AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER
Exam Practice: Efficiency
Efficiency · Energy · Lesson 6 of 7 · 16 marks · 17 minutes
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Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Use the mark allocation as a guide to how much to write.
Question 1 CALCULATE (2 marks)
A device transfers 400 J of energy as input and 100 J of useful energy. Calculate the efficiency of the device. Use the equation: efficiency = useful output energy transfer ÷ total input energy transfer
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(Total for Question 1 = 2 marks)
Question 2 USE THE DIAGRAM (3 marks)
The Sankey diagram shows the energy transfers in an electric motor. (a) Calculate the energy wasted every second. (b) Calculate the efficiency of the motor.
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(Total for Question 2 = 3 marks)
Question 3 CALCULATE (3 marks)
A lamp has an efficiency of 0.20 and a total power input of 60 W. Calculate the useful power output of the lamp.
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(Total for Question 3 = 3 marks)
Question 4 SUGGEST (2 marks)
A car engine wastes a lot of energy. Suggest two ways to increase the efficiency of the car.
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(Total for Question 4 = 2 marks)
Question 5 COMPARE (4 marks)
An LED lamp has an efficiency of 0.85 and an input power of 8 W. A filament lamp has an efficiency of 0.10 and an input power of 60 W. Compare the useful power output of the two lamps. Which is the better lamp? Show your calculations.
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(Total for Question 5 = 4 marks)
Question 6 EXPLAIN (2 marks)
A student says that a machine has an efficiency of 120%. Explain why this cannot be correct.
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(Total for Question 6 = 2 marks)
TOTAL FOR PAPER = 16 MARKS
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Answers and mark scheme
Check your answer only once you have written one.
Question 1 (2 marks)
100 ÷ 400 = 0.25 (25%)
• Correct substitution 1 mark
• 0.25 or 25% 1 mark
Question 2 (3 marks)
(a) 300 − 240 = 60 J. (b) 240 ÷ 300 = 0.80 (80%).
• 60 J 1 mark
• Correct substitution 1 mark
• 0.80 or 80% 1 mark
Question 3 (3 marks)
0.20 × 60 = 12 W
• Rearranges to power output = efficiency × input 1 mark
• Correct substitution 1 mark
• 12 W 1 mark
Question 4 (2 marks)
Any two from: lubricate moving parts to reduce friction; streamline the shape to reduce air resistance; improve insulation or design of the engine to reduce thermal losses.
• First way 1 mark
• Second way 1 mark
Question 5 (4 marks)
LED: 0.85 × 8 = 6.8 W. Filament: 0.10 × 60 = 6.0 W. The LED lamp transfers more useful power while using far less input power, so it is the better lamp.
• 6.8 W 1 mark
• 6.0 W 1 mark
• LED gives more useful power 1 mark
• With much less input power 1 mark
Question 6 (2 marks)
Useful output cannot be greater than the input because energy cannot be created.
• Energy cannot be created 1 mark
• Useful output cannot exceed the total input 1 mark