Radiation

1. Overview

Thermal radiation is the transfer of energy by infrared waves. Unlike conduction and convection, radiation does not require a medium (matter) to travel through, making it the only way heat can reach Earth from the Sun through the vacuum of space.

Key Definitions

• Thermal Radiation: Infrared radiation (electromagnetic waves) emitted by the surface of an object due to its temperature.
• Infrared (IR): A type of electromagnetic radiation with longer wavelengths than visible light.
• Emitter: An object that sends out (emits) radiation.
• Absorber: An object that takes in (absorbs) radiation.
• Reflector: An object that bounces radiation off its surface.
• Vacuum: A space entirely devoid of matter.

Core Content

Nature of Thermal Radiation

• Thermal radiation is Infrared radiation.
• All objects, regardless of their temperature, emit some thermal radiation.
• The hotter an object is, the more infrared radiation it radiates per second.

Transfer without a Medium

• Radiation does not require particles. It can travel through a vacuum (empty space) at the speed of light.
• This is how energy from the Sun reaches the Earth.

Effect of Surface Colour and Texture The ability of a surface to emit, absorb, or reflect radiation depends on its appearance:

Surface	Absorption	Emission	Reflection
Black	Excellent	Excellent	Poor
Dull / Matt	Good	Good	Poor
White	Poor	Poor	Good
Shiny / Silvery	Very Poor	Very Poor	Excellent

Extended Content (Extended Curriculum Only)

Constant Temperature and Energy Balance

• For an object to remain at a constant temperature, it must transfer energy away (emit) at the same rate it receives energy (absorbs).
• Heating Up: If the rate of absorption > rate of emission, the object’s temperature increases.
• Cooling Down: If the rate of absorption < rate of emission, the object’s temperature decreases.

Earth’s Temperature Balance The Earth’s temperature is determined by the balance between:

1. Incoming Radiation: Short-wavelength radiation from the Sun.
2. Outgoing Radiation: Long-wavelength infrared radiation emitted by the Earth’s surface back into space.

• Factors such as atmosphere composition (greenhouse gases) and surface cover (ice reflects more, dark oceans absorb more) affect this balance.

Experiments: Distinguishing Emitters and Absorbers

• To test Emission (Leslie Cube): Fill a hollow metal cube (with different surface sides: shiny silver, matt black, etc.) with boiling water. Use an infrared detector to measure the radiation from each side at the same distance. The matt black side will show the highest reading.
• To test Absorption: Place two metal plates (one matt black, one shiny silver) at equal distances from a heater. Attach a thermometer or a small coin with wax to the back of each. The wax on the matt black plate will melt first because it absorbs radiation faster.

Factors Affecting Rate of Emission

1. Surface Temperature: The higher the temperature of an object relative to its surroundings, the greater the rate at which it emits radiation.
2. Surface Area: The larger the surface area, the greater the rate of emission (more "space" for the waves to leave the object).

Key Equations

Note: In IGCSE Physics, radiation is largely descriptive. However, understand the relationship: $\text{Power (Rate of emission)} \propto \text{Surface Area} \times \text{Temperature}^4$ (You do not usually need to calculate $T^4$, but you must know that a small increase in temperature leads to a massive increase in the rate of radiation.)

Common Mistakes to Avoid

• ❌ Wrong: Two objects at the same temperature must emit radiation at the same rate.
• ✓ Right: Even at the same temperature, a matt black object will emit radiation faster than a shiny silver one.
• ❌ Wrong: Radiation only works in a vacuum.
• ✓ Right: Radiation travels through vacuums, but it also travels through transparent solids, liquids, and gases (like air).
• ❌ Wrong: Using "Shiny Black" as a good absorber.
• ✓ Right: Any degree of shininess increases reflection. A "matt black" surface is the best absorber/emitter.
• ❌ Wrong: Thinking black surfaces "hold onto" heat better.
• ✓ Right: If a surface is a good absorber, it is always a good emitter. Black objects cool down faster than silver ones if they start at the same high temperature.

Exam Tips

1. Context Matters: If a question asks why a house is painted white in a hot country, focus on it being a poor absorber (to keep it cool). If it asks why a cooling fin on a computer is black, focus on it being a good emitter (to get rid of heat).
2. Vacuum Keyword: If the question mentions "vacuum" or "space," the answer is almost certainly radiation, as conduction and convection require particles.
3. Two-Way Street: Always remember that if a surface is "good" at taking heat in (absorbing), it is also "good" at letting heat out (emitting).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student places two identical metal containers, one with a shiny silver surface and the other with a dull black surface, equidistant from a radiant heater. Both containers are filled with the same mass of water at the same initial temperature.

