heat, and do not claim it gives out only heat. It does both, and the infrared share is the larger.
- Do not explain the coiled extension lead by saying coiling raises the resistance. The resistance is unchanged, and the problem is that the trapped energy cannot escape.
- Do not offer more insulation as the cure for an overheating device. Overheating is cured by letting the energy out, not by holding it in.
Judging useful and wasted
- Decide first what the device is for, then say whether the thermal store that fills is the one the user wanted.
- Name the specific store, such as the thermal store of the water in a kettle, instead of writing heat energy.
- Give a named device when an example is asked for. A kettle, a toaster or a fuse scores where a general phrase about appliances does not.
- Pair every disadvantage with its consequence, for example hotter windings mean a shorter working life, or a smaller share of the input transferred usefully.
- Describe a fuse operating as an ordered sequence: fault, large current, thin wire heats, wire melts, circuit breaks.
- Reserve the word wasted for energy filling a store nobody wanted, and use dissipated for the transfer itself.
- Name four appliances in which the heating effect is the useful output, and state the store that fills in each.
- Explain why a heating element is made of nichrome while its supply leads are made of copper.
- Explain why the heating effect in a filament lamp is both necessary and wasteful.
- Describe, step by step, how a fuse uses the heating effect to protect a circuit.
- Give three problems that unwanted heating causes in a laptop or an electric motor.
10.3.4 Energy transferred by an electrical circuit
Energy, power and time
Joule
One joule is the energy transferred when a charge of one coulomb moves through a potential difference of one volt.
- Energy transferred by an electrical circuit is measured in joules, and the symbol for the unit is J\text{J}J.
- The first of the two energy equations is E=PtE=PtE=Pt, where EEE is the energy transferred in joules, PPP is the power in watts and ttt is the time in seconds.
- Both quantities on the right must already be in those units. A power quoted in kilowatts is converted to watts, and a time quoted in minutes or hours is converted to seconds, before either is substituted.
- The equation rearranges to P=EtP=\dfrac{E}{t}P=tE and t=EPt=\dfrac{E}{P}t=PE.
- Read directly from the equation, a device running at 1 W1\ \text{W}1 W for 1 s1\ \text{s}1 s transfers 1 J1\ \text{J}1 J.
- The joule is a small unit on household scales. Boiling a full kettle transfers a few hundred thousand joules, so answers are often written in standard form or in kilojoules, where 1 kJ=1000 J1\ \text{kJ}=1000\ \text{J}1 kJ=1000 J.
- The value calculated is the total energy transferred out of the supply and into the component during that time, so it is always positive and it always grows as the time grows.
Energy from power and time
- A kettle of power 2.2 kW2.2\ \text{kW}2.2 kW is switched on for 3.03.03.0 minutes.
- Converting the power gives P=2.2×1000=2200 WP=2.2\times1000=2200\ \text{W}P=2.2×1000=2200 W.
- Converting the time gives t=3.0×60=180 st=3.0\times60=180\ \text{s}t=3.0×60=180 s.
- The equation is E=PtE=PtE=Pt.
- Substituting gives E=2200×180=396 000 JE=2200\times180=396\,000\ \text{J}E=2200×180=396000 J.
- The energy transferred is 3.96×105 J3.96\times10^{5}\ \text{J}3.96×105 J, which is 396 kJ396\ \text{kJ}396 kJ.
Building the energy equation
- The second equation is built from two results that are already available, so it does not have to be memorised on its own.
- The charge that passes a point in a circuit is given by Q=ItQ=ItQ=It, with QQQ in coulombs, III in amperes and ttt in seconds.
- The energy that charge transfers is given by E=QVE=QVE=QV, where VVV is the potential difference in volts across the component the charge passes through.
- Substituting the first result into the second replaces QQQ with ItItIt, which gives E=(It)VE=(It)VE=(It)V.
- Written in the usual order this is E=IVtE=IVtE=IVt, with EEE in joules, III in amperes, VVV in volts and ttt in seconds.
- The chain reads as a sentence. The current and the time together fix how much charge passes, and the potential difference fixes how much energy each coulomb of that charge delivers.
