Conduction of Heat

Admin / March 13, 2026

Conduction of Heat

Understanding how thermal energy transfers through materials without the material itself moving

Heat Transfer through Solids

What is Conduction?

Particle Vibration

Heat conduction occurs when vibrating particles pass their kinetic energy to neighboring particles.

Energy Transfer

Thermal energy moves from the hotter end to the colder end of an object.

Temperature Gradient

The greater the temperature difference, the faster the rate of conduction.

Key Concept

Conduction is the main method of heat transfer in solids. In metals, free electrons also help transfer heat very efficiently.

Particle Model of Conduction

How Heat Travels Through Materials

Cold End

Particles vibrate slowly

Heat Source →

Particles vibrate rapidly

Energy transfers from hot to cold as vibrating particles collide with their neighbors

Good vs Poor Conductors

Good Conductors

Materials that transfer heat quickly:

Silver - Best conductor
Copper - Used in pans
Aluminum - Cooking foil
Iron - Radiators

High Thermal Conductivity

Poor Conductors (Insulators)

Materials that transfer heat slowly:

Wood - Handles
Plastic - Coatings
Glass - Windows
Air - Trapped in insulation

Low Thermal Conductivity

Why are metals good conductors?

Metals have free electrons that can move through the metal and transfer kinetic energy quickly from the hot end to the cold end. This makes them much better conductors than non-metals.

Thermal Conductivity Values

Material	Thermal Conductivity (W/mK)	Type	Common Use
Silver	429	Excellent Conductor	Electronics
Copper	401	Excellent Conductor	Cooking pans, wires
Aluminum	237	Good Conductor	Foil, pans
Iron	80	Moderate Conductor	Radiators
Glass	0.8	Poor Conductor	Windows
Wood	0.15	Insulator	Handles, furniture
Air	0.024	Excellent Insulator	Double glazing

Higher values mean better conduction of heat

Interactive Conduction Experiment

See how different materials conduct heat at different rates:

20°C

Heat Intensity

50%

Observations:

Copper

Heats fastest

Iron

Heats quickly

Glass

Heats slowly

Wood

Heats very slowly

Factors Affecting Conduction

Temperature Difference

Greater temperature difference = faster heat transfer

Cross-sectional Area

Thicker materials conduct more heat

Length

Shorter distance = faster conduction

Material Type

Different materials have different conductivity

Time

Longer time = more heat transferred

Rate of conduction ∝ (Area × Temperature Difference) ÷ (Length × Material Resistance)

Real-World Applications

Copper or aluminum bases conduct heat quickly and evenly to food.

Metal conducts heat from hob to food
Wood/plastic handles are insulators

Materials like fiberglass trap air to reduce heat loss through walls and roofs.

Loft insulation
Cavity wall insulation
Double glazing

Vacuum flasks use a vacuum to prevent conduction and keep drinks hot or cold.

Double-walled construction
Vacuum prevents conduction

Metal fins on electronics conduct heat away from components to prevent overheating.

Made of copper or aluminum
Large surface area helps cooling

Insulation in Daily Life

Clothing

Trapped air between fibers insulates our bodies

Upholstery

Foam and fabric reduce heat loss

Double Glazing

Air gap between glass panes

Coolers

Expanded polystyrene keeps food cold

Pizza Boxes

Corrugated cardboard traps air

Safety Gear

Heat-resistant gloves for handling hot objects

Solved Examples

GCSE Foundation

Example 1: Identifying Conductors

Question: Explain why a metal spoon in a hot drink becomes hot to touch, but a plastic spoon does not.

GCSE Higher

Example 2: Comparing Conductivity

Question: A copper rod and a glass rod of the same dimensions are heated at one end. After 2 minutes, the copper rod is hot along its entire length, but the glass rod is only hot at the heated end. Explain why.

Grade 10 Challenge

Example 3: Practical Application

Question: A house has single-glazed windows. Suggest two ways to reduce heat loss through the windows and explain how they work.