(a) State which container will heat up faster. [1]

(b) Explain your answer to part (a) in terms of radiation. [2]

(c) State one way to improve the reliability of this experiment. [1]

(d) Suggest why it is important that the containers are the same distance from the heater. [1]

Worked Solution:

(a)

1. The container with the dull black surface will heat up faster. [Black surfaces are better absorbers of radiation]

How to earn full marks:

• State the correct surface colour.

(b)

1. The dull black surface is a better absorber of infrared radiation. [Absorption explanation]
2. Therefore, it absorbs more of the thermal radiation emitted by the heater per unit time compared to the shiny silver surface. [Rate of absorption explanation]

How to earn full marks:

• Mention that the black surface absorbs radiation better than the silver surface.
• Relate the rate of absorption to the rate of heating.

(c)

1. Use a larger temperature difference between the heater and the containers. [Improves sensitivity]
2. Repeat the experiment multiple times and calculate an average for each container. [Improves reliability]
3. Use a lid on each container to minimise heat loss by convection. [Controls variables]

How to earn full marks:

• State a valid method to improve reliability or control variables.

(d)

1. To ensure that both containers receive the same amount of thermal radiation from the heater. [Ensures equal exposure]

How to earn full marks:

• State that both containers need to receive the same radiation.

Common Pitfall: Many students confuse absorption and emission. Remember that dull black surfaces are both good absorbers AND good emitters of thermal radiation, while shiny surfaces are poor absorbers and emitters. Also, be specific about infrared radiation when discussing thermal effects.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A solar water heater on a house roof uses a black metal panel to absorb solar radiation. Water flows through pipes attached to the panel, absorbing thermal energy.

(a) Explain why the outer surface of the solar panel is painted black. [2]

(b) State two other features of the solar water heater, other than its colour, that are designed to maximise the absorption of solar radiation. [2]

(c) On a particular day, $5.0 \times 10^6 \text{ J}$ of energy from solar radiation falls on the panel in 1 hour. Only 30% of this energy is transferred to the water. Calculate the power output to the water, in watts. [4]

Worked Solution:

(a)

1. Black surfaces are good absorbers of thermal radiation. [Absorption explanation]
2. Painting the panel black maximises the amount of solar radiation absorbed, which increases the thermal energy transferred to the water. [Maximises energy transfer]

How to earn full marks:

• Mention that black surfaces are good absorbers.
• Relate the absorption to the thermal energy transferred.

(b)

1. The panel is made of metal, which is a good conductor of heat. [Conduction increases heat transfer]
2. The panel has a large surface area to increase the amount of radiation absorbed. [Area increases rate of absorption]
3. The panel is covered by a glass sheet to reduce heat loss by convection. [Glass reduces convection]

How to earn full marks:

• State two valid features of the solar water heater.

(c)

1. Calculate the energy transferred to the water: $E = 0.30 \times 5.0 \times 10^6 \text{ J} = 1.5 \times 10^6 \text{ J}$. [Calculates energy]

2. Calculate the time in seconds: $t = 1 \text{ hour} = 60 \times 60 \text{ s} = 3600 \text{ s}$. [Converts time to seconds]

3. Calculate the power output: $P = \frac{E}{t} = \frac{1.5 \times 10^6 \text{ J}}{3600 \text{ s}} = 416.67 \text{ W}$. [Applies power formula]

4. Final Answer: $P = \boxed{417 \text{ W}}$. [Rounds to three significant figures]

How to earn full marks:

• Correctly calculate the energy transferred to the water.
• Correctly convert the time to seconds.
• Correctly apply the power formula.
• State the correct final answer with the correct units.

Common Pitfall: Don't forget to convert time to seconds when calculating power. Also, make sure you understand the difference between energy absorbed and energy transferred. The percentage given in the question applies only to the energy transferred to the water, not the total solar energy.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

Two identical cups of hot coffee are placed on a table. One cup is white and shiny, the other is black and dull.