- Both equations give the same energy for the same component over the same time, so E=PtE=PtE=Pt and E=IVtE=IVtE=IVt must agree.
- The data decide which equation to use. A power and a time point to E=PtE=PtE=Pt, while a current, a potential difference and a time point to E=IVtE=IVtE=IVt.
Using the equations
- The reliable method has four stages: write the equation, convert every quantity into its base unit, substitute, then state the answer with its unit.
- Time conversions cost more marks than anything else in this work. 1 minute=60 s1\ \text{minute}=60\ \text{s}1 minute=60 s, 1 hour=3600 s1\ \text{hour}=3600\ \text{s}1 hour=3600 s and 2.5 hours=9000 s2.5\ \text{hours}=9000\ \text{s}2.5 hours=9000 s.
- Power conversions follow the same pattern, since 1 kW=1000 W1\ \text{kW}=1000\ \text{W}1 kW=1000 W, so 2.2 kW=2200 W2.2\ \text{kW}=2200\ \text{W}2.2 kW=2200 W and 0.80 kW=800 W0.80\ \text{kW}=800\ \text{W}0.80 kW=800 W.
- A question that supplies a charge rather than a current is answered through E=QVE=QVE=QV instead, and any charge in millicoulombs is converted first, since 1 mC=1×10−3 C1\ \text{mC}=1\times10^{-3}\ \text{C}1 mC=1×10−3 C.
- Large answers are normally given in standard form to two or three significant figures, matching the precision of the least precise value in the question.
- A size check catches most slips. A mains appliance running for a few minutes transfers of the order of 105 J10^{5}\ \text{J}105 J, so an answer of a few hundred joules for a kettle points to a missed time conversion.
Energy from current and voltage
- A lamp on a 230 V230\ \text{V}230 V supply carries a current of 0.26 A0.26\ \text{A}0.26 A and is left on for 2.02.02.0 hours.
- Converting the time gives t=2.0×3600=7200 st=2.0\times3600=7200\ \text{s}t=2.0×3600=7200 s.
- The equation is E=IVtE=IVtE=IVt.
- Substituting gives E=0.26×230×7200E=0.26\times230\times7200E=0.26×230×7200.
- Taking the first product, 0.26×230=59.80.26\times230=59.80.26×230=59.8, and then 59.8×7200=430 56059.8\times7200=430\,56059.8×7200=430560.
- The energy transferred is 4.3×105 J4.3\times10^{5}\ \text{J}4.3×105 J to two significant figures.
Finding a missing quantity
- Either equation can be rearranged to make any one of its quantities the subject, and the energy is then the value that is given rather than the value that is wanted.
- From E=PtE=PtE=Pt the rearrangements are P=EtP=\dfrac{E}{t}P=tE and t=EPt=\dfrac{E}{P}t=PE.
- From E=IVtE=IVtE=IVt the rearrangements are I=EVtI=\dfrac{E}{Vt}I=VtE, V=EItV=\dfrac{E}{It}V=ItE and t=EIVt=\dfrac{E}{IV}t=IVE.
- The two quantities that are not wanted are multiplied together first and the energy is then divided by that product, which keeps the arithmetic to a single division.
- A question often hides the time in another unit, so a running time of 2.52.52.5 minutes must become 150 s150\ \text{s}150 s before it is used.
- The answer carries the unit of the quantity that was found: A\text{A}A for a current, V\text{V}V for a potential difference, W\text{W}W for a power and s\text{s}s for a time.
Finding the current
- A motor connected to a 12 V12\ \text{V}12 V supply transfers 9.0×103 J9.0\times10^{3}\ \text{J}9.0×103 J in 2.52.52.5 minutes.
- Converting the time gives t=2.5×60=150 st=2.5\times60=150\ \text{s}t=2.5×60=150 s.
- The equation is E=IVtE=IVtE=IVt, which rearranges to I=EVtI=\dfrac{E}{Vt}I=VtE.
- The denominator is Vt=12×150=1800Vt=12\times150=1800Vt=12×150=1800.
- Substituting gives I=90001800=5.0 AI=\dfrac{9000}{1800}=5.0\ \text{A}I=18009000=5.0 A.
