Convert units and rearrange power calculations
Use consistent joules, watts and seconds to find a transferred amount or duration.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028, section 1.7.4. Power is work done or energy transferred per unit time: P = W/t = ΔE/t. A whole-interval quotient gives average power unless a constant rate is stated. Keep system, units and interval consistent. Written practice is self-checked. Human subject-teacher review, supervised practical experience and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
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.
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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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.
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