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NSWMaths Standard 2Syllabus dot point

How do you measure energy in joules and kilowatt-hours, work out what an appliance costs to run, and balance the kilojoules you eat against the energy your body needs each day?

Solve problems involving energy and mass, including the joule and the watt, energy consumption measured in kilowatt-hours, the energy content of food measured in kilojoules, and daily energy requirements based on basal metabolic rate and level of activity

A focused answer to the HSC Maths Standard 2 dot point on energy and nutrition. The joule and the watt, measuring electricity use in kilowatt-hours, costing an appliance from its power rating and tariff, reading food energy in kilojoules, and estimating daily energy needs from basal metabolic rate times an activity factor, with worked Australian examples.

Generated by Claude Opus 4.815 min answer

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What this dot point is asking

NESA wants you to measure energy and use it in everyday decisions about electricity and food. Energy is measured in joules, and the rate at which energy is used is power, measured in watts. From there you need to do three practical jobs: work out how much electricity an appliance uses in kilowatt-hours and what it costs from a tariff; read the energy content of food in kilojoules and total it for a meal; and estimate a person's daily energy needs from their basal metabolic rate and how active they are. The arithmetic is gentle - mostly multiply and divide. The marks come from picking the right formula, keeping power in kilowatts and time in hours, and writing the unit on every answer.

The answer

There are two ideas to keep separate. Energy is the total amount of "work" done or heat produced, measured in joules (J\text{J}). Power is how fast that energy is used, measured in watts (W\text{W}), where 11 watt is 11 joule per second. An appliance's power rating (printed on it as a number of watts or kilowatts) tells you its power, and the longer it runs the more energy it uses. That single relationship runs the whole topic:

energy=power×time.\text{energy} = \text{power} \times \text{time}.

For electricity we measure energy in kilowatt-hours rather than joules, because joules are tiny for household amounts. Putting power in kilowatts and time in hours gives the energy straight in kilowatt-hours, and that is the "unit" you are billed for.

Daily running cost of four household appliancesA bar chart of the cost to run four appliances for one day at thirty cents per kilowatt-hour. The heater costs about two dollars sixteen, the fridge about one dollar eight, the dryer about seventy-five cents and the television about fifteen cents. A taller bar means a higher daily cost.Daily running cost (at $0.30 per kWh)$0$0.50$1.00$1.50$2.00$2.50HeaterFridgeDryerTV$2.16$1.08$0.75$0.15cost = power (kW) × hours × price per kWh

Joules and watts: energy and power

The base unit of energy is the joule (J\text{J}). It is a small amount - lifting an apple about a metre is roughly one joule - so for everyday energy we use larger units like the kilojoule (11 kJ =1000= 1000 J) for food, and the kilowatt-hour for electricity.

Power is the rate of using energy, measured in watts (W\text{W}), where

1 W=1 joule per second.1 \text{ W} = 1 \text{ joule per second}.

So a 10001000 W appliance uses 10001000 joules every second. Because watts are small for whole appliances, ratings are often given in kilowatts (11 kW =1000= 1000 W). A 24002400 W heater and a 2.42.4 kW heater are the same thing. The key skill is moving between watts and kilowatts: divide by 10001000 going to kilowatts, multiply by 10001000 going back to watts.

Energy in kilowatt-hours

A kilowatt-hour (kWh\text{kWh}) is the energy a 11 kW appliance uses in 11 hour. It is the "unit" of electricity on a bill. To find the energy an appliance uses, multiply its power in kilowatts by the time it runs in hours:

energy (kWh)=power (kW)×time (h).\text{energy (kWh)} = \text{power (kW)} \times \text{time (h)}.

The two units must match the formula: power in kilowatts, time in hours. A 22 kW heater for 33 hours uses 2×3=62 \times 3 = 6 kWh. A 150150 W (0.150.15 kW) fridge running all day uses 0.15×24=3.60.15 \times 24 = 3.6 kWh. If a power is given in watts, convert to kilowatts first; if a time is given in minutes, convert to hours first (divide by 6060).

The cost of running an appliance

Electricity is priced as a tariff: so many cents or dollars per kilowatt-hour. Once you have the energy in kilowatt-hours, the cost is just the energy times the price per unit:

cost=energy (kWh)×price per kWh.\text{cost} = \text{energy (kWh)} \times \text{price per kWh}.