Comparing two appliances
- A comparison of two devices means something only if the same time is used for both, so the first step is to fix that time and convert it to seconds once.
- With E=PtE=PtE=Pt and the same ttt for each device, the energy transferred is directly proportional to the power, so a device of ten times the power transfers ten times the energy.
- With E=IVtE=IVtE=IVt and two devices on the same mains supply, VVV and ttt are the same for both, so the energy transferred is directly proportional to the current each device draws.
- A ratio settles such a question faster than two full calculations, although the data for each device must still be checked for the units they are quoted in.
- A low-power device left on for a long time can transfer more energy in total than a high-power device used briefly, because the time enters the equation exactly as the power does.
Two devices, same time
- A hair dryer of power 1.5 kW1.5\ \text{kW}1.5 kW and a lamp of power 60 W60\ \text{W}60 W each run for 101010 minutes.
- The shared time is t=10×60=600 st=10\times60=600\ \text{s}t=10×60=600 s, and the dryer's power is 1500 W1500\ \text{W}1500 W.
- For the dryer, E=Pt=1500×600=9.0×105 JE=Pt=1500\times600=9.0\times10^{5}\ \text{J}E=Pt=1500×600=9.0×105 J.
- For the lamp, E=Pt=60×600=3.6×104 JE=Pt=60\times600=3.6\times10^{4}\ \text{J}E=Pt=60×600=3.6×104 J.
- The ratio is 9.0×1053.6×104=25\dfrac{9.0\times10^{5}}{3.6\times10^{4}}=253.6×1049.0×105=25.
- The dryer transfers 252525 times as much energy as the lamp in the same time.
- Do not substitute a time in minutes or hours into either equation. Convert it to seconds first, every time.
- Do not substitute a power in kilowatts. Convert to watts, since the joule pairs with the watt and the second.
- Do not pair a current from one component with the potential difference across a different one. Both values must belong to the same component.
- Do not mix the two equations in one substitution by using a power and a current together. Each equation is complete as it stands.
- Do not leave an energy without its unit. The unit here is always the joule.
- Do not round an intermediate value heavily. Carry the extra figures through and round only the final answer.
Setting out the calculation
- Quote the equation, E=PtE=PtE=Pt or E=IVtE=IVtE=IVt, before substituting any numbers.
- Show each unit conversion on its own line, such as t=4.0×60=240 st=4.0\times60=240\ \text{s}t=4.0×60=240 s, so a later slip still leaves the method visible.
- Rearrange first and substitute second when a missing quantity is wanted, keeping the two steps apart.
- Choose the equation from the data given rather than from habit, and say which one is being used.
- Give the answer with its unit and to a sensible number of significant figures, using standard form above about 104 J10^{4}\ \text{J}104 J.
- Use directly proportional only where one quantity is held fixed, for example energy against time at constant power.
- State the equation linking energy, power and time, and give the unit of each quantity in it.
- Build E=IVtE=IVtE=IVt from Q=ItQ=ItQ=It and E=QVE=QVE=QV.
- Calculate the energy transferred by a 1.8 kW1.8\ \text{kW}1.8 kW heater in 5.05.05.0 minutes.
- Find the current in a 12 V12\ \text{V}12 V lamp that transfers 1440 J1440\ \text{J}1440 J in 60 s60\ \text{s}60 s.
- Explain how a 60 W60\ \text{W}60 W lamp left on all day can transfer more energy than a 2 kW2\ \text{kW}2 kW kettle used twice.
10.3.5 Electrical power
Power and the watt
Power
Power is the energy transferred per second.
Watt
One watt is one joule of energy transferred per second, so 1 W=1 J s−11\ \text{W}=1\ \text{J s}^{-1}1 W=1 J s−1.
- The power of a component is the rate at which it transfers energy, which is the energy it transfers every second.
- The unit of power is the watt, symbol W\text{W}W. A power of 1 W1\ \text{W}1 W means an energy transfer of 1 J1\ \text{J}1 J every second, so 1 W=1 J s−11\ \text{W}=1\ \text{J s}^{-1}1 W=1 J s−1.
