A student tests how well four different surfaces absorb infrared radiation. She uses four identical aluminium plates, each coated with a different surface f...

Assessment: Physics 0625 | Paper 4 Mock 01 | Theory (Extended) Subject: Physics - 0625

Question 1 Report

A student tests how well four different surfaces absorb infrared radiation. She uses four identical aluminium plates, each coated with a different surface finish. Each plate has the same area and thickness, and a temperature sensor is taped to the back. An identical radiant heater is placed 20 cm from each plate in turn. The starting temperature is 20 °C. The heater is switched on for exactly 4 minutes and the final temperature is recorded. The results are shown in the table below.

Surface finishStarting temp / °CFinal temp / °CTemp rise / °C
Matt black203919
Shiny black203313
Matt white20299
Polished silver20233

(a) State which surface is the best absorber of infrared radiation. [1]

(b) Explain why the matt black surface shows the largest temperature rise. [2]

(c) State which surface is the best reflector of infrared radiation. Explain your reasoning. [2]

(d) Suggest why the student uses plates that are all the same size, thickness and metal. [1]

(e) State the relationship between a surface that is a good absorber of infrared radiation and a good emitter. [1]

Answer Details

Marking Scheme

  • (a) [1] Matt black.
  • (b) [1] Matt black surfaces absorb the largest proportion of the infrared radiation that falls on them. [1] More energy is absorbed per second, so the temperature rises faster and reaches a higher value in the same time.
  • (c) [1] Polished silver. [1] It has the smallest temperature rise, meaning it absorbed the least radiation; the radiation not absorbed was reflected.
  • (d) [1] To ensure a fair test: the only variable that changes is the surface finish, so any difference in temperature rise must be due to the absorbing properties of the surface.
  • (e) [1] A good absorber of infrared radiation is also a good emitter of infrared radiation.

Explanation

(a) The matt black plate shows the greatest temperature rise (19 °C), which means it absorbed the most infrared radiation in the 4-minute heating period.

(b) When infrared radiation hits a matt black surface, almost all of it is absorbed and converted to internal energy of the plate. A larger fraction of incident energy absorbed per second means the plate's temperature climbs more quickly. The matt texture scatters any reflection rather than bouncing it away cleanly, and the black colour absorbs across all visible and infrared wavelengths.

(c) The polished silver plate rose only 3 °C. Since every plate received the same amount of radiation, the one that heated least must have reflected most of the radiation away. A polished (smooth, shiny) silver surface is an excellent reflector of infrared radiation.

(d) This is a controlled experiment. Variables such as plate material, area, thickness, starting temperature, heater distance, and heating time are all kept the same. The only factor that changes is the surface finish (the independent variable). This ensures the temperature rise (dependent variable) can be attributed solely to differences in absorption properties rather than any other factor.

(e) This is a fundamental principle of thermal radiation. A surface that is good at absorbing a particular wavelength of radiation is equally good at emitting that wavelength when hot. Matt black is the best absorber and the best emitter. Polished silver is the worst absorber and the worst emitter. This symmetry arises because absorption and emission involve the same physical interaction between radiation and the surface material.

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