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Heat Transfer - Radiation

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Radiation is the transfer of heat energy by electromagnetic waves (specifically infrared radiation) and does not require a medium to travel, meaning it can occur in a vacuum.

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All objects emit and absorb thermal radiation. The hotter an object is, the more infrared radiation it emits per second.

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The nature of a surface determines its effectiveness: Dark, matte, or dull black surfaces are the best absorbers and best emitters of radiation.

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Light-colored, shiny, or metallic (silvery) surfaces are poor absorbers and poor emitters because they reflect most of the incident radiation.

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The rate of heat transfer by radiation increases significantly with an increase in absolute temperature, following the relationship where power P∝T4P \propto T^4.

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Thermal equilibrium is reached when an object absorbs radiation at the same rate it emits it, keeping its temperature constant.

📐Formulae

P=eσAT4P = e \sigma A T^4

TKelvin=TCelsius+273T_{Kelvin} = T_{Celsius} + 273

c=3×108 m/sc = 3 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

Two identical metal cans are filled with the same amount of boiling water at 100∘C100^{\circ}C. Can A is painted matte black, and Can B is polished silver. Which can will cool down faster, and why?

Solution:

Can A (the matte black can) will cool down faster.

Explanation:

Matte black surfaces are much better emitters of infrared radiation than shiny, silver surfaces. Since both cans start at the same temperature, Can A will lose thermal energy to the surroundings via radiation at a much higher rate than Can B, causing its temperature to drop more quickly.

Problem 2:

A scientist measures the temperature of a radiating body to be 27∘C27^{\circ}C. Convert this temperature to the absolute scale (Kelvin) required for radiation calculations.

Solution:

TKelvin=27+273=300 KT_{Kelvin} = 27 + 273 = 300\text{ K}

Explanation:

In physics, especially when dealing with the Stefan-Boltzmann law for radiation, temperature must be expressed in Kelvin. The conversion is done by adding 273273 to the Celsius value.

Problem 3:

If the absolute temperature of a black body is doubled from TT to 2T2T, by what factor does the radiated power increase?

Solution:

Power ratio=(2T)4T4=16T4T4=16\text{Power ratio} = \frac{(2T)^4}{T^4} = \frac{16T^4}{T^4} = 16

Explanation:

According to the Stefan-Boltzmann law, the power radiated is proportional to the fourth power of the absolute temperature (P∝T4P \propto T^4). Therefore, if the temperature doubles, the power increases by 242^4, which is 1616 times.