Power: The Rate of Energy Transfer — Practice · Grade 8 Science

Chapter: Work & Machines

What this practice covers

By the end of Power: The Rate of Energy Transfer you should be able to:

12 questions, written answer, each with a worked explanation.

Vocabulary used: power, watt, average power, kilowatt, kilowatt-hour, model, evidence, system boundary, assumption.

Problem 1

Does the slower motor necessarily transfer less energy?
Answer and explanation

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

No. It transfers the same stated 120 J over a longer interval, so its average power is smaller.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 2

What information is missing to find input power?
Answer and explanation

The input energy or input power and the losses. Useful lifting power alone does not describe all energy entering the device.

The input energy or input power and the losses. Useful lifting power alone does not describe all energy entering the device.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 3

Does 60% efficiency mean the input power is 60 W?
Answer and explanation

No. Efficiency is a ratio. The power also depends on the energy amount and time interval.

No. Efficiency is a ratio. The power also depends on the energy amount and time interval.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 4

Why can rated power alone fail to predict a daily energy bill?
Answer and explanation

The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 5

Illustrative lifting devices each deliver 240 J; A takes 8 s and B 12 s. A receives 400 J and B receives 300 J. Compare useful power and efficiency, then explain why the higher-power device is not automatically more efficient.
Answer and explanation

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

Useful powers are 30 W for A and 20 W for B. Efficiencies are 240/400=60% and 240/300=80%. A delivers useful energy faster but B directs a larger fraction of input to that output. Power and efficiency answer different questions, and both require the same defined useful output.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 6

A 50 W model device operates steadily for 20 s. Find the transferred energy and identify an assumption.
Answer and explanation

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 7

Evaluate and correct: Watts measure the total energy a device uses.
Answer and explanation

Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy.

Watts measure energy per time; multiply an appropriate average power by operating time to obtain energy.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 8

Choose one prediction from this unit. Explain a comparison that could test it, specifying observations, changed and matched conditions, and a result that would challenge it.
Answer and explanation

A hypothetical 0.5 kW device operating at that average power for 2 h transfers 1 kWh, or 3.6 million J. This is an illustrative arithmetic case, not a current price or an actual appliance guarantee. The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

A hypothetical 0.5 kW device operating at that average power for 2 h transfers 1 kWh, or 3.6 million J. This is an illustrative arithmetic case, not a current price or an actual appliance guarantee. The operating duration, actual average power, cycling and energy price are also needed. A rating describes a condition rather than every day’s use.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 9

A model device has average input power 80 W over 10 s and supplies 600 J useful work. Find input energy and efficiency; state a condition behind the arithmetic.
Hints

Distinguish the observed quantities, the proposed mechanism and its conditions.

Answer and explanation

Input is 800 J and useful efficiency 600/800=75%. The 80 W must be the applicable average over the interval; the other 200 J needs an energy pathway or stored-change account.

Input is 800 J and useful efficiency 600/800=75%. The 80 W must be the applicable average over the interval; the other 200 J needs an energy pathway or stored-change account.

Problem 10

A 50 W model device operates steadily for 20 s. Find the transferred energy and identify an assumption. Explain a condition that would change this conclusion.
Hints

Distinguish the observed quantities, the proposed mechanism and its conditions.

Answer and explanation

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Energy is 50 J/s × 20 s = 1,000 J. This uses a constant or appropriately averaged 50 W over the entire interval; real varying operation requires interval data.

Problem 11

Same lifting work, different average powerSame lifting work, different average power3 m height gain4 kg object; g = 10 N/kg; ideal steady liftingWork = 4×10×3 = 120 J. If time = 2 s, average power = 60 W.Another device taking 6 s does the same 120 J at only 20 W.
The stated 4 kg object gains 3 m in height, receiving 120 J of gravitational energy in the ideal steady-lift model. Completing that lift in 2 s gives average useful power 60 W; taking 6 s gives 20 W. More power describes a faster energy transfer, not more energy required for the same ideal height change.
For this unit’s illustrated case, if the lift takes 4 s instead, what are work and average useful power?
Answer and explanation

The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.

A watt is one joule per second. Separate how much useful energy changes from how long the change takes. Do not infer efficiency from timing alone without measuring input energy as well.

The ideal lifting work stays 120 J; average useful power is 120/4 = 30 W. Conditions and any additional acceleration or losses must be accounted for separately.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 12

A device transfers a useful 180 J in 3 s while receiving 240 J in that interval. Find useful average power and efficiency, then explain which changes if the same energy transfers take 6 s.
Answer and explanation

Useful power is 60 W and efficiency 75%. Taking 6 s reduces useful power to 30 W while the same energy ratio stays 75%.

Use energy/time for power and useful/input for efficiency.

Change only the interval in the second comparison.

Useful power is 60 W and efficiency 75%. Taking 6 s reduces useful power to 30 W while the same energy ratio stays 75%.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

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