Quick Facts

Best Conductor Silver

Best Insulator Vacuum

Metals conduct via Free electrons

Non-metals conduct via Particle vibration

Conservation and dissipation of energy

Admin / March 13, 2026

Conservation & Dissipation of Energy

Understanding how energy is transferred, stored, and wasted in physical systems

Energy cannot be created or destroyed

The Law of Conservation of Energy

"Energy cannot be created or destroyed, only transferred from one store to another or dissipated to the surroundings."

⚡

Input Energy

Total energy going into a system

✓

Useful Energy

Energy transferred to where it's wanted

✗

Dissipated Energy

Energy wasted to surroundings

Key Principle

Total energy input = Useful energy output + Wasted energy output

No energy is ever lost - it just becomes less useful!

Energy Stores & Transfers

Kinetic Energy

Energy of moving objects

Example: A moving car, running athlete

Thermal Energy

Energy of hot objects

Example: Hot coffee, warm radiator

Gravitational Potential

Energy due to height

Example: Water behind a dam, lifted weight

Chemical Energy

Energy stored in chemical bonds

Example: Food, batteries, fuel

Elastic Potential

Energy in stretched/compressed objects

Example: Stretched spring, drawn bow

Nuclear Energy

Energy stored in atomic nuclei

Example: Nuclear power, sun

Dissipation of Energy

When energy is transferred, some is always dissipated (wasted) to the surroundings, usually as thermal energy.

Energy Flow in a Light Bulb

10 J/s

Light Energy (Useful)

90 J/s

Heat Energy (Wasted)

100 J/s

Total Input

Only 10% of electrical energy is converted to light - the rest is dissipated as heat!

Common Ways Energy is Dissipated:

Friction - Kinetic energy → Thermal energy
Sound - Energy transferred as noise
Air Resistance - Objects heating the air
Electrical Resistance - Wires heating up

Efficiency

Efficiency tells us how much of the input energy is transferred usefully.

As a Decimal

Efficiency =
Useful Output ÷ Total Input

As a Percentage

Efficiency =
(Useful ÷ Total) × 100%

Example: LED Bulb (90% Efficient)

90% Useful

Example: Filament Bulb (10% Efficient)

10% Useful

Efficiency of Common Devices:

LED Bulb

90%

Electric Motor

85%

Car Engine

25%

Filament Bulb

10%

Solar Panel

20%

Sankey Diagrams

Sankey diagrams show energy flow - the width of the arrows represents the amount of energy.

LED Light Bulb

100 J Input

90 J Light

10 J Heat

Efficiency: 90%

Car Engine

100 J Fuel

25 J Motion

75 J Heat + Sound

Efficiency: 25%

The thicker the arrow, the more energy is transferred that way!

Interactive Efficiency Calculator

Calculate Efficiency

Useful Energy Output (J)

75 J

Total Energy Input (J)

100 J

Efficiency

75%

Wasted Energy: 25 J

Real-World Applications

Electric Vehicles

Electric cars are much more efficient than petrol cars.

85%

Electric Motor

25%

Petrol Engine

Electric cars waste less energy as heat

Home Insulation

Reduces energy dissipation from houses.

30%

Heat Saved

£300

Yearly Saving

Loft insulation, double glazing, cavity walls

LED Lighting

LED bulbs are 90% efficient vs 10% for filament bulbs.

90%

LED

10%

Filament

Uses 90% less electricity for same light

Regenerative Braking

Captures kinetic energy that would otherwise be wasted as heat.

70%

Energy Recovered

30%

Range Increase

Used in electric and hybrid vehicles

Solved Examples

GCSE Foundation

Example 1: Calculating Efficiency

Question: A motor transfers 500J of electrical energy. It produces 400J of kinetic energy. Calculate its efficiency.

GCSE Higher

Example 2: Finding Wasted Energy

Question: A television with 250W input power is 65% efficient. Calculate the power wasted.

Grade 10 Challenge

Example 3: Sankey Diagram

Question: A kettle uses 2000J of electrical energy. 1500J heats the water, the rest is wasted. Draw a Sankey diagram and calculate efficiency.

