Define the kilowatt-hour (kWh) and calculate the cost of using electrical appliances where the energy unit is the kWh

Published by Patrick Mutisya · 8 days ago

Cambridge IGCSE Physics 0625 – 4.2.4 Resistance: kWh and Cost Calculations

4.2.4 Resistance – Energy Consumption and Cost

Objective

Students will be able to:

  • Define the kilowatt‑hour (kWh) as a unit of energy.
  • Convert between kWh and joules.
  • Calculate the energy used by electrical appliances.
  • Determine the cost of running an appliance when the electricity price is given in £/kWh (or any other currency).

What is a kilowatt‑hour?

The kilowatt‑hour (kWh) is the amount of energy transferred when a power of 1 kilowatt (1000 W) is used for 1 hour.

Mathematically,

\$E = P \times t\$

where

  • \$E\$ = energy (in joules, J)
  • \$P\$ = power (in watts, W)
  • \$t\$ = time (in seconds, s)

Because 1 kW = 1000 W and 1 h = 3600 s, the conversion to joules is

\$1\;\text{kWh}=1000\;\text{W}\times3600\;\text{s}=3.6\times10^{6}\;\text{J}\$

Cost of electricity

If the electricity supplier charges \$C\$ (currency per kWh), the cost \$K\$ of using an appliance is

\$K = E_{\text{kWh}} \times C\$

where \$E_{\text{kWh}}\$ is the energy consumed expressed in kilowatt‑hours.

Step‑by‑step calculation

  1. Find the power rating of the appliance (usually given in watts, W).
  2. Convert the power to kilowatts: \$P{\text{kW}} = \dfrac{P{\text{W}}}{1000}\$.
  3. Determine how long the appliance is used (in hours), \$t_{\text{h}}\$.
  4. Calculate the energy used: \$E{\text{kWh}} = P{\text{kW}} \times t_{\text{h}}\$.
  5. Multiply by the unit price of electricity to obtain the cost.

Example

A 1500 W electric heater is switched on for 3 hours each day. The electricity price is £0.20 per kWh.

  1. Power in kilowatts: \$P_{\text{kW}} = 1500/1000 = 1.5\;\text{kW}\$.
  2. Energy per day: \$E_{\text{kWh}} = 1.5\;\text{kW} \times 3\;\text{h} = 4.5\;\text{kWh}\$.
  3. Cost per day: \$K = 4.5\;\text{kWh} \times £0.20/\text{kWh} = £0.90\$.
  4. Cost per month (30 days): \$£0.90 \times 30 = £27.00\$.

Typical household appliances

AppliancePower rating (W)Daily use (h)Energy per day (kWh)Cost per day (at £0.20/kWh)
LED lamp (5 W)55\$\dfrac{5}{1000}\times5 = 0.025\$£0.005
Refrigerator (150 W)15024\$\dfrac{150}{1000}\times24 = 3.6\$£0.72
Electric kettle (2000 W)20000.5\$\dfrac{2000}{1000}\times0.5 = 1.0\$£0.20
Television (100 W)1004\$\dfrac{100}{1000}\times4 = 0.40\$£0.08
Washing machine (800 W)8001\$\dfrac{800}{1000}\times1 = 0.80\$£0.16

Suggested diagram: A simple schematic showing a power rating label on an appliance, a clock indicating usage time, and a calculator displaying the cost calculation.

Key points to remember

  • 1 kWh = 3.6 MJ (megajoules).
  • Always convert watts to kilowatts before using the kWh formula.
  • The cost of running an appliance depends on both its power rating and the duration of use.
  • Reducing the time an appliance is on, or using lower‑power alternatives, directly lowers the electricity bill.