Energy in the Home

This lesson covers the domestic use of energy, including power, energy calculations, kilowatt-hours, and reducing energy use in the home, as required by the Edexcel GCSE Physics specification (1PH0), Topic 3: Conservation of Energy. You need to be able to calculate energy costs and evaluate energy-saving measures.

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Power

Power is the rate at which energy is transferred (or the rate at which work is done). It tells you how quickly a device transfers energy.

The Equation

$P = \frac{E}{t}$P=tE​

Where:

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

Units

• 1 watt (W) = 1 joule per second (1 J/s).
• A 100 W light bulb transfers 100 J of energy every second.
• 1 kilowatt (kW) = 1000 W.

Exam Tip: Always check whether the question gives power in watts or kilowatts, and time in seconds or hours. Convert to consistent units before calculating.

Rearranging

Find this	Equation
Power	$P = \frac{E}{t}$P=tE​
Energy	$E = P \times t$E=P×t
Time	$t = \frac{E}{P}$t=PE​

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Worked Examples — Power and Energy

Example 1: Finding Power

Question: A motor transfers 6000 J of energy in 30 seconds. What is its power?

Solution:

$P = \frac{E}{t} = \frac{6000}{30} = 200 \text{ W}$P=tE​=306000​=200 W

Example 2: Finding Energy

Question: A 2000 W kettle is on for 3 minutes. How much energy does it transfer?

Solution:

Convert time: 3 minutes = 3 × 60 = 180 s.

$E = P \times t = 2000 \times 180 = 360{,}000 \text{ J} = 360 \text{ kJ}$E=P×t=2000×180=360,000 J=360 kJ

Example 3: Finding Time

Question: A 60 W light bulb transfers 216 000 J of energy. How long was it on for?

Solution:

$t = \frac{E}{P} = \frac{216{,}000}{60} = 3600 \text{ s} = 1 \text{ hour}$t=PE​=60216,000​=3600 s=1 hour

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The Kilowatt-Hour (kWh)

Electricity companies do not measure energy in joules — the joule is too small a unit for domestic use. Instead, they use the kilowatt-hour (kWh).

1 kilowatt-hour is the amount of energy transferred by a 1 kW device operating for 1 hour.

Calculating Energy in kWh

$E \text{ (kWh)} = P \text{ (kW)} \times t \text{ (hours)}$E (kWh)=P (kW)×t (hours)

Important: When using this equation, power must be in kW and time must be in hours.

Converting Between kWh and Joules

$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3{,}600{,}000 \text{ J} = 3.6 \text{ MJ}$1 kWh=1000 W×3600 s=3,600,000 J=3.6 MJ

Worked Example

Question: A 3 kW oven is used for 2 hours. How many kWh of energy does it use?

Solution:

$E = P \times t = 3 \times 2 = 6 \text{ kWh}$E=P×t=3×2=6 kWh

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Calculating the Cost of Electricity

$\text{Cost} = \text{Energy (kWh)} \times \text{Price per kWh}$Cost=Energy (kWh)×Price per kWh

Or, combining with the energy equation:

$\text{Cost} = P \text{ (kW)} \times t \text{ (hours)} \times \text{Price per kWh}$Cost=P (kW)×t (hours)×Price per kWh

Worked Examples

Example 1: Basic Cost Calculation

Question: A 2 kW heater is used for 5 hours. Electricity costs 30p per kWh. What is the cost?

Solution:

$E = 2 \times 5 = 10 \text{ kWh}$E=2×5=10 kWh $\text{Cost} = 10 \times 30 = 300 \text{p} = £3.00$Cost=10×30=300p=£3.00

Example 2: Comparing Appliances

Question: Which costs more to run — a 100 W light bulb on for 10 hours, or a 3 kW kettle on for 10 minutes? (Price = 30p per kWh)

Solution:

Light bulb: $E = 0.1 \text{ kW} \times 10 \text{ h} = 1 \text{ kWh}$E=0.1 kW×10 h=1 kWh $\text{Cost} = 1 \times 30 = 30 \text{p}$Cost=1×30=30p

Kettle: $E = 3 \text{ kW} \times \frac{10}{60} \text{ h} = 3 \times 0.167 = 0.5 \text{ kWh}$E=3 kW×6010​ h=3×0.167=0.5 kWh $\text{Cost} = 0.5 \times 30 = 15 \text{p}$Cost=0.5×30=15p

The light bulb costs more (30p vs 15p) because it is on for much longer.

Example 3: Finding Cost of Daily Use

Question: A household uses the following appliances daily:

Appliance	Power	Time Used
LED lights	0.05 kW	8 hours
TV	0.15 kW	4 hours
Washing machine	2 kW	1 hour
Oven	3 kW	1.5 hours

Electricity costs 30p per kWh. Calculate the daily cost.

Solution:

Appliance	Energy (kWh)	Cost (p)
LED lights	0.05 × 8 = 0.4	0.4 × 30 = 12p
TV	0.15 × 4 = 0.6	0.6 × 30 = 18p
Washing machine	2 × 1 = 2.0	2.0 × 30 = 60p
Oven	3 × 1.5 = 4.5	4.5 × 30 = 135p
Total	7.5 kWh	225p = £2.25

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Reducing Energy Use in the Home

Reducing energy use saves money and reduces CO₂ emissions. Here are the main methods:

Insulation Methods

Method	How It Works	Typical Cost	Annual Saving	Payback Time
Loft insulation	Reduces thermal energy transfer through the roof by trapping air	£300	£150	2 years
Cavity wall insulation	Foam injected into gap between inner and outer walls — reduces convection	£500	£100	5 years
Double glazing	Two panes of glass with trapped air/vacuum between — reduces conduction and convection	£5000	£100	50 years
Draught excluders	Strips around doors/windows — reduce convection (draughts)	£50	£50	1 year
Hot water tank jacket	Insulation around hot water tank — reduces thermal energy transfer	£20	£30	8 months

Exam Tip: The values in the table above are approximate and used for illustration. In the exam, you will be given specific values. The key is to be able to calculate payback time and evaluate which measures are most cost-effective.

Other Energy-Saving Measures

• LED bulbs instead of filament bulbs — LEDs are much more efficient (about 90% vs 10%).
• Smart meters — show real-time energy use, encouraging reduction.
• Energy-efficient appliances (A+++ rated) — use less energy for the same job.
• Turning off appliances when not in use — reduces standby power consumption.
• Lower thermostat setting — reducing room temperature by 1 °C can save about 10% on heating bills.

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Payback Time