# (a) Calculate the rate in watts at which heat transfer through radiation occurs (almost entirely...

## Question:

(a) Calculate the rate in watts at which heat transfer through radiation occurs (almost entirely in the infrared) from 1.0 {eq}m^{2}
{/eq} of the Earth s surface at night. Assume the emissivity is 0.90, the temperature of the Earth is {eq}15^{\circ}C
{/eq}, and that of outer space is 2.7 *K*.

(b) Compare the intensity of this radiation with that coming to the Earth from the Sun during the day, which averages about 800 {eq}W/m^{2} {/eq}, only half of which is absorbed.

(c) What is the maximum magnetic field strength in the outgoing radiation, assuming it is a continuous wave?

## Heat transfer through radiation:

- After the solar energy is absorbed in the form of electromagnetic waves, the earth emits longer wavelength which will be observed by the traces present in the atmosphere. This is also described by the term 'black body radiation'. Perfect black bodies do not exist only stars are considered as the approximate black bodies.
- Intensity is calculated as the energy per unit time per unit area. It is a measure of brightness and one can observe colors when lights of different wavelengths strike eyes. The formula for intensity of light can be given as,

{eq}Intensity=\frac{Power}{area} {/eq}

It is measured in watts per square meter.

## Answer and Explanation: 1

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View this answer(a) The heat transfer through radiation can be given by the following equation,

{eq}P=\epsilon\sigma A(T_{2}^{4}-T_{1}^{4}) {/eq}

where,

- P =...

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Chapter 8 / Lesson 15A blackbody is an object that is called a perfect absorber of radiation while blackbody radiation is electromagnetic radiation emitted by a perfect radiator. Discover how these two concepts function while exploring electromagnetic radiation.