Energy Efficiency of Common Devices

Device	Input Energy	Useful Output	Wasted Output	Efficiency
LED Light Bulb	100 J	90 J (light)	10 J (heat)	90%
Electric Motor	100 J	85 J (movement)	15 J (heat)	85%
Solar Panel	100 J	20 J (electricity)	80 J (heat)	20%
Car Engine	100 J	25 J (movement)	75 J (heat, sound)	25%
Filament Bulb	100 J	10 J (light)	90 J (heat)	10%

Reducing Energy Dissipation

Lubrication

Reduces friction between moving parts, reducing heat dissipation.

Example: Oil in car engines

Insulation

Traps air to reduce thermal energy transfer.

Example: Loft insulation, double glazing

Streamlining

Reduces air resistance, saving energy.

Example: Aerodynamic cars

Low Resistance

Using materials with lower electrical resistance.

Example: Superconductors, thicker wires

Key Facts

Conservation Law Energy can't be destroyed

Dissipation Energy spreads out

Maximum Efficiency 100% (theoretically)

Wasted Energy Usually heat

power

Admin / February 16, 2026

Power

Understanding the rate of energy transfer - from light bulbs to power plants

P = E/t = W/t = VI

What is Power?

Definition

Power is the rate at which energy is transferred or work is done. It tells us how quickly energy is being used or generated.

Key Concept

A high power device transfers energy quickly, while a low power device transfers energy slowly.

Unit

Power is measured in watts (W). 1 watt = 1 joule per second (1 W = 1 J/s)

Real-World Understanding

A 100W light bulb converts 100 joules of electrical energy into light and heat every second. A 2000W kettle converts 2000 joules per second - that's why it heats water much faster!

Power Formulas

Basic Power Formula

P = E / t

Power (W)

Energy (J)

Time (s)

Work Formula

P = W / t

Power (W)

Work (J)

Time (s)

Electrical Power

P = V × I

Power (W)

Voltage (V)

Current (A)

Alternative Electrical

P = I²R

Power (W)

Current (A)

Resistance (Ω)

Formula Rearrangements

Find Energy

E = P × t

Find Time

t = E ÷ P

Find Current

I = P ÷ V

Find Voltage

V = P ÷ I

Interactive Power Calculator

Adjust the sliders to see how power, energy, and time are related:

Energy Transfer Simulator

Power (Watts)

100 W

Time (seconds)

60 s

Energy Used

6000 J

E = P × t

Energy Transfer Rate

100 J/s

Your device transfers 100 joules of energy every second

In 60 seconds, total energy = 6000 J

Household Appliance Power Ratings

Different appliances have different power ratings - see how they compare:

LED Light Bulb

10W

Uses 10 J/s
Cost: ~£0.03 per day
100 hours on 1 kWh

Laptop

50W

Uses 50 J/s
Cost: ~£0.15 per day
20 hours on 1 kWh

LED TV (50")

100W

Uses 100 J/s
Cost: ~£0.30 per day
10 hours on 1 kWh

Refrigerator

150W

Uses 150 J/s (avg)
Cost: ~£0.45 per day
Runs 30% of time

Kettle

2000W

Uses 2000 J/s
Cost: ~£0.10 per boil
30 minutes on 1 kWh

Electric Shower

9000W

Uses 9000 J/s
Cost: ~£0.45 per 10 mins
6.7 minutes on 1 kWh

Key Point: High power appliances (like kettles and showers) transfer energy very quickly, which is why they heat things rapidly but also cost more to run.

Real-World Power Examples

Solar Panel

Typical residential solar panel produces about 300W of power in full sunlight.

300W

Peak Power

1.5kWh

Daily Energy

Car Engine

A typical family car engine produces around 100,000W (100kW) of power.

100kW

Engine Power

134hp

Horsepower

Wind Turbine

Modern wind turbines can generate up to 3,000,000W (3MW) of power.

3MW

Peak Power

2000 homes

Can Power

Human Power

An average healthy human can sustain about 100W of mechanical power output.

