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Year 12 Maths Standard 2 (2027) Ratios and rates

Energy Rates & Kilowatt Hours

20 practice questions 0 video lessons Theory + worked examples

Learn energy rates and kilowatt-hours for NSW Year 12 Mathematics Standard 2. Power is measured in watts and kilowatts (\(1000\text{ W}=1\text{ kW}\)), and the energy an appliance uses in kilowatt-hours (kWh) is its power in kilowatts multiplied by the hours it runs.

You will convert watts to kilowatts, use energy \(=\) power \(\times\) time to find the kWh used, and calculate the cost of running an appliance from the electricity tariff in cents or dollars per kWh β€” a core rates skill for Standard 2 that lets you compare appliances and read an electricity bill.

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Theory

Energy rates and kilowatt-hours are part of the rates topic in Year 12 Standard 2 (NSW). This guide shows how to convert watts to kilowatts, find the energy an appliance uses in kilowatt-hours with energy \(=\) power \(\times\) time, and cost that energy using the electricity tariff in cents or dollars per kWh.

Power measures how fast an appliance uses energy. It is given in watts (W) or kilowatts (kW), where \(1000\text{ W}=1\text{ kW}\). To convert watts to kilowatts, divide by \(1000\) (for example \(2500\text{ W}=2.5\text{ kW}\)).

A kilowatt-hour (kWh) is a unit of energy β€” the energy used by a \(1\text{ kW}\) appliance running for one hour. The energy an appliance uses is \(\text{energy}=\text{power}\times\text{time}\), with power in kW and time in hours.

In this Year 12 Standard 2 (NSW) topic the cost of running an appliance is the energy in kWh multiplied by the electricity tariff (the price per kWh). A tariff quoted in cents is changed to dollars by dividing by \(100\), so \(30\)c per kWh is \(\$0.30\) per kWh.

Energy used vs running timeA straight line E=2t through the origin for a 2 kW appliance; 5 hours uses 10 kWh t E 2 4 6 8 4 8 12 16 20 5 10 E=2t
Energy \(E=2t\) for a \(2\text{ kW}\) appliance β€” the gradient is the power.
Running cost vs energy for two tariffsTwo lines through the origin; the 45c/kWh tariff is steeper than the 30c/kWh tariff E C 20 40 60 80 100 10 20 30 40 45c/kWh 30c/kWh
Cost rises with energy; a steeper line is a dearer tariff (c/kWh).

Convert the power rating to kilowatts:

\[P\ (\text{kW}) = \dfrac{P\ (\text{W})}{1000}\]
P(kW)=P(W)1000

Energy used, in kilowatt-hours, is power (kW) times time (hours):

\[E = P \times t\]
E=P×t

The cost of that energy is the kilowatt-hours times the tariff (price per kWh):

\[\text{Cost} = E \times \text{tariff}\]
Cost=E×tariff
Watch the units. Power must be in kW and time in hours to get kWh, and a tariff in cents must be turned into dollars (\(30\)c \(=\$0.30\)) before you multiply.

How to cost the energy an appliance uses

  1. Power to kW. If the rating is in watts, divide by \(1000\) to get kilowatts.
  2. Energy. Multiply power (kW) by the running time in hours: \(E=P\times t\). Add up all the hours (per day \(\times\) number of days) if needed.
  3. Cost. Multiply the energy in kWh by the tariff. Convert a cents tariff to dollars first by dividing by \(100\).
  4. Interpret. State the answer with units (kWh or \(\$\)); to compare appliances, cost each one and subtract to find the saving.
Example 1 β€” Watts to energy
A clothes dryer is rated at \(2500\) W. How much energy does it use if it runs for \(3\) hours?
Solution

Convert the power to kilowatts, then use energy \(=\) power \(\times\) time.

\(P\)\(=\)\(\dfrac{2500}{1000} = 2.5\text{ kW}\)
\(E\)\(=\)\(P\times t = 2.5\times 3\)
\(\)\(=\)\(7.5\text{ kWh}\)
2.5×3=7.5

The dryer uses \(7.5\text{ kWh}\).

