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Conservation of Energy and Efficiency Calculations — Answers

Answer sheet with model answers and success criteria for the Year 11 energy, light, sound & electricity worksheet.

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Success TutoringWorksheetsAnswer sheetYear 11 · Science · Energy, Light, Sound & Electricity

Conservation of Energy and Efficiency Calculations — Answers

Part A: Calculating efficiency

Show the formula, your working and units. Give efficiencies as percentages.

The law of conservation of energy: energy cannot be created or destroyed, only transferred or transformed. The total energy before a change equals the total energy after it.

In any device, the total energy input = useful energy output + wasted energy. Wasted energy is usually thermal energy (and sometimes sound) that spreads out into the surroundings, where it is hard to use again.

Efficiency = useful energy output ÷ total energy input × 100%. It can also be calculated with power (in watts) in place of energy. Rearranged: useful output = efficiency × total input and total input = useful output ÷ efficiency, with the efficiency written as a decimal (for example 75% = 0.75).

1.An electric kettle uses 2000 J of electrical energy, and 1800 J of this heats the water. How much energy is wasted, and what is the efficiency of the kettle?

Answer: 200 J wasted; 90% efficient

  1. Wasted = 2000 J − 1800 J = 200 J
  2. Efficiency = 1800 ÷ 2000 × 100% = 90%
2.An old-style filament light bulb uses 60 J of electrical energy each second and gives out 3 J of light. What is its efficiency?

Answer: 5%

  1. Efficiency = 3 ÷ 60 × 100% = 5%
  2. The other 57 J each second is transferred as thermal energy, which is why these bulbs get very hot.
3.A fan has an input power of 60 W and produces 45 W of useful kinetic energy (moving air). What is its efficiency?

Answer: 75%

  1. Power is energy per second, so the formula works with watts too
  2. Efficiency = 45 ÷ 60 × 100% = 75%
4.A solar panel receives 1000 W of sunlight and produces 200 W of electrical power. What is its efficiency?

Answer: 20%

  1. Efficiency = 200 ÷ 1000 × 100% = 20%
5.An electric motor is 80% efficient. It is supplied with 2500 J of electrical energy. How much useful kinetic energy does it produce, and how much is wasted?

Answer: 2000 J useful; 500 J wasted

  1. Useful output = efficiency × total input
  2. = 0.80 × 2500 J = 2000 J
  3. Wasted = 2500 J − 2000 J = 500 J
6.A crane motor is 75% efficient. It must produce 600 J of useful energy to lift a load. How much electrical energy must be supplied?

Answer: 800 J

  1. Total input = useful output ÷ efficiency
  2. = 600 J ÷ 0.75 = 800 J
7.An inventor claims a machine that is 120% efficient. This is impossible.

Answer: True

An efficiency above 100% would mean the machine puts out more energy than it takes in, creating energy. This breaks the law of conservation of energy. In practice every real machine is less than 100% efficient.

Part B: Energy accounts and Sankey diagrams

8.A drill is supplied with 1000 J of electrical energy. It produces 250 J of kinetic energy, 650 J of thermal energy and 100 J of sound. Show that energy is conserved, and find the efficiency of the drill.

Answer: 250 + 650 + 100 = 1000 J, so energy is conserved; efficiency 25%

  1. Total output = 250 J + 650 J + 100 J = 1000 J = total input, so energy is conserved
  2. Useful output is the kinetic energy: 250 J
  3. Efficiency = 250 ÷ 1000 × 100% = 25%
9.A car engine receives 40 000 J of chemical energy from its fuel. It produces 12 000 J of useful kinetic energy and 2000 J of sound. The rest is thermal energy. How much thermal energy is produced, and what is the efficiency of the engine?

Answer: 26 000 J of thermal energy; 30% efficient

  1. Thermal energy = 40 000 − 12 000 − 2000 = 26 000 J
  2. Efficiency = 12 000 ÷ 40 000 × 100% = 30%
10.In an energy flow (Sankey) diagram, what does the width of each arrow represent?

Answer: The amount of energy. The wider the arrow, the more energy it represents, and the widths of the output arrows add up to the width of the input arrow.

11.In the drill above, which energy transfer is the largest?
  1. AThermal energy
  2. BKinetic energy
  3. CSound
  4. DElectrical energy output

Answer: Thermal energy

650 J of the 1000 J input becomes thermal energy, more than the 250 J of useful kinetic energy.

Part C: Potential and kinetic energy

Gravitational potential energy: Ep = mgh, where m is mass (kg), g = 9.8 N/kg on Earth and h is height (m).

Kinetic energy: Ek = ½mv[sup 2], where v is speed (m/s). Energy is measured in joules (J).