A 2.42.4 kW air conditioner running 55 hours uses 2.4×5=122.4 \times 5 = 12 kWh, and at $0.30 per kWh that costs 12×0.30=3.6012 \times 0.30 = 3.60, that is $3.60. Keep the tariff in dollars (a 3030 c tariff is $0.30) so the answer comes out in dollars. Many bills also have a fixed daily supply charge added on top; unless a question gives one, the usage cost is the energy times the tariff.

Food energy in kilojoules

The energy in food is measured in kilojoules (kJ\text{kJ}); it is the chemical energy your body can release from the food. (You may see the old unit, the Calorie, where 11 Cal 4.2\approx 4.2 kJ, but Australian labels and NESA use kilojoules.) Food energy is additive: the energy of a meal is the sum of the energy of its parts. Labels often quote energy "per 100100 g", so for a different amount you scale:

energy=amount100×(energy per 100 g).\text{energy} = \frac{\text{amount}}{100} \times (\text{energy per }100\text{ g}).

For example, butter at 30003000 kJ per 100100 g gives 10100×3000=300\tfrac{10}{100} \times 3000 = 300 kJ in a 1010 g serve.

Daily energy needs: BMR times activity

Your body uses energy even at complete rest - to breathe, pump blood and stay warm. That baseline rate is the basal metabolic rate (BMR), in kilojoules per day. Being active uses more, so your total daily need is the BMR scaled up by a physical activity level (PAL), also called an activity factor:

daily energy need=BMR×PAL.\text{daily energy need} = \text{BMR} \times \text{PAL}.

A sedentary person has a PAL near 1.41.4; someone moderately active is around 1.61.6 to 1.751.75; a very active person can be over 22. So a BMR of 70007000 kJ with a PAL of 1.61.6 gives 7000×1.6=112007000 \times 1.6 = 11\,200 kJ per day. The average-adult reference figure you see on packaging is 87008700 kJ, which is a population average, not a personal target. Comparing a meal or a day's needs to 87008700 kJ as a percentage is a common exam step: divide by 87008700 and multiply by 100100.

Weight stays steady when the energy you eat balances the energy you use. The schematic compares the energy coming in as food with the energy going out as your basal metabolic rate plus activity. When the two columns match, energy is balanced; if the food column is taller the surplus is stored, and if it is shorter the body draws on its stores.

Energy balance: energy in from food versus energy outTwo columns of equal height. The left column is energy in from food, eleven thousand two hundred kilojoules. The right column is energy out, made of a basal metabolic rate of seven thousand kilojoules plus activity of four thousand two hundred kilojoules, also totalling eleven thousand two hundred. Because the columns match, energy is balanced and weight is steady.Energy balance: in vs out (kJ per day)11 200foodEnergy INBMR 7000+ activity 4200Energy OUTcolumns match → balanced → steady weight

How exam questions ask about energy and nutrition

The wording tells you which formula to reach for:

  • "What is the power rating ... in kilowatts / in watts" is a watt-kilowatt conversion: divide by 10001000 to kilowatts, multiply by 10001000 to watts.
  • "How much energy ... in kWh" or "how many units of electricity" wants power (kW)×time (h)\text{power (kW)} \times \text{time (h)} - convert watts to kilowatts and minutes to hours first.
  • "How much does it cost to run ..." wants the energy in kWh times the tariff; write the tariff in dollars and watch for a per-day, per-week or per-year time span.
  • "How much would they save ..." asks for the difference between two costs (a higher and a lower tariff, or longer and shorter run time).
  • "Find the total energy of the meal / how many kilojoules" wants you to add the kilojoule figures, scaling any per-100100-g item first.
  • "Estimate the daily energy requirement" wants BMR×PAL\text{BMR} \times \text{PAL}; if a PAL is described in words ("moderately active"), use the value the question supplies.
  • "As a percentage of the recommended intake" means divide by the reference (usually 87008700 kJ) and multiply by 100100.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC-style4 marksA dishwasher has a power rating of 1.81.8 kW and takes 1.51.5 hours to complete one wash cycle. A household runs 44 cycles each week, for all 5252 weeks of the year. Electricity costs $0.29 per kWh. (a) Find the energy used in one cycle, in kWh. (b) Find the total cost of running the dishwasher for the year.
Show worked answer →

Award one mark for energy per cycle, energy = power times time = 1.8 times 1.5 = 2.7 kWh. For part (b), one mark for the total energy, 2.7 times 4 times 52 = 561.6 kWh; one mark for multiplying by the tariff, 561.6 times 0.29; and one mark for the correct cost $162.86 (accept $162.864 rounded to the nearest cent). A common error is leaving the power in watts, giving an answer 1000 times too large, or costing one cycle instead of the year, drop the relevant marks but follow through if the method is otherwise correct.

