Efficiency.pptx

AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Efficiency

Energy

Lesson 6 of 7

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Warm-up

Energy

Lesson 6 of 7

Answer each one, then check.

1

What is dissipated energy?

2

What is 0.35 as a percentage?

3

Write 60% as a decimal.

4

Give the unit of power.

5

What does 'useful' energy mean?

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Warm-up - Answers

Energy

Lesson 6 of 7

1

What is dissipated energy?

Energy spread to the surroundings in less useful stores

2

What is 0.35 as a percentage?

35%

3

Write 60% as a decimal.

0.60

4

Give the unit of power.

Watt

5

What does 'useful' energy mean?

Energy transferred in the way we want

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Learning Objectives

Energy

Lesson 6 of 7

1

Recall and apply the efficiency equations.

2

Give efficiency as a decimal or a percentage.

3

Interpret Sankey diagrams.

4

Describe ways to increase the efficiency of an energy transfer (Higher tier).

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Energy

Lesson 6 of 7

EFFICIENCY

efficiency = (useful output energy transfer)/(total input energy transfer)

Also: efficiency = (useful power output)/(total power input). Efficiency has no units and is never more than 1 (100%).

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A Sankey Diagram

Energy

Lesson 6 of 7

The width of each arrow shows the amount of energy.

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Efficiency of a Lamp

Energy

Lesson 6 of 7

A lamp transfers 100 J of electrical energy and 8 J of it is transferred as light. Calculate its efficiency.

1

Write the equation

efficiency = (useful output)/(total input)

2

Substitute

8/100

3

Decimal

0.08

4

Percentage

0.08 × 100 = 8%

ANSWER

Efficiency = 0.08 or 8%.

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Using Power

Energy

Lesson 6 of 7

A motor has a total power input of 500 W and a useful power output of 350 W. Calculate its efficiency as a percentage.

1

Substitute

350/500

2

Decimal

0.70

3

Percentage

70%

ANSWER

70%

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Finding the Useful Output

Energy

Lesson 6 of 7

A machine is 85% efficient. How much useful energy does it transfer from a 400 J input?

1

Efficiency as a decimal

0.85

2

Multiply

0.85 × 400 = 340 J

ANSWER

340 J

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Ways to Increase Efficiency (Higher Tier)

Energy

Lesson 6 of 7

Reduce the energy that is dissipated.

Lubrication

Reduces friction between moving parts.

Thermal insulation

Reduces energy lost by heating, for example around a boiler or a house.

Streamlining

Reduces air resistance so less energy is wasted.

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Ways to Increase Efficiency (Higher Tier) (cont.)

Energy

Lesson 6 of 7

Better design

For example LED lamps transfer more input energy as light.

Higher tier

The question may ask you to suggest a way suitable for a described machine.

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Common Mistakes

Energy

Lesson 6 of 7

Check your answer.

Efficiency above 1

It cannot be greater than 1 (100%): useful output cannot be more than the input.

Wrong way up

Useful output goes on top.

Units

Do not write J or W after an efficiency.

Percent or decimal

Read the question for the form required.

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Key Terms

Energy

Lesson 6 of 7

Efficiency

The fraction of the input energy that is transferred usefully.

Useful output

Energy transferred in the way that is wanted.

Wasted energy

Energy transferred to stores that are not useful.

Sankey diagram

A diagram with arrow widths that show energy amounts.

Streamlining

Shaping an object to reduce air resistance.

Input

The energy or power supplied to a device.

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Energy

Lesson 6 of 7

YOUR TASK

Compare the Lamps

12 minutes

LED lamp: efficiency 0.85, input power 8 W. Filament lamp: efficiency 0.10, input power 60 W. Calculate the useful power output of each and say which lamp is better.

1

Rearrange the equation for useful power.

2

Compare the two answers.

WHAT A GOOD ANSWER SHOWS

LED: 0.85 × 8 = 6.8 W. Filament: 0.10 × 60 = 6.0 W. The LED gives more useful output from far less input, so it is better.

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Can I...?

Energy

Lesson 6 of 7

Recall both efficiency equations.

Calculate efficiency.

Convert decimal and percentage.

Interpret a Sankey diagram.

Explain efficiency cannot exceed 1.

Find useful output from an efficiency.

Describe ways to increase efficiency.

Compare devices.

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AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER

Summary & Exam Focus

Efficiency = useful output ÷ total input.

Also useful power ÷ total power.

Multiply by 100 for a percentage.

Efficiency is never more than 1.

EXAM FOCUS

A device has an input of 400 J and a useful output of 100 J. Calculate the efficiency. (2 marks)

Put useful over total.

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