Lesson 2 of 4
Energy use and cost
I can recall and apply P = VI and E = VIt in new situations.Syllabus
Syllabus K223 / K224, 12(b). Recall and apply P = VI and E = VIt in new situations.
I can calculate appliance running costs using energy in kW h.Syllabus
Syllabus K223 / K224, 12(c). Calculate appliance running costs using energy in kW h.
In short
- The electricity bill counts energy in kilowatt-hours, kW h.
- 1 kW h is a 1 kW appliance running for 1 hour. It is a unit of energy.
- Energy in kW h = power in kW x time in hours.
- Cost = energy in kW h x price per kW h.
An electricity charge depends on energy used. Work out the power and operating time in matching units before applying the price per unit of energy.
Power is energy transferred per second. At constant power, E = Pt; use the interval's average power if it varies. A larger power means a faster energy transfer, but the total also depends on time.
A kilowatt-hour is an energy unit
One kilowatt-hour, written kW h or kWh, is the energy transferred by a power of 1 kW maintained for 1 hour.
A kilowatt is a unit of power; a kilowatt-hour is a unit of energy. The hour multiplies the power. Writing kW/h would mean a rate of change of power and does not describe the energy used here.
Use hours with kilowatts
The same heater, with a supplied tariff
The heater uses 1.15 kW for 12 minutes. First, t = 12/60 = 0.20 h.
Energy = 1.15 x 0.20 = 0.230 kW h. This is also 0.230 x 3 600 000 = 828 000 J.
Use an example tariff of $0.30 per kW h. Cost = 0.230 x $0.30 = $0.069, or 6.9 cents.
If the answer is required to the nearest cent, this is $0.07. Keep the unrounded energy and cost until the final step. The tariff here is a supplied example price.
Try it step by step
Work it out in steps: Power in kW, then Energy in kW h, then Cost in cents.
Show the answer
Problem 1.
Power in kW: P = 2000 / 1000 = 2.0 kW
Energy in kW h: E = 2.0 x 3.0 = 6.0 kW h
Cost in cents: Cost = 6.0 x 30 = 180 cents, which is $1.80
Problem 2.
Power in kW: P = 1500 / 1000 = 1.5 kW
Energy in kW h: E = 1.5 x 2.0 = 3.0 kW h
Cost in cents: Cost = 3.0 x 30 = 90 cents
Problem 3.
Power in kW: P = 800 / 1000 = 0.80 kW
Energy in kW h: E = 0.80 x 5.0 = 4.0 kW h
Cost in cents: Cost = 4.0 x 30 = 120 cents, which is $1.20
Compare the complete use pattern
Appliance A
0.10 kW for 5.0 h
Energy = 0.10 x 5.0 = 0.50 kW h.
Appliance B
1.0 kW for 0.25 h
Energy = 1.0 x 0.25 = 0.25 kW h.
B has ten times the power but runs for much less time. It uses half as much energy as A in these stated patterns. At the same tariff, its energy cost is also half as much.
More detail (optional)
A thermostat can turn a heater on and off. If a 1 kW heater is powered for only part of an hour, its energy use is not automatically 1 kW h. Use its total powered-on time at that power, or a supplied average power over the whole interval. Include any other energy use if the question specifies it.
Check the time unit. Twelve minutes is 720 s or 0.20 h. It is neither 12 h nor 0.12 h. Choose seconds for a calculation in joules and hours for a calculation using kilowatts and kilowatt-hours.
Check your understanding
A heater uses a constant 1.15 kW for 12 minutes. Using a supplied example tariff of $0.30 per kW h, what is the energy cost before rounding to the nearest cent?
Show the answer
$0.069, or 6.9 cents
12 minutes is 0.20 h. The energy is 1.15 x 0.20 = 0.230 kW h, so cost = 0.230 x $0.30 = $0.069, or 6.9 cents.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
A 2.0 kW heater is used for 3.0 hours each day for 30 days. Electricity costs 30 cents per kW h. Calculate the energy used in kW h and the total cost. [3]
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Exam-style question
Data-based question: Choosing a kettle [8]
A family compares three kettles. For each one they boil 1.0 L of water, starting from tap temperature, and time how long it takes. Electricity costs $0.30 per kW h.
| kettle | power / W | time to boil / s |
|---|---|---|
| A | 2200 | 150 |
| B | 3000 | 115 |
| C | 1800 | 190 |
- (a) Calculate the energy kettle A uses to boil 1.0 L of water. Give your answer in J. [2]
- (b) Convert your answer to (a) into kW h. [1]
- (c) The family boils 1.0 L five times a day. Calculate the daily cost of using kettle A. [2]
- (d) State one assumption you made in (c). [1]
- (e) A salesperson says: "Kettle B costs the most to run because it has the highest power." Use the data to evaluate this claim. [2]
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