100W

Sustained

400W

Peak (short)

Energy Efficiency & Cost

Appliance Energy Cost Calculator

Appliance Power (W)

Hours Used per Day

Electricity Rate (p/kWh)

Efficiency Ratings (A+++ to G)

A+++ D G

A+++

Most Efficient

Saves ~60% energy

Average

Standard efficiency

Least Efficient

Costs 3x more to run

Solved Examples

GCSE Foundation

Example 1: Energy Transfer

Question: A 2.5kW kettle is used for 3 minutes. How much energy does it transfer?

GCSE Higher

Example 2: Electrical Power

Question: A hairdryer operates at 230V and draws a current of 4.35A. Calculate its power and the energy used in 5 minutes.

Grade 10 Challenge

Example 3: Cost Calculation

Question: A 2kW electric heater runs for 4 hours daily. If electricity costs 28p per kWh, calculate the weekly running cost.

Power Calculator

Solve Power Problems

Energy (J)

Time (s)

Power (W)

Time (s)

Energy (J)

Power (W)

Voltage (V)

Current (A)

Result:

Power Rating Comparison

Device	Typical Power	Energy in 1 hour	Cost per hour*
LED Light Bulb	10W	0.01 kWh	0.28p
Laptop	50W	0.05 kWh	1.4p
LED TV (50")	100W	0.1 kWh	2.8p
Fridge Freezer	150W	0.15 kWh	4.2p
Desktop Computer	300W	0.3 kWh	8.4p
Microwave	800W	0.8 kWh	22.4p
Kettle	2000W	2.0 kWh	56p
Electric Shower	9000W	9.0 kWh	£2.52

*Based on 28p per kWh electricity rate

Quick Facts

1 kW 1000 watts

1 MW 1,000,000 watts

1 GW 1,000,000,000 watts

1 hp 746 watts

1 kWh 3.6 million J

Common Power Ratings

LED Bulb 5-15W

Phone Charger 5-20W

Laptop 30-100W

TV 50-250W

Kettle 2000-3000W

Book Free Demo Class

Common Mistake

Don't confuse power (watts) with energy (joules or kWh)! Power is how FAST you use energy, energy is HOW MUCH you use.

Analogy: Power is like speed (mph), Energy is like distance (miles).

Physics

Specific Heat Capacity

Admin / February 16, 2026

Specific Heat Capacity

Understanding why different materials heat up at different rates - essential for GCSE Physics

What is Specific Heat Capacity?

Specific heat capacity (c) is the amount of energy required to raise the temperature of 1 kg of a substance by 1°C. It's a measure of how much energy a material can store.

High c

Lots of energy needed to change temperature. Material heats slowly, cools slowly. Example: Water (4180 J/kg°C)

Low c

Little energy needed to change temperature. Material heats quickly, cools quickly. Example: Copper (385 J/kg°C)

Key Formula

Q = m × c × ΔT
Energy = mass × specific heat capacity × temperature change

Real-World Analogy

Think of specific heat capacity as a material's "thermal inertia." Water has high thermal inertia (like a heavy object that's hard to push) - it resists temperature changes. Metals have low thermal inertia (like a light object that's easy to push) - their temperature changes easily.

The Specific Heat Capacity Formula

Specific Heat Capacity Formula

Q = m × c × ΔT

Where energy transferred (Q) equals mass (m) times specific heat capacity (c) times temperature change (ΔT)

Energy (joules, J)

Mass (kilograms, kg)

Specific Heat Capacity (J/kg°C)

ΔT

Temperature Change (°C)

Rearranging the Formula

We can rearrange the formula to solve for any variable:

Find Energy (Q)

Q = m × c × ΔT

Find Mass (m)

m = Q ÷ (c × ΔT)

Find c

c = Q ÷ (m × ΔT)

Find ΔT

ΔT = Q ÷ (m × c)

Interactive Heating Simulation

Watch how different materials heat up at different rates due to their specific heat capacities.