- A component of power 50 W50\ \text{W}50 W therefore takes 50 J50\ \text{J}50 J out of the electrical pathway during every second it is switched on.
- Power says nothing at all about how long a device runs for. It is a rate, so a large power sustained for a short time and a small power sustained for a long time can transfer the same total energy.
- The usual prefixes apply, with 1 kW=1000 W1\ \text{kW}=1000\ \text{W}1 kW=1000 W and 1 MW=1×106 W1\ \text{MW}=1\times10^{6}\ \text{W}1 MW=1×106 W, so 0.75 kW=750 W0.75\ \text{kW}=750\ \text{W}0.75 kW=750 W and 2400 W=2.4 kW2400\ \text{W}=2.4\ \text{kW}2400 W=2.4 kW.
- Values in a laboratory circuit run from about 0.10 W0.10\ \text{W}0.10 W for a small LED to a few watts for a filament lamp, while a mains heating element is measured in kilowatts.
- In a circuit the energy is carried by the moving charge, so the power of a component depends on two things: how much charge passes through it each second, and how much energy each coulomb of that charge delivers.
Power from current and voltage
- Each coulomb of charge passing through a component delivers a number of joules equal to the potential difference across it, measured in volts.
- The current states how many coulombs pass through the component every second, measured in amperes.
- Multiplying coulombs per second by joules per coulomb leaves joules per second, which is the power, so P=IVP=IVP=IV.
- The same result follows from the energy relationship already established, because dividing the energy transferred by the time it took removes the time and leaves P=IVP=IVP=IV.
- In this equation PPP is the power in watts, III is the current in amperes and VVV is the potential difference in volts across the component.
- The rearrangements are I=PVI=\dfrac{P}{V}I=VP and V=PIV=\dfrac{P}{I}V=IP.
- The current and the potential difference must both belong to the same component, because the equation gives the power of whatever component those two readings describe.
Power of a lamp
- A lamp carries a current of 0.35 A0.35\ \text{A}0.35 A when the potential difference across it is 230 V230\ \text{V}230 V.
- The equation is P=IVP=IVP=IV.
- Substituting gives P=0.35×230P=0.35\times230P=0.35×230.
- The product is 80.5 W80.5\ \text{W}80.5 W.
- The power of the lamp is 81 W81\ \text{W}81 W to two significant figures.
Finding a current
- A motor of power 36 W36\ \text{W}36 W runs from a 12 V12\ \text{V}12 V supply.
- The equation P=IVP=IVP=IV rearranges to I=PVI=\dfrac{P}{V}I=VP.
- Substituting gives I=3612I=\dfrac{36}{12}I=1236.
- The current in the motor is 3.0 A3.0\ \text{A}3.0 A.
Power in a resistance
- When the resistance of a component is known but the potential difference across it is not, a second form of the power equation is more direct.
- The potential difference across a resistance RRR carrying a current III is given by V=IRV=IRV=IR.
- Substituting that expression for VVV into P=IVP=IVP=IV gives P=I×IRP=I\times IRP=I×IR.
- Collecting the two current terms gives P=I2RP=I^{2}RP=I2R, where PPP is in watts, III is in amperes and RRR is in ohms.
- Only the current is squared in this equation. The resistance is not squared, and the current must be squared before it is multiplied by the resistance.
- The rearrangements are R=PI2R=\dfrac{P}{I^{2}}R=I2P and I=PRI=\sqrt{\dfrac{P}{R}}I=RP.
- This form is the natural one for a resistor, a heating element or a length of cable, because for those components it is the resistance that is quoted rather than the potential difference across them.
Power dissipated in a resistor
- A resistor of resistance 15 Ω15\ \Omega15 Ω carries a current of 0.40 A0.40\ \text{A}0.40 A.
- The equation is P=I2RP=I^{2}RP=I2R.
- Squaring the current gives I2=0.402=0.16I^{2}=0.40^{2}=0.16I2=0.402=0.16.
- Substituting gives P=0.16×15P=0.16\times15P=0.16×15.
- The power dissipated in the resistor is 2.4 W2.4\ \text{W}2.4 W.
