GCSE Physics (9-1)

1.2 Conservation, dissipation and energy resources — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

  1. A device takes in 800 J of energy and transfers 600 J usefully. Calculate its efficiency as a decimal and as a percentage.
  2. A lamp is supplied with 50 J of energy and transfers 10 J of it as useful light. Calculate the efficiency as a percentage.
  3. A motor is 80% efficient and its useful output is 320 J. Calculate the total energy supplied to the motor.
  4. A kettle is 90% efficient and takes in 6.0×105 J6.0 \times 10^5\,\text{J} of energy. Calculate the energy transferred usefully to the water.
  5. A ball is dropped from a height of 2.0 m and bounces back to 1.5 m. Explain the energy changes, in terms of stores.
  6. A hot water tank loses energy to its surroundings. Describe two ways to reduce this unwanted energy transfer.
  7. Give one advantage of using wind rather than coal to generate electricity, one advantage of using coal rather than wind, and one disadvantage of a wind farm.
  8. Explain why the share of electricity generated from renewable resources has increased in many countries.
  9. In the chapter problem the falling water gives up 3528 J each second and the generator gives out 2100 J each second. Calculate the efficiency of the generator.
  10. A device takes in 1500 J of energy and transfers 600 J usefully. Calculate the energy dissipated, and name the store that receives it.
  11. A 2000 W heater runs for 300 s. The water gains 480000 J of useful energy. Work out the efficiency. Say which two quantities you need first, and why.
  12. State the principle of conservation of energy for a closed system.
Answers and concise solutions

1. Efficiency of the device

Answer: efficiency=600800=0.75=75%\text{efficiency} = \dfrac{600}{800} = 0.75 = 75\%.

2. Efficiency of the lamp

Answer: efficiency=1050=0.20=20%\text{efficiency} = \dfrac{10}{50} = 0.20 = 20\%.

3. Total energy supplied

Answer: 0.80=320total input0.80 = \dfrac{320}{\text{total input}}, so total input =3200.80=400 J=4.0×102 J= \dfrac{320}{0.80} = 400\,\text{J} = 4.0 \times 10^2\,\text{J}.

4. Useful energy transferred to the water

Answer: useful output=0.90×6.0×105=5.4×105 J\text{useful output} = 0.90 \times 6.0 \times 10^5 = 5.4 \times 10^5\,\text{J}.

5. The energy changes in the bounce

  • At the top of the drop the ball has energy in its gravitational potential store.
  • As the ball falls, that energy moves to the kinetic store, which is largest just before the bounce.
  • During the bounce, some energy is transferred to the thermal store of the ball, the floor and the air, so the ball rises to a lower height.

6. Two ways to reduce the transfer

  • Wrap the tank in a thick insulating jacket, which traps air and lowers the rate at which energy is transferred.
  • Insulate the hot pipes, because a material with a low thermal conductivity transfers energy more slowly.

7. Comparing wind and coal

  • Wind is renewable, so it is replaced as it is used and will not run out.
  • Coal is reliable, because a coal power station can generate at any time, while the wind does not always blow.
  • A wind farm changes the look of the landscape, which is an environmental impact.

8. Why the renewable share has increased

  • Renewable resources are replaced as they are used, so they will not run out.
  • Burning fossil fuels adds carbon dioxide to the air, so many countries choose renewable resources instead to reduce this effect.

9. Efficiency of the generator

Answer: efficiency=21003528=0.60=60%\text{efficiency} = \dfrac{2100}{3528} = 0.60 = 60\% to 2 significant figures.

10. Energy dissipated

Answer: dissipated=1500−600=900 J=9.0×102 J\text{dissipated} = 1500 - 600 = 900\,\text{J} = 9.0 \times 10^2\,\text{J}. The energy is transferred to the thermal store of the surroundings.

11. Efficiency of the heater

Answer: First find the total energy in: E=Pt=2000×300=600000 JE = Pt = 2000 \times 300 = 600000\,\text{J}. Then efficiency=480000600000=0.80=80%\text{efficiency} = \dfrac{480000}{600000} = 0.80 = 80\%.

12. The principle of conservation of energy

  • In a closed system the total energy stored does not change.
  • Energy can be transferred between stores, but it is not created and it is not destroyed.

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