Material:

Mass (kg): 1.0

1.0 kg

Heater Power (W): 1000

1000 W

Time (s): 60

60 s

Heating Calculations

Energy Supplied

60,000 J

Q = Power × Time

Temperature Rise

15.6 °C

ΔT = Q ÷ (m × c)

Final Temperature

35.6 °C

Starting at 20°C

ΔT = Q ÷ (m × c) = 60,000 J ÷ (1.0 kg × 385 J/kg°C) = 15.6 °C

Notice how materials with lower specific heat capacity (like copper) heat up much faster than water with the same energy input!

Comparing Specific Heat Capacities

Different materials have very different specific heat capacities. Here's how they compare:

Material	Specific Heat Capacity (J/kg°C)	Heating Rate*	Practical Significance
Water	4180	Very Slow	Excellent heat store, moderates climate
Wood	1700	Slow	Good insulator, feels warm to touch
Aluminum	900	Moderate	Cooking pans, heats evenly
Iron	450	Fast	Radiators, heats quickly
Copper	385	Very Fast	Electrical wires, heat exchangers
Lead	130	Extremely Fast	Lowest common metal, feels cold

*Heating rate comparison assumes same mass and energy input

Temperature Rise vs. Specific Heat Capacity

For the same energy input, materials with lower specific heat capacity experience greater temperature increases.

Practical Applications

Specific heat capacity has important real-world applications:

Central Heating Systems

Water is used in radiators because its high specific heat capacity means it can carry lots of heat around a building without cooling down too quickly.

Comparison: 1 kg of water cooling 1°C releases 4180 J, while 1 kg of iron cooling 1°C releases only 450 J.

Cooking & Food

Different materials in cookware heat differently. Copper-bottom pans heat quickly (low c), while cast iron retains heat well (moderate c but high density).

Example: Water in food prevents burning - it absorbs heat without getting too hot (high c).

Climate Moderation

Oceans and large lakes moderate coastal climates because water's high c means it heats and cools slowly, preventing extreme temperature changes.

Fact: Coastal areas have milder winters and cooler summers than inland areas at the same latitude.

Experiment: Comparing Water and Oil

A common GCSE experiment compares the specific heat capacity of water and oil by heating equal masses with identical heaters and measuring temperature rise.

Water Beaker

Specific Heat: 4180 J/kg°C

Temperature Rise: 5.0 °C

Observation: Heats slowly

Oil Beaker

Specific Heat: ~2000 J/kg°C

Temperature Rise: 10.5 °C

Observation: Heats quickly

With the same heater and same time, oil heats about twice as much as water because it has about half the specific heat capacity.

Solved Example Problems

Example 1: Heating Water

GCSE Foundation

Calculate the energy required to heat 2.5 kg of water from 20°C to 80°C. The specific heat capacity of water is 4200 J/kg°C.

Step 1: Write the formula

Q = m × c × ΔT

Step 2: Identify known values

m = 2.5 kg, c = 4200 J/kg°C, T₁ = 20°C, T₂ = 80°C

Step 3: Calculate temperature change

ΔT = T₂ - T₁ = 80°C - 20°C = 60°C

Step 4: Substitute values into formula

Q = 2.5 × 4200 × 60

Step 5: Perform multiplication

2.5 × 4200 = 10,500

10,500 × 60 = 630,000

Step 6: State the answer with units

Q = 630,000 J or 630 kJ

Step 7: Interpretation

It takes 630 kJ of energy to heat 2.5 kg of water by 60°C. Water's high specific heat capacity means it requires a lot of energy to heat up.

Example 2: Finding Specific Heat Capacity

GCSE Higher

In an experiment, a 0.8 kg block of metal is heated using a 500 W heater for 3 minutes. Its temperature increases from 25°C to 85°C. Calculate the specific heat capacity of the metal.

