Physics, 23.12.2019 21:31 ICAMARON632

The intensity of a sunspot is found to be 3 times smaller than the intensity emitted by the solar surface. what is the approximate temperature of this sunspot if the temperature of the solar surfaceis 5800 k?
a. 4400 k
b. 470,000 k
c. 1900 k
d. 7600 k
e. 1400 k

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Answer from: queenbroookk

If the temperature of the solar surface is 5800 K then the approximate temperature of the sunspot is a) 4400 K.

Explanation:

The most straightforward way to solve this is using Stefan-Boltzmann law that states that I the energy radiated per unit surface area per unit time (watt per unit area $\frac{W}{m^{2}}$ ) of a black body is proportional to the fourth power of the temperature T of the body:

$I=\sigma T^{4}$

with $\sigma=5.67x10^{-8} Wm^{-2} K^{-4}$ being the Stefan constant.

A black body is an idealized physical body that is a perfect absorber because it absorbs all incident electromagnetic radiation and is also an ideal emitter. The Sun is considered to be a black body at different layers and different temperatures.

We are told that the intensity of a sunspot $I_{sunspot}$ is found to be 3 times smaller than the intensity emitted by the solar surface $I_{surface}$ , that means that:

$I_{sunspot}=\frac{I_{surface}}{3}$

then using the expression of Stefan-Boltzmann law we get that

$\sigma T_{sunspot} ^{4}=\sigma T_{surface} ^{4}$

we cross out $\sigma$ and use the fourth root in each side of the equation

$\sqrt[4]{T_{sunspot} ^{4}}=\frac{\sqrt[4]{T_{surface} ^{4}}}{\sqrt[4]{3}}$

$T_{sunspot}=\frac{T_{surface}}{\sqrt[4]{3} }$

then we use that

$T_{surface}=5800 K$ $\sqrt[4]{3}\approx1,316$

$T_{sunspot}=\frac{5800 K}{1,316}$

$T_{sunspot}=4407,3 K$

So finally we get that

$T_{sunspot}\approx4400 K$

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