Convert units and rearrange power calculations
Use consistent joules, watts and seconds to find a transferred amount or duration.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Define power as energy per time.
- Use metric prefixes and minute-to-second conversion.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many seconds is 3.0 min?
What you will learn
- Use E = Pt and t = E/P for a stated constant or average rate.
- Convert kJ, kW and minutes before calculating.
Match the energy and time units
For 9.0 kJ transferred in 3.0 min, convert to 9000 J and 180 s. Average power is 9000/180 = 50 W. Dividing the displayed numbers 9 by 3 without conversions produces 3 kJ/min, not 3 W. Use the requested final unit explicitly rather than changing only its label.
A 0.40 kW constant or interval-average power over 2.0 min transfers 400 × 120 = 48 000 J or 48 kJ. The equation E = Pt applies to that supplied constant rate or the appropriate average over the same interval. An unrelated peak reading cannot replace the average without a matching transfer profile.
Find a duration from the required amount
From P = E/t, rearrange t = E/P. A 6000 J transfer at constant 200 W takes 30 s. A positive duration needs a physically consistent amount and rate. Zero power maintained throughout cannot supply a non-zero transfer by this model; dividing a non-zero required amount by zero does not produce a finite valid time.
Keep a quantity's meaning after rearranging: 30 s is a transfer duration, not the absolute clock time of completion unless the start time is supplied. Likewise, an amount computed from input power is input energy, not automatically useful output when a device has other destinations. An efficiency account can later allocate the useful share.
Sources for this lesson
Pause and explain
Can a device with zero power throughout deliver a non-zero required energy in finite time under E = Pt?
Worked example
Find average power for 9.0 kJ in 3.0 min, then energy transferred at 0.40 kW for 2.0 min.
Show the worked solution
- 9000 J / 180 s = 50 W.
- 0.40 kW = 400 W and 2.0 min = 120 s.
- E = 400 × 120 = 48 000 J = 48 kJ.
Answer 50 W; 48 kJ.
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Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find average power for 9.0 kJ transferred in 3.0 min.
Give me a hint
Convert both quantities before the quotient.
Compare my reasoning
- Energy = 9000 J.
- Time = 180 s.
- P = 9000/180 = 50 W.
50 W.
Look for these in your work
- Use seconds.
- Convert the energy prefix.
Find energy for 0.40 kW maintained for 2.0 min, and time for 6000 J at constant 200 W.
Give me a hint
Use E = Pt and t = E/P with consistent units.
Compare my reasoning
- 400 W for 120 s gives 48 000 J or 48 kJ.
- 6000/200 gives 30 s.
- The second value is duration, not a clock reading.
48 kJ; 30 s.
Look for these in your work
- Choose the correct rearrangement.
- State the quantity and unit.
A calculation divides 9.0 kJ by 3.0 min and labels the answer 3 W. Correct the unit and the requested watt result.
Give me a hint
The displayed-number quotient retains kJ/min until converted.
Compare my reasoning
- 9/3 = 3 kJ/min.
- 3 kJ/min = 3000 J per 60 s.
- This is 50 J/s = 50 W.
3 kJ/min equals 50 W, not 3 W.
Look for these in your work
- Preserve units through the first quotient.
- Convert the rate correctly.
Common mistakes
- Convert prefixes and minutes.
- Use the same interval's average or a stated constant rate.
- Input energy is not necessarily useful output.
What can you explain now?
How long does 6000 J take at constant 200 W?
Compare with the explanation
30 s.
t = E/P = 6000/200 = 30 s.
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Make it stick
Rearrange power for energy and time, convert a kJ/min rate and distinguish duration, clock time and useful output.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.