Doubling the current
- Because the current appears squared in P=I2RP=I^{2}RP=I2R, a change in the current has a far larger effect on the power than the same fractional change in the resistance.
- Doubling the current in a fixed resistance multiplies the power dissipated by 22=42^{2}=422=4, so the power becomes four times as large.
- Trebling the current multiplies the power by 32=93^{2}=932=9, and halving the current reduces the power to 14\dfrac{1}{4}41 of its earlier value.
- Doubling the resistance at a fixed current only doubles the power, because the resistance is not squared.
- This squared dependence is why the current is the quantity watched most closely wherever dissipation matters: a modest rise in current brings a steep rise in the power dissipated.
- Squaring also removes any sign, so the power calculated is positive whichever direction the current is in.
Effect of doubling the current
- The same 15 Ω15\ \Omega15 Ω resistor now carries a current of 0.80 A0.80\ \text{A}0.80 A, twice the earlier value.
- The equation is P=I2RP=I^{2}RP=I2R.
- Squaring the new current gives I2=0.802=0.64I^{2}=0.80^{2}=0.64I2=0.802=0.64.
- Substituting gives P=0.64×15=9.6 WP=0.64\times15=9.6\ \text{W}P=0.64×15=9.6 W.
- Comparing the two results, 9.62.4=4.0\dfrac{9.6}{2.4}=4.02.49.6=4.0, so doubling the current has quadrupled the power.
Choosing the right equation
- The data in a question decide which form to use, so the first step is to list the quantities that are known with their units.
- A current together with a potential difference points straight to P=IVP=IVP=IV.
- A current together with a resistance points straight to P=I2RP=I^{2}RP=I2R.
- A potential difference together with a resistance is handled by finding the current first from I=VRI=\dfrac{V}{R}I=RV, after which either form can be used.
- The two forms describe the same physical quantity, so a power found one way can be confirmed the other way, which is a fast check on the arithmetic.
- Units are converted before substitution: a power in kilowatts becomes watts, a resistance in kilohms becomes ohms, and a current in milliamperes becomes amperes using 1 mA=1×10−3 A1\ \text{mA}=1\times10^{-3}\ \text{A}1 mA=1×10−3 A.
- The answer is normally quoted in watts to two or three significant figures, and converted to kilowatts only when the question asks for that unit.
Two routes to the power
- A component of resistance 8.0 Ω8.0\ \Omega8.0 Ω has a potential difference of 6.0 V6.0\ \text{V}6.0 V across it.
- The current is found first from I=VR=6.08.0=0.75 AI=\dfrac{V}{R}=\dfrac{6.0}{8.0}=0.75\ \text{A}I=RV=8.06.0=0.75 A.
- Using the first form, P=IV=0.75×6.0=4.5 WP=IV=0.75\times6.0=4.5\ \text{W}P=IV=0.75×6.0=4.5 W.
- Using the second form, P=I2R=0.752×8.0=0.5625×8.0=4.5 WP=I^{2}R=0.75^{2}\times8.0=0.5625\times8.0=4.5\ \text{W}P=I2R=0.752×8.0=0.5625×8.0=4.5 W.
- The two routes agree, so the power dissipated is 4.5 W4.5\ \text{W}4.5 W.
- Do not square the resistance in P=I2RP=I^{2}RP=I2R. Only the current is squared.
- Do not put a potential difference where the resistance belongs. The two power equations are not interchangeable term by term.
- Do not pair a current in one component with the potential difference across a different one. Both readings must describe the same component.
- Do not square a current that is still in milliamperes. Convert it to amperes first, or the error is squared as well.
- Do not treat a power as a quantity of energy. A power is energy per second, so a value in watts can never answer a question that asks for joules.
- Do not assume that doubling the current doubles the power. In a fixed resistance it quadruples it.
Power calculation method
- Quote the power equation being used before substituting any numbers into it.
- List the known quantities with their units, then convert every one into amperes, volts or ohms.
- Square the current as a separate step and write the squared value down, so the method still earns credit if the arithmetic slips.
- Show the rearrangement on its own line whenever a current or a resistance is the quantity wanted.
