Energy-use questions combine power, time and price. The physics is one multiplication, so the marks depend on putting every quantity into the right unit first. This lesson gives you a checking routine.
It follows calculating electrical energy from ratings, which covers the joule version. Here the unit is usually the kilowatt-hour.
Which conversions come up?
- Watts to kilowatts: divide by 1000. 2000 W = 2 kW.
- Minutes to hours: divide by 60. 45 min = 0.75 h.
- kWh to joules: multiply by 3 600 000, because 1 kW × 1 h = 1000 W × 3600 s.
- Joules to kWh: divide by 3 600 000.
Then: energy (kWh) = power (kW) × time (h), and cost = energy (kWh) × price per kWh.
How do you check the conversion?
Write each quantity with its unit, convert, then ask whether the answer is plausible.
A 2 kW heater for 3 hours cannot be more than a few tens of kWh. If your number looks huge, check for a forgotten division by 1000 or by 60. The bounds and rounding explainer also helps you decide how many figures to keep in a final cost.
Worked example
An invented heater is rated 800 W. It runs for 45 minutes.
The price is RM0.40 per kWh (invented for practice). Find the energy in kWh, in joules, and the cost.
Step 1, power in kW: 800 ÷ 1000 = 0.8 kW.
Step 2, time in hours: 45 ÷ 60 = 0.75 h.
Step 3, energy: 0.8 × 0.75 = 0.6 kWh.
Step 4, cost: 0.6 × 0.40 = RM0.24.
Step 5, joules, as a cross-check: 800 W × (45 × 60 = 2700 s) = 2 160 000 J. And 0.6 kWh × 3 600 000 = 2 160 000 J. The two routes agree.
The mistake to watch for
A common slip is to use power in kilowatts but time in minutes.
Mistaken working: E = 0.8 × 45 = 36 kWh
The student converted the power to kilowatts but left the time in minutes, so the answer is 60 times too big.
A plausibility check would have caught it: 36 kWh from an 800 W heater would need about 45 hours of running, not 45 minutes. The correction is to convert the time to hours first: 0.8 × 0.75 = 0.6 kWh.
Check yourself
Use RM0.40 per kWh (invented) where a cost is asked.
1. A 1500 W heater runs for 20 minutes. Find the energy in kWh.
Show answer
P = 1.5 kW. t = 20 ÷ 60 = 1/3 h. E = 1.5 × 1/3 = 0.5 kWh.
2. A 250 W computer runs for 8 hours. Find the energy in kWh and the cost.
Show answer
P = 0.25 kW. E = 0.25 × 8 = 2 kWh. Cost = 2 × 0.40 = RM0.80.
3. Convert 0.5 kWh to joules.
Show answer
0.5 × 3 600 000 = 1 800 000 J (1.8 × 10⁶ J).
Where this leads next
Next, test all five lessons together in the electrical energy practice set. The module overview is electrical energy and safe interpretation.
If unit checks still feel slow, a teacher giving online one-to-one Physics tuition can watch your working live and point to the exact line where a conversion slips.