To compare two energy systems fairly, keep the conditions the same and change only the system. Then calculate the same quantity for both, state the difference and say what the comparison assumes.
This lesson closes the electricity and energy cluster before its practice set. It uses the formula from calculating energy from a device table.
What does a fair comparison need?
Write down what is held constant before you calculate. For two classroom lamps the usual items are the hours used per day, the number of days and, if cost is asked, the rate.
If a question leaves one of these out, state it as your assumption. That sentence is often worth a mark.
Worked example: two invented lamps
Classroom A uses an LED lamp rated 10 W. Classroom B uses a fluorescent lamp rated 24 W. Assume both run 6 hours a day for 5 days, and the rate is RM0.45 per kWh.
Step 1: calculate the hours. 6 × 5 = 30 h for each lamp.
Step 2: calculate the energy.
- LED: 10 × 30 = 300 Wh = 0.30 kWh
- Fluorescent: 24 × 30 = 720 Wh = 0.72 kWh
Step 3: find the difference. 0.72 − 0.30 = 0.42 kWh a week.
Step 4: find the cost under the assumed rate. LED: 0.30 × 0.45 = RM0.135. Fluorescent: 0.72 × 0.45 = RM0.324. The fluorescent lamp costs about RM0.19 more a week.
Step 5: conclude with assumptions. The LED lamp uses 0.42 kWh less energy each week, assuming the same hours, the same days and the same rate. Brightness was not compared.
The last sentence is the honest limit: the data does not say the two lamps give the same light.
The mistake that costs marks
The common slip is an unfair comparison: “The LED lamp uses 300 Wh and the fluorescent uses 1 440 Wh, so the LED is much better”. The fluorescent figure came from running it 60 hours, twice as long.
The corrected approach gives both lamps the same 30 hours. Only then is the difference in energy due to the lamps alone, which is the point of a comparison.
Check yourself
An invented fan in Room P is rated 50 W and in Room Q 80 W. Assume both run 4 hours a day for 5 days. Find the weekly energy for each in kWh and the difference.
Answer
Hours: 4 × 5 = 20 h each.
Room P: 50 × 20 = 1 000 Wh = 1.0 kWh. Room Q: 80 × 20 = 1 600 Wh = 1.6 kWh.
Difference = 1.6 − 1.0 = 0.6 kWh, assuming the same hours and days, and that the fans give the same airflow.
What to study next
Test all four lessons in the integrated practice set. The graph evidence and fair-comparison lab gives more fair-test scenarios.
For a teacher to read your assumption statements with you, see online one-to-one Science tuition.