Question

Let us assume that the sun and another star A are perfect spheres and perfect black bodies. The radius of star A is four times the radius of the sun and it radiates heat energy at exactly the same rate as the sun. If the surface temperature of the sun is 6000 K, the surface temperature (in Kelvin) of the star A is:

Answer 1

Answer:

Power absorbed by earth

power emitted by earth. <br>

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<br>

Answer 2

Answer:

The surface temperature of star A is 3000K.

Explanation:

Black Body Radiation:

• A Black Body is non-reflective. It absorbs all radiations that are falling on it, at all wavelengths.
• These bodies emit thermal electromagnetic radiation and the intensity of these radiations depends only on the temperature of the black body.
• The power emitted by a black body is given by the Stefan-Boltzmann Law. Let 'P' be the power emitted, 'A' be the area of the black body and 'T' its temperature. Then,

                               $P=A\epsilon \sigma T^{4}$

Where $\epsilon$ is the emissivity of the black body ($0<\epsilon<1$) and $\sigma$ is called the Stefan-Boltzmann constant.

• For a perfect black body, the emissivity is 1, $\epsilon = 1$

Step 1:

Given sun and star A are perfect spheres and that the radius of star A ($R_{A}$) is four times the radius of the sun ($R_{S}$). It is also given that the rate of radiating heat energy which is power is equal for star A and the sun. The surface temperature of the sun is $T_{S}=6000K$ and we are required to find the surface temperature of star A ($T_{A}$).

Writing the radius relation mathematically,

                            $R_{A} = 4R_{S}$

Step 2:

From the formula written for power and considering that both the sun and the star are perfect black bodies, the power depends only on the surface area (A) and the surface temperature (T). Therefore,

                       $P_{A} = P_{S} \\A_{A} T_{A} ^{4} = A_{S}T_{S} ^{4}$

Step 3:

Substituting the values we have,

                      $(\pi R_{S})^{2} T_{S} ^{4} = (\pi R_{A})^{2}T_{A} ^{4} \\( R_{S})^{2} T_{S} ^{4} = ( R_{A})^{2}T_{A} ^{4}\\( R_{S})^{2} T_{S} ^{4} = ( 4R_{S})^{2}T_{A} ^{4} \\T_{A} ^{4} = \frac{T_{S} ^{4}}{16} \\T_{A} = \frac{T_{S}}{2}\\T_{A} = 3000K$

Therefore, the surface temperature of star A is 3000K.