- State the answer in watts with the unit written, and give kilowatts only when they are asked for.
- Quote a factor when describing a change, such as the power becomes four times as large, rather than saying the power increases.
- State what power means and say what 1 W1\ \text{W}1 W means in joules and seconds.
- Calculate the power of a component carrying 2.5 A2.5\ \text{A}2.5 A with 12 V12\ \text{V}12 V across it.
- Calculate the power dissipated in a 22 Ω22\ \Omega22 Ω resistor carrying 0.50 A0.50\ \text{A}0.50 A.
- Show how P=I2RP=I^{2}RP=I2R follows from P=IVP=IVP=IV and V=IRV=IRV=IR.
- State the factor by which the power changes when the current in a fixed resistance is trebled.
10.3.6 Energy transfers in domestic devices
Stores and pathways
Energy store
An energy store is a way of accounting for where the energy in a system is held, such as the thermal, kinetic, chemical or gravitational store.
- An energy store says where energy is being held: the thermal store of a warm object, the kinetic store of a moving one, the chemical store of a battery or a fuel, the elastic store of a stretched spring, and the gravitational store of a raised mass.
- Energy is never held in a pathway. A pathway is the mechanism that moves energy from one store to another, and there are four of them: electrical working, mechanical working, heating and radiation.
- Every mains device begins in the same way. The supply drives a current through the device, so energy is transferred into it along the electrical working pathway.
- A complete description of a device names three things: the store that empties, the pathway used, and the store or stores that fill.
- For a device plugged into the mains the store that empties is far away and out of sight, so the description starts at the supply doing electrical work on the device itself.
- One device usually fills several stores at the same time, and only some of that filling is the filling the user wanted.
- Every mains device starts with the electrical working pathway, because a current in the device is what carries the energy in.
- A description is finished only when the store that fills has been named, not when the pathway has been named.
- Every device fills at least one store that nobody wanted, which is why no device transfers all of its input usefully.
Filling a thermal store
- An electric kettle is described like this: the mains supply does electrical work on the heating element, and the thermal store of the water fills, so the temperature of the water rises.
- The kettle also fills the thermal store of its own body and of the surrounding air, and more energy leaves with the escaping steam. Those transfers are wasted, because the user wanted hot water and nothing else.
- A hair dryer carries out two transfers at once. Electrical working on the element fills the thermal store of the air, and electrical working on the motor fills the kinetic store of the fan blades.
- Both of those transfers are useful in a hair dryer, since warm air is wanted and moving air is needed to carry it to the hair. The wasted part is the energy filling the thermal store of the casing, together with the noise of the motor.
- An electric iron, an electric oven and a toaster all follow the kettle's pattern, with electrical working filling the thermal store of the soleplate, of the oven air, or of the bread.
- An immersion heater is the same transfer stretched over hours, filling the thermal store of the water in a tank so that the water is still hot when it is wanted.
Filling a kinetic store
- A food mixer is described like this: the supply does electrical work on the motor, the kinetic store of the rotating blades fills, and the blades then do mechanical work on the food.
- Not all of the input reaches the blades. Energy dissipated in the resistance of the motor windings fills the thermal store of the motor, and friction at the bearings fills the thermal store of the bearings and the air around them.
- The mixer is also noisy, and the energy carried away by those sound waves finishes in the thermal store of the surroundings once the walls and the air absorb them.
- A desk fan makes the same transfer, filling the kinetic store of the blades and then the kinetic store of the moving air, which is the output the user actually wants.
- A washing machine fills two different stores in turn: the kinetic store of the drum while it tumbles and spins, and the thermal store of the water while its element heats the wash.
- An electric drill fills the kinetic store of the rotating bit, and that store then does mechanical work on the wood or masonry, filling the thermal store of the material and of the bit. This is why a drill bit is hot to the touch after use.
- A vacuum cleaner fills the kinetic store of a fan, and the moving air it creates carries the dust. Its motor and its rushing air both fill the thermal store of the surroundings as well.
Light and sound
- A filament lamp is described in two steps. Electrical working fills the thermal store of the filament, and the hot filament then transfers energy to the surroundings by radiation, part of which is visible light.