2021 HSC-style3 marksA breakfast consists of two slices of toast at 650650 kJ each, one boiled egg at 300300 kJ, and a glass of juice at 450450 kJ. (a) Find the total energy of the breakfast, in kilojoules. (b) The average adult daily energy intake is 87008700 kJ. Express the breakfast as a percentage of this figure, correct to one decimal place.
Show worked answer →

Part (a): one mark for the correct total, 2 times 650 + 300 + 450 = 2050 kJ. Part (b): one mark for setting up the percentage, (2050 divided by 8700) times 100, and one mark for the answer 23.6 percent (accept 23.56 to 23.6). Markers look for the two slices of toast being doubled; using a single slice (1400 kJ total) loses the first mark but can still earn the percentage mark by follow-through.

2020 HSC-style5 marksA swimming pool pump is rated at 1.11.1 kW and runs for 88 hours each day over a 3030-day month. Electricity costs $0.28 per kWh. (a) Find the total energy used in the month, in kWh. (b) Find the cost for the month. (c) The owner reduces the daily running time to 55 hours. Find how much they save over the month.
Show worked answer →

Part (a): one mark for daily energy 1.1 times 8 = 8.8 kWh, one mark for the monthly total 8.8 times 30 = 264 kWh. Part (b): one mark for 264 times 0.28 = $73.92. Part (c): one mark for the reduced cost (1.1 times 5 times 30 = 165 kWh, 165 times 0.28 = $46.20) and one mark for the saving $73.92 minus $46.20 = $27.72. Accept the shortcut for part (c) of costing the 3-hour daily reduction directly: 1.1 times 3 times 30 times 0.28 = $27.72. Keeping the tariff in cents instead of dollars is the most common fatal slip.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksAn appliance has a power rating of 15001500 W. (a) Express this power in kilowatts. (b) A second appliance is rated at 2.42.4 kW. Express its power in watts.
Show worked solution →

Part (a) - watts to kilowatts divides by 10001000. The prefix kilo means 10001000, so 11 kW =1000= 1000 W and a watt is the smaller unit. Changing to the larger unit makes the number smaller, so divide:

1500÷1000=1.51500 \div 1000 = 1.5

so the appliance is rated at 1.51.5 kW.

Part (b) - kilowatts to watts multiplies by 10001000. Changing to the smaller unit makes the number bigger, so multiply:

2.4×1000=24002.4 \times 1000 = 2400

so the second appliance is rated at 24002400 W. (Power, in watts, measures how fast energy is used; the kilowatt is just 10001000 watts.)

foundation2 marksA 22 kW fan heater runs for 33 hours. How much energy does it use, in kilowatt-hours (kWh)?
Show worked solution →

Use energy == power ×\times time. Energy in kilowatt-hours is the power in kilowatts multiplied by the time in hours:

E=2×3E = 2 \times 3

Evaluate.

E=6 kWhE = 6 \text{ kWh}

so the heater uses 66 kWh. (One kilowatt-hour is the energy of a 11 kW appliance running for 11 hour, so a 22 kW heater for 33 hours is 2×3=62 \times 3 = 6 of them.)

foundation2 marksA breakfast is made up of a bowl of cereal (800800 kJ), a glass of milk (500500 kJ) and a banana (400400 kJ). Find the total energy of the breakfast in kilojoules.
Show worked solution →

Add the energy of each item. Food energy is additive, so total the three kilojoule figures:

800+500+400=1700800 + 500 + 400 = 1700

so the breakfast provides 17001700 kJ. (Energy on a food label is measured in kilojoules; you simply add the components of a meal.)

core3 marksA 2.42.4 kW reverse-cycle air conditioner runs for 55 hours on a hot day. Electricity costs $0.30 per kilowatt-hour. (a) Find the energy used in kilowatt-hours. (b) Find the cost of running it for the day.
Show worked solution →

Part (a) - energy is power times time.

