Mathematics Standard • Year 11 • Module 2 • Lesson 18
Units of Energy and Mass — Problem Set
Apply mass, food-energy and electrical-energy conversions to realistic Australian household and nutrition scenarios.
Problem 1 — Pool pump annual cost
A household pool pump is rated at 750 W. The owner runs it for 8 hours per day, every day of the year. Electricity costs 32 cents per kWh.
Set up: What are we solving for?
(i) Calculate the daily energy use of the pump in kWh. 2 marks
(ii) Calculate the daily cost in dollars, to the nearest cent. 1 mark
(iii) Calculate the annual cost (365 days), to the nearest dollar. 2 marks
Stuck? Convert W → kW first, then use E = P × t per day, then multiply by 365.Problem 2 — Lunchbox energy audit
A student's lunchbox contains:
1 sandwich (energy = 1850 kJ)
1 apple (energy = 360 kJ)
1 muesli bar (energy = 756 kJ)
1 small drink (energy = 420 Cal — note the food Calorie unit)
Set up: What are we solving for?
(i) Convert the drink's energy from Calories to kilojoules, to 1 d.p. 1 mark
(ii) Calculate the total lunchbox energy in kilojoules. 2 marks
(iii) The student's recommended daily energy intake is 8700 kJ. Express the lunchbox total as a percentage of the daily intake, to 1 d.p. 2 marks
Stuck? Use 1 Cal (kcal) = 4.184 kJ. Add all four items in kJ before doing the percentage.Problem 3 — Medication course mass
A doctor prescribes a 250 mg antibiotic capsule three times a day for 7 days, followed by 125 mg twice a day for a further 5 days as a tapering dose.
Set up: What are we solving for?
(i) Calculate the total mass of antibiotic, in milligrams, taken during the 7-day full-dose phase. 1 mark
(ii) Calculate the total mass, in milligrams, during the 5-day tapering phase. 1 mark
(iii) Find the total mass of antibiotic taken across the whole course. Express your answer in grams. 3 marks
Stuck? Calculate each phase in mg, sum them, then divide by 1000 for grams.Problem 4 — Choosing a more efficient heater
A family is choosing between two electric heaters.
Heater A: rated 2400 W; needs to run 4 hours per day to warm the room.
Heater B: rated 1800 W; needs to run 6 hours per day for the same warming effect.
Set up: What are we solving for?
(i) Calculate the daily energy used by each heater in kWh. 2 marks
(ii) If electricity costs 34 cents per kWh, calculate the daily cost of each heater (to the nearest cent). 1 mark
(iii) Over a winter period of 90 days, which heater is cheaper to run, and by how much (to the nearest dollar)? State your conclusion clearly. 2 marks
Stuck? Higher power × less time can equal lower power × more time. Calculate each total — don't assume.Problem 5 — Calorie tracking and conversion
A fitness app tells a user they have burnt 480 Calories (kcal) in a 45-minute spin class. Their daily energy intake is 9200 kJ.
Set up: What are we solving for?
(i) Convert the 480 Calories burnt to kilojoules. 1 mark
(ii) Express the calories burnt as a percentage of the user's daily intake (in kJ), to 1 d.p. 2 marks
(iii) If the user wanted to burn 2500 kJ in one session at the same rate, how many minutes would they need to spin? (Round to the nearest minute.) 3 marks
Stuck on (iii)? First find the burn rate in kJ per minute from the original session, then divide 2500 by that rate.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Pool pump annual cost
Set up. Convert W → kW, find daily kWh, daily cost, then scale to annual.
(i) P = 750 ÷ 1000 = 0.75 kW; daily E = 0.75 × 8 = 6 kWh.
(ii) Daily cost = 6 × 32 = 192 cents = $1.92.
(iii) Annual cost = 1.92 × 365 = 700.8 ≈ $701.
Problem 2 — Lunchbox energy audit
Set up. Convert all items to a common unit (kJ), sum the total, then express as a percentage of 8700 kJ.
(i) Drink = 420 × 4.184 = 1757.28 ≈ 1757.3 kJ.
(ii) Total = 1850 + 360 + 756 + 1757.3 = 4723.3 kJ.
(iii) Percentage = 4723.3 ÷ 8700 × 100 ≈ 54.3% of daily intake.
Problem 3 — Medication course
Set up. Phase 1 total + Phase 2 total, then convert mg → g.
(i) Full dose: 250 × 3 × 7 = 5250 mg → 5250 mg.
(ii) Taper: 125 × 2 × 5 = 1250 mg → 1250 mg.
(iii) Total = 5250 + 1250 = 6500 mg = 6500 ÷ 1000 = 6.5 g.
Problem 4 — Heater comparison
Set up. Both heaters use the same daily energy if P × t is equal; calculate to confirm, then cost it out across the winter period.
(i) Heater A: 2.4 × 4 = 9.6 kWh per day. Heater B: 1.8 × 6 = 10.8 kWh per day.
(ii) A: 9.6 × 34 = 326.4 cents ≈ $3.26. B: 10.8 × 34 = 367.2 cents ≈ $3.67.
(iii) 90-day cost: A = 3.264 × 90 = $293.76; B = 3.672 × 90 = $330.48. Difference = $330.48 − $293.76 = $36.72 ≈ $37. Heater A is cheaper by about $37 over winter. (Common slip: assuming the higher-wattage heater is always more expensive — here it runs fewer hours and wins.)
Problem 5 — Calorie tracking
Set up. Convert calories to kJ; compare to daily intake; then derive a burn rate to invert the question.
(i) 480 × 4.184 = 2008.32 kJ (≈ 2008.3 kJ).
(ii) 2008.32 ÷ 9200 × 100 ≈ 21.8% of daily intake.
(iii) Burn rate = 2008.32 ÷ 45 ≈ 44.63 kJ/min. To burn 2500 kJ: 2500 ÷ 44.63 ≈ 56.02 ≈ 56 minutes.