Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

6.5.3Chemical Reactions

Kilowatt-Hours — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 16 of 21 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
0%read

Section 1 of 5The Kilowatt-Hour

Kilowatt-hour (kWh)
  • Kilowatt-hour (kWh) A unit of energy equal to the energy transferred by a 1 kW appliance operating for one hour.
Conversion
  • Conversion 1 kWh = 1000 W × 3600 s = 3 600 000 J, which is 3.6 × 10⁶ J.
Use in homes
  • Use in homes Electricity meters measure energy in kilowatt-hours, which is the unit used on household electricity bills.
Why not joules
  • Why not joules The kilowatt-hour gives much smaller and more practical numbers for household energy use than the joule.

Section 2 of 5Calculating Energy and Cost

1
Convert the power of the appliance into kilowatts by dividing its power in watts by 1000.
2
Convert the time into hours, so that 30 minutes becomes 0.5 hours.
3
Multiply the power in kilowatts by the time in hours to find the energy in kilowatt-hours.
4
Multiply the energy in kilowatt-hours by the cost per kilowatt-hour to find the total cost.

Section 3 of 5Energy Equations

Energy Equations
Energy in kWh
  • Energy in kWh Energy (kWh) = power (kW) × time (h), so a 2 kW appliance used for 3 hours uses 6 kWh.
Cost
  • Cost Cost = energy (kWh) × cost per kWh, usually given in pence or pounds.
Comparing appliances
  • Comparing appliances Appliances with equal power use energy in proportion to how long they are switched on.
Saving money
  • Saving money Reducing the power rating or the time an appliance is used reduces the energy and the cost of electricity.

Section 4 of 5Worked Example

Question

  • A 2.0 kW electric heater is used for 3.0 hours.
  • Electricity costs 30 p per kWh.
  • Calculate (a) the energy used in kWh and (b) the total cost in pounds.

  1. (a) Energy = power × time = 2.0 kW × 3.0 h = 6.0 kWh.
  2. (b) Cost = 6.0 kWh × 30 p per kWh = 180 p.
  3. Convert to pounds180 p ÷ 100 = £1.80.

Answer The heater uses 6.0 kWh of energy, which costs £1.80.

Section 5 of 5Exam Checklist & Tips

6.5.3 – Key Points for the Exam

  • Kilowatt-hour (kWh) A unit of energy equal to 3.6 × 10⁶ J.
  • Energy meter A device that measures the energy supplied to a home in kilowatt-hours.
  • Cost per unit The price charged for each kilowatt-hour of electricity.
  • Joule (J) The SI unit of energy.
  • Power The rate of energy transfer, measured in kilowatts or watts.

Common Mistakes

  • Using watts with hours directly, which gives watt-hours, without dividing by 1000 to get kilowatt-hours.
  • Confusing kilowatt-hours with kilowatts; kWh is a unit of energy and kW is a unit of power.
  • Forgetting to convert minutes into hours before multiplying by power.
  • Quoting the cost in pence and then in pounds without converting correctly.

Last updated:

\n