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Question: Two bodies of the same shape, size, and radiating power have emissivity 0.2 and 0.8. the ratio of th...

Two bodies of the same shape, size, and radiating power have emissivity 0.2 and 0.8. the ratio of their temperature is:
(A) 3:1\sqrt{3}:1
(B) 2:1\sqrt{2}:1
(C) 1:51:\sqrt{5}
(D) 1:31:\sqrt{3}

Explanation

Solution

Hint Use the Stefan-Boltzmann equation for both the bodies and form respective equations for each body. Either equate the values of radiative power from both the equations or simply divide one equation by the other. Cancel all common terms and substitute the given values into the final equation and simplify to get the answer.
Equation used
Stefan-Boltzmann law of radiation:
P=Qt=σeAT4P=\dfrac{Q}{t}=\sigma eA{{T}^{4}}
Where PP is the radiative power,
QQ is the heat transferred,
t is the time taken,
σ\sigma is the Stefan-Boltzmann constant with a value of 5.67×108J/sm2K45.67\times {{10}^{-8}}J/s\cdot {{m}^{2}}\cdot {{K}^{4}},
ee is the emissivity of the object,
AA is the surface area of the object,
TT is the temperature.

Complete Step by step solution
Let the emissivity of the first object be e1{{e}_{1}} and that of the second object be e2{{e}_{2}}, and let the temperature of the first and second object be T1{{T}_{1}} and T2{{T}_{2}} respectively.
By using the Stefan-Boltzmann equation that gives the radiative power of a body for first and second bodies respectively, we get
P=σe1AT14P=\sigma {{e}_{1}}A{{T}_{1}}^{4}
P=σe2AT24P=\sigma {{e}_{2}}A{{T}_{2}}^{4}
As it is given in the question that the radiative powers of both the bodies are equal and that they have the same shape and size, we can take the variable for radiative power and surface area of both the bodies to be PP and AA respectively.
Dividing the radiative powers of both the bodies, we get
PP=σe1AT14σe2AT24\dfrac{P}{P}=\dfrac{\sigma {{e}_{1}}A{{T}_{1}}^{4}}{\sigma {{e}_{2}}A{{T}_{2}}^{4}}
Canceling the common terms and rearranging the above equation, we get
T14T24=e2e1\dfrac{{{T}_{1}}^{4}}{{{T}_{2}}^{4}}=\dfrac{{{e}_{2}}}{{{e}_{1}}}
Substituting the given values of emissivity, we get
T14T24=0.80.2\dfrac{{{T}_{1}}^{4}}{{{T}_{2}}^{4}}=\dfrac{0.8}{0.2}
Taking 1/4th{{{}^{1}/{}_{4}}^{th}} power on both sides, we get
T1T2=(0.80.2)1/4\dfrac{{{T}_{1}}}{{{T}_{2}}}={{\left( \dfrac{0.8}{0.2} \right)}^{{}^{1}/{}_{4}}}
T1T2=21\Rightarrow \dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{\sqrt{2}}{1}
Or T1:T2=2:1{{T}_{1}}:{{T}_{2}}=\sqrt{2}:1
\therefore the ratio of the temperature of the first body to the second body is 2:1\sqrt{2}:1 .

That is, option (B) is the correct option.

Note
The Stefan-Boltzmann equation is an equation that tells us about the power radiated out of an object. We can find the luminosity of a star by using the Stefan-Boltzmann equation by taking the value of emissivity to be 11 as stars are examples of a blackbody and black bodies have emissivity equal to 11.