When an object falls with no air resistance, all of the potential energy it loses becomes kinetic energy. In real situations some energy is transferred to the surroundings as thermal energy and sound.

12.A 0.5 kg ball is held 5.0 m above the ground. Calculate its gravitational potential energy.

Answer: 24.5 J

  1. Ep = mgh
  2. = 0.5 kg × 9.8 N/kg × 5.0 m = 24.5 J
13.The ball is dropped. Ignoring air resistance, what is its kinetic energy just before it hits the ground, and how fast is it travelling?

Answer: 24.5 J; about 9.9 m/s

  1. All the potential energy becomes kinetic energy: Ek = 24.5 J
  2. Ek = ½mv2, so v2 = 2Ek ÷ m = 2 × 24.5 ÷ 0.5 = 98
  3. v = √98 ≈ 9.9 m/s
14.A roller coaster car of mass 200 kg starts at rest at the top of a 20 m high hill. Calculate its gravitational potential energy at the top, and its maximum possible speed at the bottom if no energy is transferred to the surroundings.

Answer: 39 200 J; about 19.8 m/s

  1. Ep = mgh = 200 × 9.8 × 20 = 39 200 J
  2. If all of it becomes kinetic energy: ½ × 200 × v2 = 39 200, so v2 = 392
  3. v = √392 ≈ 19.8 m/s
15.The car is actually measured travelling at 18 m/s at the bottom of the hill. Calculate its kinetic energy, the energy transferred to the surroundings, and the efficiency of the energy conversion.

Answer: 32 400 J; 6800 J transferred; efficiency about 82.7%

  1. Ek = ½mv2 = ½ × 200 × 182 = 100 × 324 = 32 400 J
  2. Energy transferred to the surroundings = 39 200 − 32 400 = 6800 J (thermal energy and sound, due to friction and air resistance)
  3. Efficiency = 32 400 ÷ 39 200 × 100% ≈ 82.7%
16.A 0.1 kg ball is dropped from a height of 2.0 m and bounces back up to 1.5 m. Calculate its potential energy at the start and at the top of the bounce, and the efficiency of the bounce.

Answer: 1.96 J and 1.47 J; 75%

  1. Start: Ep = 0.1 × 9.8 × 2.0 = 1.96 J
  2. Top of bounce: Ep = 0.1 × 9.8 × 1.5 = 1.47 J
  3. Efficiency = 1.47 ÷ 1.96 × 100% = 75% (the same as 1.5 m ÷ 2.0 m, because Ep is proportional to height)
17.An electric lift motor raises a 50 kg load through a height of 12 m and uses 8000 J of electrical energy. What is the efficiency of the lift?

Answer: 73.5%

  1. Useful output = gain in Ep = mgh = 50 × 9.8 × 12 = 5880 J
  2. Efficiency = 5880 ÷ 8000 × 100% = 73.5%

Part D: Improving efficiency

18.Give three ways to make machines and devices more energy efficient.

Answer: Any three, for example: lubricate moving parts to reduce friction; use streamlined shapes to reduce air resistance; insulate to reduce unwanted thermal energy transfer; replace filament bulbs with LED lamps; use more efficient motors; keep machines well maintained.

19.When a car brakes, its kinetic energy seems to disappear. Where does it go? Describe one way some cars recover part of it.

Answer: Friction in the brakes transforms the kinetic energy into thermal energy (and some sound), which spreads into the surroundings. Hybrid and electric cars use regenerative braking, in which the motor works as a generator to turn some of the kinetic energy back into stored electrical energy.

Part E: Explain it

20.A student says: Energy is lost when a ball bounces, so the law of conservation of energy must be wrong. Write a reply that explains where the energy goes, why the ball never bounces back to its starting height, and why no machine can be 100% efficient.[5 marks]

Model answer: Energy is not destroyed; it is transferred. As the ball falls, its gravitational potential energy becomes kinetic energy. When it hits the ground, some of the energy is stored as elastic potential energy as the ball squashes and is returned as kinetic energy, but some is transferred to the surroundings as thermal energy (the ball and ground warm slightly) and sound, and some is transferred to air particles by air resistance. Because less energy is available as kinetic energy after the bounce, the ball cannot rise as high. If all the energy were added up, including the thermal energy and sound, the total would equal the starting energy. In every real machine some energy is always transferred to the surroundings, mainly by friction, air resistance and electrical resistance, and this energy spreads out and is not useful, so the useful output is always less than the input and efficiency is always below 100%.

Success criteria:

  • States that energy is transferred, not destroyed
  • Describes the chain potential → kinetic → elastic → kinetic
  • Identifies thermal energy and sound as the energy transferred to the surroundings
  • Explains why the rebound height is lower
  • Explains why efficiency is always less than 100%
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