**1. Definition of the heat transfer by radiation**

How does it feel if you wear black clothes on a hot day or when you are exercising during the day? Compare with when you wear white clothes? If you wear black clothes during the day, then you feel hot. Why is that? The distance between the sun and the earth in the morning is almost the same as the distance between the sun and the earth during the day and evening. Then why are the mornings and evenings cooler, hotter during the day? The answers to these questions are related to heat transfer by radiation.

Heat transfer by radiation is heat transfer in the form of electromagnetic waves. Examples of heat transfer by radiation are the warmth of your body when you are near a furnace and heat transfer from the sun to the earth. The sun has a higher temperature (around 6000 Kelvin), while the earth has a lower temperature. The temperature difference between the sun and the earth causes heat to move from the sun (higher temperature) to the earth (lower temperature). If heat transfer from the sun to the earth requires an intermediary or medium, as heat transfer by conduction and convection, heat cannot arrive on earth; heat must pass through a vacuum (or almost empty). If there is no heat from the sun, life on earth will never exist because life requires energy from the sun.

Another example of heat transfer by radiation is the heat felt when we are near a flame. The heat we feel is not caused by overheated air due to a flame. As previously explained, hot air will expand so that the density decreases. As a result, the air with reduced density moves vertically upwards, not moving horizontally towards us. Our body feels warm or hot when it is near the flame because heat moves by way of radiation from the flame (higher temperature) to our body (lower temperature).

Heat transfer by radiation is slightly different compared to heat transfer by conduction and convection. Heat transfer by conduction and convection occurs when objects that have temperature differences touch each other. Conversely, heat transfer by radiation can occur without touch.

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**2. The equation of the heat transfer by radiation**

The rate of heat transfer by radiation is proportional to the area of the object and the absolute temperature (Kelvin Scale) of the object. Objects that have a larger surface area have a higher heat transfer rate than objects that have a smaller surface area.

Likewise, objects with a temperature of 2000 Kelvin, for example, have a heat transfer rate of 2^{4} = 16 times greater than objects with a temperature of 1000 Kelvin. This result was discovered by Josef Stefan in 1879 and was theoretically derived by Ludwig Boltzmann about five years later.

*Q = heat, t = time, A = surface area of object (m ^{2}), T = the absolute temperature of object (K) e = emissivity (dimensionless number whose size ranges from 0 to 1), σ = 5.67 x 10^{-8} W/m^{2}.K^{4} (Universal constant. Also referred to as the Stefan-Boltzmann constant), Q/t = the rate of heat transfer by radiation*

Objects with dark surfaces (black) have an emissivity close to 1, while objects with light color have an emissivity close to 0. The greater the emissivity of an object (e approaches 1), the greater the heat rate emitted by the object. Conversely, the smaller the emissivity of an object (e approaches 0), the smaller the heat rate emitted. We can say that dark-colored objects (black) usually emit more heat than light-colored objects (white).

The amount of emissivity not only determines the ability of an object to emit heat but also the ability of an object to absorb heat emitted by other objects. Objects that have an emissivity close to 1 (dark colored objects) absorb almost all the heat emitted on them. Only a small portion is reflected. Conversely, objects that have an emissivity close to 0 (light-colored objects) absorb a little heat emitted on them. Most of the heat is reflected by the object. An object that absorbs all the heat emitted on it has emissivity = 1. This type of object is known as the “black object.” The term black object does not explain that objects are black but explains the ability of objects to absorb all the heat emitted on them.

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**3. ****The rate of heat emitted by the sun**

Based on calculations (according to reality), was found that there was a heat of 1350 Joules per second per square meter that moved from the sun to the earth. On a bright day (no clouds), there is a heat of 1000 Joules per second per square meter that arrives on the surface of the earth. On a sunny day (lots of clouds), around 70% of the heat is absorbed by the earth’s atmosphere, only 30% of heat arrives on the earth’s surface. The magnitude of heat that disappears in the earth’s atmosphere depends on many or the few clouds. The heat amount of 1350 Joules per second per square meter is known as the solar constant. Because Joule per second (J/s) = Watt, then we can rewrite the solar constant to 1350 Watts per square meter = 1350 W/m^{2}

When the heat emitted by the sun arrives at the surface of the earth, the heat is absorbed by the earth and objects on the surface of the earth. The heat absorption rate depends on the emissivity (e) of the object, the surface area of the object and the angle formed by sunlight with a line perpendicular to the surface of the object.

Information :

Q/t = heat absorption rate, 1000 W/m^{2} = solar constant, e = emissivity of the object, A = surface area of the object, θ = angle formed by sunlight with lines perpendicular to the surface of the object, A cos θ = effective area (component surface area of objects perpendicular to sunlight)

During the day, the sun rays parallel or coincide with a line perpendicular to the earth’s surface (formed angle = 0).

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The lower formed = 0^{o}, therefore, the heat absorption rate is:

The heat absorption rate (Q/t) is maximum if the angle formed by sunlight with a line perpendicular to the earth’s surface = 0^{o }(cos 0 = 1). Usually, this occurs during the day, where the sun is perpendicular to the surface of the earth. So there is no need to be surprised if daylight is hotter than morning or evening. In the morning and evening, the angle formed is close to 90^{o}.

If the angle formed is 80^{o}, then the heat absorption rate is:

The heat absorption rate (Q/t) in the morning and evening is minimum because cos θ is close to zero. The smaller cos θ, the smaller the heat absorption rate.

At sunset on the western horizon or about to rise on the eastern horizon, the angle formed = 90^{o}.

Because the angle formed is 90^{o}, the heat absorption rate is:

Heat absorption rate (Q / t) at sunset or about to rise = 0.

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