E=2.4×5=12 kWhE = 2.4 \times 5 = 12 \text{ kWh}

so the air conditioner uses 1212 kWh.

Part (b) - cost is energy times the price per unit. Each kilowatt-hour costs $0.30, so multiply the energy by the tariff:

cost=12×0.30=3.60\text{cost} = 12 \times 0.30 = 3.60

so it costs $3.60 to run for the day. (The unit on an electricity bill is the kilowatt-hour; the bill is the number of units times the price per unit.)

core3 marksA woman has a basal metabolic rate (BMR) of 70007000 kJ per day. Her physical activity level (PAL) is 1.61.6. (a) Estimate her total daily energy requirement. (b) The average adult reference intake is 87008700 kJ. Express her requirement as a percentage of this reference, correct to the nearest whole percent.
Show worked solution →

Part (a) - multiply BMR by the activity factor. Daily energy need is the basal metabolic rate multiplied by the physical activity level:

energy=7000×1.6=11200 kJ\text{energy} = 7000 \times 1.6 = 11\,200 \text{ kJ}

so she needs about 1120011\,200 kJ per day.

Part (b) - express as a percentage of 87008700. Divide by the reference and multiply by 100100:

112008700×100=128.7\frac{11\,200}{8700} \times 100 = 128.7\ldots

which rounds to 129%129\%. (Her needs are above the average reference because the 87008700 kJ figure is a population average, not a personalised target.)

exam5 marksA 3.53.5 kW ducted air conditioner runs for 66 hours every day across a 9090-day summer. (a) Find the total energy used over the summer, in kilowatt-hours. (b) On a peak tariff of $0.33 per kWh, find the total cost. (c) The household could instead run it overnight on an off-peak tariff of $0.18 per kWh. How much would they save over the summer by using the off-peak tariff?
Show worked solution →

Part (a) - energy per day, then over the summer. Energy each day is power times hours:

3.5×6=21 kWh per day3.5 \times 6 = 21 \text{ kWh per day}

Over 9090 days:

21×90=1890 kWh21 \times 90 = 1890 \text{ kWh}

so the air conditioner uses 18901890 kWh across the summer.

Part (b) - cost on the peak tariff. Multiply the energy by the tariff of $0.33:

1890×0.33=623.701890 \times 0.33 = 623.70

so the peak-tariff cost is $623.70.

Part (c) - saving on the off-peak tariff. The off-peak cost is

1890×0.18=340.201890 \times 0.18 = 340.20

so the off-peak bill is $340.20. The saving is the difference:

623.70340.20=283.50623.70 - 340.20 = 283.50

so the household saves $283.50 over the summer. (A faster route for part (c) is to multiply the energy by the difference in tariff: 1890×(0.330.18)=1890×0.15=283.501890 \times (0.33 - 0.18) = 1890 \times 0.15 = 283.50, the same answer.)

exam5 marksA man has a basal metabolic rate of 72007200 kJ per day and a physical activity level of 1.751.75. (a) Estimate his daily energy requirement. (b) He actually consumes 1400014\,000 kJ each day. Find his daily energy surplus. (c) A surplus is stored as body fat, and gaining 11 kg of fat requires a surplus of about 3700037\,000 kJ. Estimate how many days of this surplus would add 11 kg, correct to one decimal place.
Show worked solution →

Part (a) - energy requirement is BMR times PAL.

7200×1.75=12600 kJ7200 \times 1.75 = 12\,600 \text{ kJ}

so he needs about 1260012\,600 kJ per day.

Part (b) - surplus is intake minus requirement.

1400012600=1400 kJ14\,000 - 12\,600 = 1400 \text{ kJ}

so he has a surplus of 14001400 kJ each day.

Part (c) - days to store 11 kg of fat. Divide the energy needed for 11 kg by the daily surplus:

370001400=26.4\frac{37\,000}{1400} = 26.4\ldots

which rounds to 26.426.4 days. (This is why a small daily surplus matters: 14001400 kJ is only a couple of snacks, yet it adds about a kilogram a month.)

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