Question 1: Define absorption in the context of black body radiation.

Answer: Absorption is the process by which a material takes in electromagnetic radiation, converting it into internal energy, thereby reducing the amount of radiation that passes through or is emitted.

Question 2: Explain how the absorption of radiation depends on the material's properties.

Answer: Absorption depends on the material's composition, thickness, and surface properties; materials with high absorption coefficients, such as dull, dark surfaces, absorb more radiation than shiny or light-coloured surfaces.

Question 3: What is the relationship between absorption and the material's absorptivity?

Answer: Absorptivity is a measure of how effectively a material absorbs radiation; a higher absorptivity indicates greater absorption of incident radiation.

Question 1: A black surface absorbs 85% of incident radiation with an incident power of 200 W. Calculate the power absorbed by the surface.

Answer: Power absorbed = incident power × absorption coefficient = 200 W × 0.85 = 170 W

Question 2: An object with an absorptivity of 0.6 is exposed to radiation of intensity 500 W/m² over an area of 0.5 m². Find the total power absorbed.

Answer: Power absorbed = intensity × area × absorptivity = 500 W/m² × 0.5 m² × 0.6 = 150 W

Question 3: Calculate the temperature of a black body if it absorbs 100 J of energy in 10 seconds and has an absorption area of 2 m². Assume the specific heat capacity is 900 J/(kg·K) and the mass is 1 kg.

Answer: Power absorbed = energy / time = 100 J / 10 s = 10 W; Temperature increase ΔT = Power / (mass × specific heat capacity) = 10 W / (1 kg × 900 J/(kg·K)) ≈ 0.011 K

Question 1: Describe a simple experiment to investigate the absorption of different materials for infrared radiation. Include variables to control and safety precautions.

Answer: A possible experiment involves exposing samples of different materials (e.g., aluminium foil, black paper, glass) to infrared radiation sources and measuring temperature change with a thermometer or thermopile. Control variables include distance from source, radiation intensity, and exposure time. Safety precautions include wearing heat-resistant gloves, avoiding direct eye exposure to infrared sources, and ensuring proper ventilation.

Question 2: Identify the key variables affecting the absorption of radiation in a material and explain their effects.

Answer: Key variables include the material's surface texture and colour, thickness, and surface reflectivity. Dark, matte, and thick materials tend to absorb more radiation, while shiny, light-coloured, and thin materials reflect or transmit more radiation.

Question 1: A scientist measures the temperature increase of two materials when exposed to the same infrared radiation source. Material A shows a larger temperature rise than Material B. What can be inferred about their relative absorptivities?

Answer: Material A has a higher absorptivity than Material B, meaning it absorbs more of the incident radiation, resulting in a greater temperature increase.

Question 2: A black surface and a white surface are both exposed to the same radiation intensity. Which surface absorbs more energy and why?

Answer: The black surface absorbs more energy because it has a higher absorptivity compared to the white surface, which reflects more incident radiation.

Question 1: Explain how the concept of absorption is utilised in designing thermal insulators for buildings.

Answer: Thermal insulators are designed with materials that have high reflectivity and low absorptivity to minimise heat absorption, thereby reducing heat transfer into buildings and maintaining indoor temperature.

Question 2: Describe how understanding absorption is important in the development of solar thermal panels.

Answer: Solar thermal panels are designed to have high absorptivity and low reflectivity in the relevant wavelengths to maximise absorption of solar energy and improve efficiency in converting sunlight into heat.