This Physics quiz is called 'Physics - Wasted Energy and Efficiency' and it has been written by teachers to help you if you are studying the subject at senior high school. Playing educational quizzes is one of the most efficienct ways to learn if you are in the 11th or 12th grade - aged 16 to 18.
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When energy is transferred only part of it is usefully transferred - the rest is ‘lost’. Only it isn't really lost. It is still there but in different forms and spread out into the surroundings, so we refer to it as 'wasted energy'. This wasted energy is often in the form of heat but could be in other forms like light and sound. One way or another, the energy that has been transferred to the surroundings makes them warmer. The wasted energy is increasingly spread out and is therefore a lot less useful.
[readmore]In mechanical systems, an important cause of wasted energy is friction. Friction transfers kinetic energy into heat and sound which dissipates directly into the surroundings. As the sound energy passes through the air (or any other medium) it is transferred back into kinetic energy as it causes the particles to vibrate. The sound energy is thus absorbed and spread through the medium.
In electrical systems, as the free electrons move through a conductor they collide with its atoms. Some of the electrical energy is therefore transferred into kinetic energy by the particles of the conductor, making them vibrate more. Vibrating particles transfer kinetic energy into heat energy and the conductor becomes warmer. This sets up a temperature gradient which transfers the thermal energy into the surroundings, wasting some of the original electrical energy. Sparks in electrical systems also waste some energy in the form of light, which spreads out into the surroundings.
No attempt to transfer one form of energy into another form of useful energy is ever 100% successful. Some come pretty close but there is always a bit of energy that is wasted. In class, you will have come across Sankey diagrams. These are simply scale drawings on graph paper that show how much energy is transferred usefully and how much is wasted. The proportion of energy that is usefully transferred is called the efficiency. Efficiency can be worked out by dividing the amount of useful energy obtained by the amount of energy at the start of the transfer, then multiplying it by 100 to express it as a percentage.
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The key useful energy transfer is to produce kinetic energy. The noise of the train and friction waste energy in the form of heat and sound
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These crop up almost every year in the exams
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Efficiency is always a percentage worked out by dividing the output energy by the input energy and multipying by 100
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Answer four would be what you got if you did the division the wrong way round. The majority of the electrical energy is wasted as heat
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Not only does LED lighting consume a fraction of the power of a filament lamp, it is also much more efficient at transferring the electrical energy to useful light energy
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Since this is a useful energy transfer, it increases the overall efficiency by using some of the heat energy that would otherwise be wasted
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It is important to remember that efficiency is a measure of the useful energy transfer. The way the question is worded could make you forget that the calculation gives you the percentage of energy that has been wasted and needs to subtracted from 100% to find the efficiency. Watch out for this sort of question in your exams
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The efficiency is 25 ÷ 29 x 100 = 86.2% efficient
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The total energy transfer losses are 1.87% so to the nearest whole number that is 2% wastage, leaving 98% efficiency. Efficiency is worse in the lower voltage sections of the National Grid because the higher currents cause more energy loss as heat
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This question could look very complicated at first. The knack is to break down complicated questions into smaller parts which makes them more manageable. So, first you need to work out the area of the solar cell which is 0.1 m2. That means that it receives 100 W of the insolation (incoming solar energy). Finally, rearranging the equation for efficiency provides the answer:
(efficiency x input wattage) ÷ 100 = output wattage |