Question 778: Thermodynamics & KTG

Question

Two bodies ' X ' and ' Y ' at temperatures ' $\mathrm{T}_1$ ' K and ' $T_2$ ' K respectively have the same dimensions. If their emissive powers are same, the relation between their temperatures is

Accepted Answer

According to Stefan-Boltzmann's law, the emissive power of a body is given by:
$ \frac{d Q}{d t} = e \left( \sigma A T^4 ight) \quad 	ext{..... (i)} $
Here, $ dQ/dt $ represents the rate of emission of radiant energy, $ e $ is the emissivity, $ \sigma $ is the Stefan-Boltzmann constant, $ A $ is the surface area, and $ T $ is the temperature in Kelvin.
Given that the dimensions of the two bodies, X and Y, are identical:
$ A_1 = A_2 \quad 	ext{..... (ii)} $
It is also stated that the emissive powers of both bodies are the same. Therefore, we have:
$ \begin{aligned} & \frac{dQ_1}{dt} = \frac{dQ_2}{dt} \ \Rightarrow & e_1 \, T_1^4 = e_2 \, T_2^4 \quad 	ext{... [From equations (i) and (ii)]} \ \Rightarrow & \left( \frac{T_1}{T_2} ight)^4 = \frac{e_2}{e_1} \end{aligned} $
This equation describes the relationship between the temperatures $ T_1 $ and $ T_2 $ of the two bodies, given their equal emissive powers.

Answer

$\frac{T_1}{T_2}=\frac{1}{3}$

Answer

$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{81}{1}$

Answer

$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{3^{\frac{1}{4}}}{1}$

Answer

$\frac{T_1}{T_2}=\frac{9^{\frac{1}{4}}}{1}$

Thermodynamics & KTG

Two bodies ' X ' and ' Y ' at temperatures ' \mathrm{T}_1 ' K and ' T_2 ' K respectively have the same dimensions. If their emissive powers are same, the relation between their temperatures is

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