Units of Energy and Mass
From the joules in a lightning bolt to the kilojoules on a food label — energy and mass have a hierarchy of units. Convert fluently between them and you can answer any real-world measurement question.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Your household electricity bill charges you by the kilowatt-hour (kWh). A standard electric kettle is rated at 2400 W. If you boil the kettle for 3 minutes, how much energy does it use — and roughly what does that cost if electricity is 30 cents per kWh?
Make a rough guess before the lesson teaches you the method.
Mass and energy both have a hierarchy of units — each step multiplies or divides by a fixed factor. Identify the direction first: going to a smaller unit means multiply; going to a larger unit means divide.
Mass chain: mg → g → kg → t, each step ×1000 going down or ÷1000 going up.
Energy chain: J → kJ → MJ (÷1000 each step). Food uses kcal: 1 kcal = 4.184 kJ. Electricity uses kWh: 1 kWh = 3600 kJ.
Key facts
- The hierarchy of mass units: mg, g, kg, tonne
- The hierarchy of energy units: J, kJ, MJ
- The calorie conversions: 1 cal = 4.184 J; 1 kcal = 4.184 kJ
- Power–energy–time: $E \text{ (kWh)} = P \text{ (kW)} \times t \text{ (h)}$
- 1 kWh = 3600 kJ
Concepts
- Why electricity is billed in kWh rather than joules
- The difference between a calorie (small) and a Calorie/kcal (food label)
- How to convert minutes to hours before using the energy formula
Skills
- Convert between mass units in multi-step problems
- Convert between joules, kilojoules, and calories
- Calculate energy consumption using $E = P \times t$
- Calculate electricity costs from energy used and a given rate
Mass units follow the same pattern as length: each step up the chain multiplies by 1000. The key is always identifying which direction you are converting.
To go smaller (e.g. kg → g): multiply by 1000. To go larger (e.g. g → kg): divide by 1000.
What to write in your book
- Mass chain: mg → g → kg → t. Each step: ÷1000 going to larger units; ×1000 going to smaller units.
- To convert 3.25 kg to g: $3.25 \times 1000 = 3250$ g (going smaller, multiply).
- To convert 4500 mg to g: $4500 \div 1000 = 4.5$ g (going larger, divide).
- Multi-step: 2.4 t → kg → g: $2.4 \times 1000 = 2400$ kg; $2400 \times 1000 = 2\,400\,000$ g.
Quick check: A truck carries 3.6 t of goods. The truck itself has a mass of 8500 kg. What is the total mass in tonnes?
Worked examples · 3 in a row, reveal as you go
A medication tablet contains 250 mg of active ingredient. A patient takes 3 tablets per day for 14 days. (a) How many grams of active ingredient does the patient take in total? (b) Express this as kilograms.
A muesli bar has an energy content of 756 kJ per serve. (a) Convert this to joules. (b) Convert this to Calories (kcal), correct to the nearest whole number.
An electric kettle has a power rating of 2400 W. It is used for 3 minutes each morning. Electricity costs 32 cents per kWh. (a) How many kWh does the kettle use each morning? (b) What is the daily cost, to the nearest tenth of a cent? (c) What is the annual cost (365 days), to the nearest dollar?
What to write in your book
- Energy formula: $E \text{ (kWh)} = P \text{ (kW)} \times t \text{ (h)}$. P in watts must be divided by 1000. t in minutes must be divided by 60.
- $1 \text{ kWh} = 3600 \text{ kJ}$ — useful for converting from electricity units to scientific energy units.
- Cost = Energy (kWh) × price per kWh. Keep track of units: if price is in cents, answer is in cents; divide by 100 for dollars.
- The capital C in "Calorie" = kilocalorie = 1000 cal = 4.184 kJ. Never confuse a Cal with a cal.
True or false: When a food label says "350 Calories", it means 350 joules of energy.
Common errors · the 3 traps that cost marks
What to write in your book
- Check: have I converted P to kW and t to hours? Do this as Step 1 every time.
- $1 \text{ kJ} = 1000 \text{ J}$; $1 \text{ MJ} = 1000 \text{ kJ}$; $1 \text{ kcal} = 4.184 \text{ kJ}$; $1 \text{ kWh} = 3600 \text{ kJ}$.