(a) Describe how you would design an experiment to determine which cup cools down faster. [3]

(b) State one variable that you would keep constant in this experiment. [1]

(c) Predict which cup will cool down faster. Explain your answer. [2]

Worked Solution:

(a)

1. Fill both cups with equal volumes of hot coffee at the same initial temperature. [Equal initial conditions]
2. Use a thermometer to measure the temperature of the coffee in each cup at regular time intervals (e.g., every 2 minutes) for a set period (e.g., 20 minutes). [Measure temperature over time]
3. Record the temperature readings for both cups in a table. [Data logging]

How to earn full marks:

• Mention keeping the initial temperature and volume the same.
• Describe how to measure the temperature change over time.
• State that the readings should be recorded.

(b)

1. The initial temperature of the coffee. [Controls initial temperature]
2. The volume of coffee in each cup. [Controls volume]
3. The surrounding temperature. [Controls room temperature]
4. The material of the cups (except for the surface colour/texture). [Controls material]

How to earn full marks:

• State a relevant variable to control.

(c)

1. The black and dull cup will cool down faster. [Prediction]
2. Dull black surfaces are better emitters of thermal radiation than shiny white surfaces. Therefore, the black cup will lose thermal energy more quickly to the surroundings via radiation. [Emission explanation]

How to earn full marks:

• Correctly predict which cup cools faster.
• Explain that black surfaces are better emitters of thermal radiation.

Common Pitfall: When describing experiments, be specific about how you would measure the temperature change and what variables you would control to ensure a fair test. Many students forget to mention the importance of keeping the initial temperature and volume of the coffee the same.

Exam-Style Question 4 — Extended Response [10 marks]

Question:

A satellite in space is orbiting the Earth. The side of the satellite facing the Sun receives solar radiation, causing its temperature to increase. The satellite also emits thermal radiation into space.

(a) Explain why the satellite needs to emit thermal radiation to maintain a constant temperature. [2]

(b) The satellite's surface is covered with a material that is a good reflector of solar radiation. Explain why this is important. [2]

(c) The satellite has a surface area of $20 \text{ m}^2$. The intensity of the solar radiation incident on the satellite is $1400 \text{ W/m}^2$. Only 40% of this radiation is absorbed by the satellite. Calculate the total power absorbed by the satellite. [3]

(d) Assuming the satellite emits radiation equally from all its surfaces, and that the rate of energy emission is equal to the rate of energy absorption, calculate the intensity of radiation emitted by the satellite. [3]

Worked Solution:

(a)

1. The satellite absorbs energy from solar radiation. [Absorption]
2. To maintain a constant temperature, the rate at which the satellite emits thermal radiation must be equal to the rate at which it absorbs energy. [Energy balance]

How to earn full marks:

• State that the satellite absorbs energy.
• Mention that the rate of emission must equal the rate of absorption.

(b)

1. Reflecting solar radiation reduces the amount of energy absorbed by the satellite. [Reflection reduces absorption]
2. This prevents the satellite from overheating and maintains a stable internal temperature for its components to function correctly. [Prevents overheating]

How to earn full marks:

• State that reflection reduces energy absorbed.
• Explain why preventing overheating is important.

(c)

1. Calculate the total solar power incident on the satellite: $P_{incident} = \text{intensity} \times \text{area} = 1400 \text{ W/m}^2 \times 20 \text{ m}^2 = 28000 \text{ W}$. [Calculates incident power]

2. Calculate the power absorbed: $P_{absorbed} = 0.40 \times 28000 \text{ W} = 11200 \text{ W}$. [Calculates absorbed power]

3. Final Answer: $P_{absorbed} = \boxed{11200 \text{ W}}$. [States final answer with units]

How to earn full marks:

• Correctly calculate the incident power.
• Correctly calculate the power absorbed.
• State the correct final answer with the correct units.

(d)

1. The power emitted is equal to the power absorbed: $P_{emitted} = 11200 \text{ W}$. [Equates emitted and absorbed power]

2. Calculate the intensity of the emitted radiation: $I_{emitted} = \frac{P_{emitted}}{\text{area}} = \frac{11200 \text{ W}}{20 \text{ m}^2} = 560 \text{ W/m}^2$. [Applies intensity formula]

3. Final Answer: $I_{emitted} = \boxed{560 \text{ W/m}^2}$. [States final answer with units]

How to earn full marks:

• State that the power emitted equals the power absorbed.
• Correctly apply the intensity formula.
• State the correct final answer with the correct units.

Common Pitfall: Many students struggle with the concept of energy balance. Remember that for an object to maintain a constant temperature, the rate of energy absorption must equal the rate of energy emission. Also, pay close attention to the units and make sure you are using the correct formula for intensity.