- The useful output of a lamp is the light radiated. The energy filling the thermal store of the bulb, the fitting and the air around them is wasted, because the user wanted a lit room and not a warm one.
- An LED lamp does the same job while filling a thermal store far less, which is why an LED stays cool in its fitting and a filament lamp cannot be touched.
- A television takes energy in by electrical working, and its useful outputs are the light radiated from the screen and the sound waves from its speakers. The thermal store of its circuit boards and case fills too, which is why the back of a television is warm.
- A loudspeaker fills the kinetic store of its vibrating cone, and the cone then sets the air vibrating so that sound waves travel out. Its coil has resistance, so the thermal store of the speaker fills as well.
- Light and sound both end up in the thermal store of the surroundings once walls, furniture, air and people have absorbed them.
Charging a battery
- A phone on charge is the everyday case where the store that fills is neither thermal nor kinetic. The supply does electrical work on the phone and the chemical store of the battery fills.
- The transfer runs the other way once the charger is unplugged. The chemical store of the battery empties and does electrical work on the screen, the processor and the speaker.
- Both the charger and the phone become warm during charging, so part of the transfer fills the thermal store of the charger, the battery and the room rather than the chemical store.
- A cordless drill on its base, an electric toothbrush in its stand and a rechargeable torch all fill a chemical store in exactly the same way.
- A battery is therefore a store and not a source, because the energy held in it was put there by an earlier transfer from the mains.
Useful and wasted transfers
Wasted energy
Wasted energy is energy transferred to a store that is not the one the device is designed to fill.
- In any device the transfer that fills the store the user wanted is the useful transfer, and every other transfer counts as wasted energy.
- The same store can be useful in one device and wasted in another. The thermal store of the surroundings is the whole point of a fan heater and a plain nuisance in a laptop.
- No device transfers all of its input usefully, and three separate reasons appear in real appliances.
- Every part that carries a current has some resistance, so a thermal store fills wherever charge flows.
- Every moving part meets friction at its bearings and drag from the air, so a thermal store fills wherever something turns.
- Every device that makes light or sound sends some of it in directions nobody is using.
- Wasted energy is not destroyed. It spreads into the surroundings and becomes shared between so many particles that it cannot be collected and used again.
- Wasted transfers can be reduced but never removed. Lubricating the bearings cuts friction, thicker copper windings cut the resistance of a motor, and a smoother fan blade cuts air resistance.
- Deciding which transfer is useful always starts from what the device is for, so identical physics is judged differently in a kettle, a fan and a lamp.
- Do not name a pathway where a store is wanted. Electrical working is how the energy travels, and it is not somewhere energy can sit.
- Do not write electrical energy, heat energy or sound energy. Name the store that fills, or the pathway that does the transferring.
- Do not say the energy is used up inside the device. It fills other stores, and the total is conserved.
- Do not call a battery a source of energy. It is a chemical store that an earlier transfer filled.
- Do not claim a device could be built to waste nothing. Resistance, friction and stray light and sound guarantee some wasted transfer.
- Do not label a transfer wasted before deciding what the device is for. Filling the thermal store of a room is the useful output of a heater.
Describing an energy transfer
- Write the answer as one sentence in three parts: the store that empties, the pathway, and the store that fills.
- Start a mains device at the supply doing electrical work on the device, since that is where the current comes from.
- Name the object as well as the store, for example the thermal store of the water rather than simply a thermal store.
- Say which transfer is useful and which is wasted, and justify the choice by what the device is for.
- Avoid the words heat, electrical energy and sound energy, all of which lose marks in this style of answer.
- Finish a wasted transfer by saying where the energy ends up, which is almost always the thermal store of the surroundings.
- Describe the energy transfer in an electric kettle, naming the pathway and the store that fills.
- Describe the transfer in a food mixer and name two stores that fill besides the useful one.
- Explain why the transfer in a phone on charge is described differently from the transfer in a kettle.
- State which transfer is useful and which is wasted in a filament lamp, and say where the wasted energy ends up.
- Give three reasons why no device transfers all of its input usefully.