Step 1: Calculate energy supplied

Power = 500 W, Time = 3 minutes = 180 seconds

Energy = Power × Time = 500 × 180 = 90,000 J

Step 2: Write the specific heat capacity formula

Q = m × c × ΔT

Step 3: Rearrange to solve for c

c = Q ÷ (m × ΔT)

Step 4: Identify known values

Q = 90,000 J, m = 0.8 kg, ΔT = 85°C - 25°C = 60°C

Step 5: Substitute values into formula

c = 90,000 ÷ (0.8 × 60)

Step 6: Calculate denominator

0.8 × 60 = 48

c = 90,000 ÷ 48

Step 7: Perform division

90,000 ÷ 48 = 1,875

Step 8: State the answer with units

c = 1875 J/kg°C

Step 9: Interpretation

The metal has a specific heat capacity of 1875 J/kg°C, which is moderately high (between aluminum and wood).

Example 3: Cooling Problem

Grade 10

A 0.5 kg copper kettle (c = 385 J/kg°C) containing 1.2 kg of water (c = 4200 J/kg°C) cools from 95°C to 25°C. Calculate the total energy released to the surroundings.

Step 1: Calculate energy released by copper kettle

Q_copper = m × c × ΔT = 0.5 × 385 × (95 - 25)

Q_copper = 0.5 × 385 × 70 = 13,475 J

Step 2: Calculate energy released by water

Q_water = m × c × ΔT = 1.2 × 4200 × (95 - 25)

Q_water = 1.2 × 4200 × 70 = 352,800 J

Step 3: Calculate total energy released

Q_total = Q_copper + Q_water

Q_total = 13,475 + 352,800 = 366,275 J

Step 4: Convert to kJ and round appropriately

366,275 J = 366.275 kJ ≈ 366 kJ

Step 5: State the answer with units

Total energy released = 366 kJ

Step 6: Interpretation

Notice that although the copper has lower specific heat capacity, the water releases much more energy because it has more mass and much higher c. The water accounts for 96% of the total energy released!

Practice Problems

Test your understanding with these specific heat capacity problems:

Problem 1: Heating Aluminum

GCSE Foundation

A 1.2 kg aluminum pan (c = 900 J/kg°C) is heated from 20°C to 180°C. Calculate the thermal energy required.

Problem 2: Finding Temperature Change

GCSE Higher

When 75,000 J of energy is supplied to 3 kg of iron (c = 450 J/kg°C), calculate the temperature increase.

Problem 3: Mixed Materials

Grade 10 Challenge

A 0.25 kg copper container (c = 385 J/kg°C) holds 0.8 kg of water (c = 4200 J/kg°C). The system is heated from 15°C to 85°C.

Calculate the energy needed to heat the water.
Calculate the energy needed to heat the copper container.
Calculate the total energy required.

Water Energy (J):

Copper Energy (J):

Total Energy (J):

Specific Heat Capacity Calculator

Use this calculator to solve for any variable in the specific heat capacity equation (Q = m × c × ΔT).

Solve Specific Heat Problems

Mass (kg)

Specific Heat (J/kg°C)

ΔT (°C)

Q (J)

Specific Heat (J/kg°C)

ΔT (°C)

Q (J)

Mass (kg)

ΔT (°C)

Q (J)

Mass (kg)

Specific Heat (J/kg°C)

Result:

Practical Applications

Free Demo Class

Master specific heat capacity calculations and applications with our expert tutors in an interactive online session

Book Free Demo

Limited spots available for Year 9-10 students

Quick Tip: Water's Special Property

Water has an unusually high specific heat capacity (4180 J/kg°C) compared to most common materials. This makes it excellent for storing heat, which is why it's used in central heating systems and why large bodies of water moderate climate.

Units Check

Always ensure your units are consistent: mass in kg, specific heat in J/kg°C, temperature change in °C, and energy in joules (J). 1 kJ = 1000 J.

Common Mistake

Don't confuse specific heat capacity with thermal conductivity! Specific heat capacity tells us how much energy is needed to change temperature. Thermal conductivity tells us how quickly heat travels through a material.

Quick Reference

Water: 4180 J/kg°C
Ice: 2100 J/kg°C
Aluminum: 900 J/kg°C
Iron: 450 J/kg°C
Copper: 385 J/kg°C
Lead: 130 J/kg°C

IGCSE physics