Electrical energy has two unit pairs that must not be mixed: watts with seconds give joules, and kilowatts with hours give kilowatt-hours. Most wrong answers come from using one from each pair.
This page shows the fix with one original example, worked two ways. The bounds and rounding explainer helps you decide how many significant figures your final answer deserves, and the reasoning board gives you practice choosing the relationship before you calculate.
Why do the units go wrong?
The equation E = P × t looks the same in every question, so the brain stops reading the units. But the question gives power in kW or W, and time in minutes, hours or seconds, in whatever combination the examiner chose.
The equation only works when the units belong to the same pair. Power in watts needs time in seconds for joules. Power in kilowatts needs time in hours for kWh. Minutes fit neither pair, so they always need converting.
A reliable routine
- Underline the power and the time in the question, with their units.
- Decide the answer unit the question wants: J, kJ, MJ or kWh.
- Convert first. Write each quantity in the same pair, for example 2.0 kW and 0.75 h, or 2000 W and 2700 s.
- Substitute with units written next to each number.
- Check the other way. Convert the answer into the other unit using 1 kWh = 3.6 MJ and see whether it agrees.
Worked example: a heater
A 2.0 kW room heater is switched on for 45 minutes. Find the energy transferred in kWh and in joules.
Step 1, time in hours: 45 min ÷ 60 = 0.75 h.
Step 2, kWh route: E = P × t = 2.0 kW × 0.75 h = 1.5 kWh.
Step 3, joules route: power is 2.0 kW = 2000 W, and time is 45 × 60 = 2700 s. E = 2000 × 2700 = 5 400 000 J = 5.4 MJ.
Step 4, check: 1.5 kWh × 3.6 MJ per kWh = 5.4 MJ. Both routes agree, so the conversions are right.
If the electricity costs 30 sen per kWh in this example, the cost is 1.5 × 0.30 = RM0.45. Cost always uses kWh, not joules.
The mistake to watch for
A student reads “2.0 kW” and “45 minutes” and writes:
E = 2.0 × 45 = 90 kWh
This puts minutes into a kilowatt-hour equation. The answer is 60 times too big, and a household heater using 90 kWh in under an hour should have looked wrong.
A second version is E = 2000 × 45 = 90 000 J. Watts are right, but minutes are not seconds, so the answer is 60 times too small.
The fix is step 3 of the routine: write the converted time with its unit before you multiply.
A final plausibility question helps too. Does 90 kWh sound like less than one hour of a heater? No, so go back.
Check yourself
1. A 60 W lamp is on for 5.0 hours. Find the energy in kWh and in joules.
Show answer
kWh: 60 W = 0.060 kW, so E = 0.060 × 5.0 = 0.30 kWh.
Joules: 5.0 h = 18 000 s, so E = 60 × 18 000 = 1 080 000 J (1.08 MJ).
Check: 0.30 × 3.6 = 1.08 MJ, which agrees.
2. A kettle rated 800 W runs for 3.0 minutes. Find the energy in joules.
Show answer
3.0 min = 180 s. E = 800 × 180 = 144 000 J.
Check in kWh: 0.8 kW × 0.050 h = 0.040 kWh, and 0.040 × 3 600 000 = 144 000 J. Both agree.
3. A 1500 W appliance is used for 2.0 hours. Electricity costs 25 sen per kWh in this question. Find the cost.
Show answer
1500 W = 1.5 kW. E = 1.5 × 2.0 = 3.0 kWh. Cost = 3.0 × 0.25 = RM0.75.
Where this leads next
Practise the exact skill in checking a unit conversion in an energy-use calculation, then build the calculation itself in calculating electrical energy from stated ratings. The electrical energy and safe interpretation module gathers the rest, and the Physics terminology guide separates power, energy and charge.
If unit errors keep appearing in your own corrected work, a teacher can go through it line by line in online one-to-one Physics tuition.