You are viewing a free preview of this lesson.
Subscribe to unlock all 7 lessons in this course and every other course on LearningBro.
No machine is perfect. Every kettle, engine, motor and light bulb wastes some of the energy it is given — usually by heating up. Efficiency is the measure of how good a device is at transferring energy usefully: a highly efficient device wastes very little, while an inefficient one throws most of its energy away (almost always as heat). Knowing how to calculate efficiency lets engineers compare devices, lets shoppers pick appliances that are cheaper to run, and explains why a modern LED bulb is so much better than an old filament one. This lesson, part of Topic P5 (Energy) of OCR Gateway Combined Science A, defines efficiency, introduces the energy and power forms of the equation, works through calculations as decimals and percentages, and explains why no device can ever be 100% efficient.
By the end of this lesson you should be able to define efficiency, use the energy and power equations for efficiency, express it as a decimal or a percentage, explain why no machine is 100% efficient, and describe ways to reduce wasted energy.
This lesson combines AO2 (calculating efficiency as a decimal or percentage from energy or power) with AO3 (interpreting Sankey diagrams and evaluating how a device wastes energy and how to reduce it).
Efficiency is the fraction (or percentage) of the energy supplied to a device that is transferred usefully — that is, to the store the device is designed to fill. The rest is wasted, transferred to stores we do not want (almost always the thermal store of the surroundings).
An efficient device transfers most of its energy usefully and wastes little; an inefficient one wastes most of it. An LED bulb, for instance, is highly efficient — most of the electrical energy becomes light — whereas an old filament bulb is inefficient, transferring most of the electrical energy to the thermal store (it gets very hot) and only a little to light.
Efficiency has no units, because it is a ratio of two energies (or two powers) and the joules cancel out. It is always a number between 0 and 1 (as a decimal) or between 0% and 100% (as a percentage).
Exam Tip: Efficiency compares useful output to total input. It has no units and can never be more than 1 (or 100%). A common misconception is that efficiency can exceed 100% — if you ever calculate a value above 100%, you have made an error, so check you have not swapped the useful and total values.
Efficiency can be worked out from energy or from power; the two forms are equivalent.
Energy form:
efficiency=total energy inputuseful energy output
Power form:
efficiency=total power inputuseful power output
Both give the same answer for a given device, because power is just energy per second — dividing the useful and total energies by the same time leaves the ratio unchanged. Use the energy form when the question gives energies in joules, and the power form when it gives powers in watts.
To express the answer as a percentage, multiply the decimal by 100:
efficiency(%)=total inputuseful output×100
A motor is supplied with 500 J of electrical energy and transfers 350 J usefully to the kinetic store. Calculate its efficiency as a decimal and as a percentage.
Step 1 — write the equation: efficiency =total inputuseful output.
Step 2 — substitute: efficiency =500350.
Step 3 — calculate (decimal): efficiency =0.7.
Step 4 — convert to a percentage: 0.7×100=70%.
Answer: the motor is 0.7 efficient, or 70% efficient.
A lamp has a power input of 60 W and a useful (light) power output of 9 W. Calculate its efficiency as a percentage.
Step 1 — write the equation: efficiency =total poweruseful power×100.
Step 2 — substitute: efficiency =609×100.
Step 3 — calculate: efficiency =0.15×100=15%.
Answer: the lamp is 15% efficient — most of the energy is wasted as heat.
A device is 80% efficient and is supplied with 2000 J of energy. How much energy is transferred usefully?
Step 1 — rearrange: useful output = efficiency × total input.
Step 2 — use the efficiency as a decimal: useful output =0.80×2000.
Step 3 — calculate: useful output =1600 J.
Answer: 1600 J is transferred usefully (and the remaining 400 J is wasted).
An electric heater is 90% efficient at heating a room and is supplied with 5000 J. How much energy is wasted?
Step 1 — useful output =0.90×5000=4500 J.
Step 2 — wasted energy = total input − useful output = 5000−4500.
Step 3 — calculate: wasted =500 J.
Answer: 500 J is wasted (transferred to stores other than the intended one).
Exam Tip: When you use an efficiency in a calculation, turn the percentage into a decimal first (divide by 100), then multiply. To find the wasted energy, subtract the useful output from the total input — by conservation of energy, the two always add up to the input.
No real device is ever 100% efficient: some energy is always wasted. The reason is that there are always unwanted energy transfers that cannot be eliminated:
In nearly every case the wasted energy finishes in the thermal store of the surroundings, warming the device and the air around it very slightly. Because these transfers can never be reduced to zero, a real machine's efficiency is always less than 1 (less than 100%). The closest we come are devices with very few moving parts and little resistance, but even they waste a little. This does not break conservation of energy — the wasted energy is not destroyed, it is simply dissipated to a less useful store.
Exam Tip: The standard reason no machine is 100% efficient is that some energy is always wasted, usually transferred to the thermal store of the surroundings by friction, air resistance or electrical resistance. Simply saying "energy is lost" is too vague — say where it goes.
Because wasted energy costs money and cuts performance, engineers try to make devices more efficient by reducing the unwanted transfers. The main methods are:
Each of these raises the fraction of energy transferred usefully, so the efficiency goes up. None of them creates energy or pushes the efficiency above 100% — they simply cut down the proportion that is wasted.
An old filament bulb is 5% efficient and an LED bulb is 80% efficient. Both give out 4 W of useful light power. Calculate the total power each draws, and comment.
Step 1 — rearrange the power form for the total input: total power =efficiencyuseful power.
Step 2 — filament bulb: total power =0.054=80 W.
Step 3 — LED bulb: total power =0.804=5 W.
Subscribe to continue reading
Get full access to this lesson and all 7 lessons in this course.