Example 2 β€” Cost over a period
A portable heater rated \(1800\) W is used \(5\) hours a day for \(14\) days. Electricity costs \(32\) cents per kWh. Find the total cost.
Solution

Find the total energy in kWh, then charge it at \(\$0.32\) per kWh.

\(P\)\(=\)\(\dfrac{1800}{1000} = 1.8\text{ kW}\)
\(E\)\(=\)\(1.8\times 5\times 14 = 126\text{ kWh}\)
\(\text{Cost}\)\(=\)\(126\times 0.32 = \$40.32\)
126×0.32=40.32

The heater costs \(\$40.32\) over the \(14\) days.

Example 3 β€” Comparing appliances
Heater A is rated \(2400\) W and Heater B \(1500\) W. Each runs \(4\) hours a day for \(90\) days at \(30\) cents per kWh. How much is saved by using Heater B?
Solution

Cost each heater for the \(90\) days, then subtract.

\(\text{cost}_A\)\(=\)\(2.4\times4\times90\times0.30 = \$259.20\)
\(\text{cost}_B\)\(=\)\(1.5\times4\times90\times0.30 = \$162.00\)
\(\text{saving}\)\(=\)\(259.20-162.00 = \$97.20\)

Heater B is cheaper to run, saving \(\$97.20\).

Example 4 β€” A monthly bill
An electric hot-water system is rated \(3600\) W and heats water for an average of \(2.5\) hours a day. Off-peak power is \(24\) cents per kWh. Find (i) the power in kW, (ii) the energy used each day, (iii) the cost for a \(30\)-day month.
Solution

Convert to kW, find the daily energy, then cost \(30\) days.

\(\text{(i) } P\)\(=\)\(\dfrac{3600}{1000} = 3.6\text{ kW}\)
\(\text{(ii) } E\)\(=\)\(3.6\times 2.5 = 9\text{ kWh/day}\)
\(\text{(iii) Cost}\)\(=\)\(9\times30\times0.24 = \$64.80\)
270×0.24=64.80

(i) \(3.6\text{ kW}\); (ii) \(9\text{ kWh}\); (iii) \(\$64.80\).

Common pitfalls

kW is power, kWh is energy. A \(2\text{ kW}\) rating is not \(2\text{ kWh}\); you must multiply by the hours it runs to get the energy.
Convert watts first. Change the rating to kilowatts before multiplying by time: \(1500\text{ W}=1.5\text{ kW}\), not \(1500\text{ kW}\).
Match the tariff units. \(30\)c per kWh is \(\$0.30\), so cost in dollars \(=\) kWh \(\times\,0.30\) β€” do not multiply by \(30\).

Frequently asked questions

What is a kilowatt-hour (kWh)?

A kilowatt-hour is a unit of energy: the amount of energy used by a 1 kilowatt appliance running for one hour. It is what your electricity bill charges you for. Energy in kWh equals the power in kilowatts multiplied by the running time in hours.

What is the difference between a kilowatt and a kilowatt-hour?

A kilowatt (kW) measures power, which is how fast energy is used. A kilowatt-hour (kWh) measures the total energy used. You get energy in kWh by multiplying the power in kW by the number of hours the appliance runs.

How do you convert watts to kilowatts?

Divide the number of watts by 1000, because 1000 watts equals 1 kilowatt. For example, a 2500 watt dryer is 2500 divided by 1000, which is 2.5 kilowatts. Always convert to kilowatts before working out energy in kWh.

How do you calculate the cost of running an appliance?

First find the energy used in kWh by multiplying the power in kilowatts by the hours it runs. Then multiply that energy by the electricity tariff, the price per kWh. Convert a tariff given in cents to dollars first, so 30 cents per kWh is $0.30 per kWh.

Why do you divide the tariff by 100?

Electricity tariffs are often quoted in cents per kilowatt-hour, but costs are usually written in dollars. Dividing the cents by 100 changes the tariff to dollars per kWh, so 24 cents per kWh becomes $0.24 per kWh before you multiply by the energy used.

How can kilowatt-hours help compare two appliances?

Work out the energy each appliance uses over the same time, then multiply by the tariff to get each running cost. The appliance that uses fewer kWh costs less to run, and subtracting the two costs gives the saving from choosing the more efficient one.