- Mass direction check: are you going from a big unit to a small unit? Multiply. Small to big? Divide.
- Food-label trap: "Calories" (capital C) = kcal. Multiply by 4.184 to get kJ.
Fill the gap: A 1200 W toaster is used for 4 minutes. Converting: $P = 1200 \div 1000 =$ kW; $t = 4 \div 60 \approx$ h; $E = 1.2 \times 0.0\overline{6} \approx$ kWh.
Quick-fire practice · 5 calculations
Convert: (a) 3.25 kg to g (b) 4500 mg to g (c) 2.4 t to kg (d) 0.5 mg to g
Three packages have masses of 450 g, 1.2 kg, and 800 g. Find the total mass in kilograms.
Convert: (a) 5000 J to kJ (b) 1050 kJ to Cal (kcal, 1 d.p.) (c) 8700 kJ to MJ
A pool pump is rated at 0.75 kW and runs for 8 hours per day. How many kWh does it use per day? If electricity costs 28 cents per kWh, find the weekly cost.
A household uses 22 kWh of electricity in a day. If electricity costs 35 cents per kWh, find the daily cost in dollars.
Odd one out: Three of these are correct statements about the energy formula. Which one is wrong?
Earlier you estimated the kettle energy and cost. Let's calculate:
$P = 2400 \text{ W} = 2.4 \text{ kW}$; $t = 3 \text{ min} = 0.05 \text{ h}$. So $E = 2.4 \times 0.05 = 0.12 \text{ kWh}$.
Cost at 30 cents/kWh: $0.12 \times 30 = 3.6 \text{ cents}$ — less than 4 cents per boil. Over a year: $3.6 \times 365 = 1314 \text{ c} \approx \$13.14$. Your kettle habit costs about $13 per year.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A portable speaker is charged using a 5 W USB cable for 2.5 hours. (a) Calculate the energy used to charge the speaker in kWh. (b) Convert this energy to kilojoules. (c) If electricity costs 30 cents per kWh, find the cost to charge the speaker. (3 marks)
Q2. A household appliance runs for 3 hours at a power of 800 W. Electricity is charged at 34 cents per kWh. (a) Find the energy used in kWh. (b) Find the cost of running the appliance. (c) If the appliance is used every day, find the annual cost to the nearest dollar. (4 marks)
Q3. An athlete burns 2800 Cal (kcal) per day through exercise and metabolism. (a) Convert 2800 Cal to kilojoules. (b) Convert your answer to megajoules, correct to 2 decimal places. (c) If the athlete's daily food provides 12 500 kJ, by how many kilojoules does their intake compare to their energy requirement? (4 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: (a) $3250$ g (b) $4.5$ g (c) $2400$ kg (d) $0.0005$ g
Drill 2: $450\text{ g} = 0.45\text{ kg}$; total $= 0.45 + 1.2 + 0.8 = 2.45$ kg
Drill 3: (a) $5$ kJ · (b) $1050 \div 4.184 \approx 250.9$ Cal · (c) $8.7$ MJ
Drill 4: $E = 0.75 \times 8 = 6$ kWh/day; weekly $= 6 \times 7 = 42$ kWh; cost $= 42 \times 28 = 1176\text{ c} = \$11.76$
Drill 5: $22 \times 0.35 = \$7.70$
Q1 (3 marks): (a) $P = 0.005$ kW; $E = 0.005 \times 2.5 = 0.0125$ kWh [1]. (b) $0.0125 \times 3600 = 45$ kJ [1]. (c) $0.0125 \times 30 = 0.375$ cents [1].
Q2 (4 marks): (a) $P = 0.8$ kW; $E = 0.8 \times 3 = 2.4$ kWh [1]. (b) $2.4 \times 34 = 81.6$ cents $= \$0.816$ [1]. (c) Annual: $0.816 \times 365 = 297.84 \approx \$298$ [2].
Q3 (4 marks): (a) $2800 \times 4.184 = 11\,715.2$ kJ [1]. (b) $11\,715.2 \div 1000 = 11.72$ MJ [1]. (c) Intake $12\,500$ kJ vs requirement $11\,715.2$ kJ; intake exceeds requirement by $12\,500 - 11\,715.2 = 784.8$ kJ [2].
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