Question 851: Thermodynamics & KTG

Question

Two bodies $$\mathrm{A}$$ and $$\mathrm{B}$$ at temperatures '$$\mathrm{T}_1$$' $$\mathrm{K}$$ and '$$\mathrm{T}_2$$' $$\mathrm{K}$$ respectively have the same dimensions. Their emissivities are in the ratio $$1: 3$$. If they radiate the same amount of heat per unit area per unit time, then the ratio of their temperatures $$\left(\mathrm{T}_1: \mathrm{T}_2\right)$$ is

Accepted Answer

From Stefan - Boltzmann's law
$$\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{e}\left(\sigma \mathrm{AT}^4ight)$$
Given $$\mathrm{A}$$ and $$\frac{\mathrm{dQ}}{\mathrm{dt}}$$ are same for both the bodies
$$\begin{aligned}
& \Rightarrow \mathrm{e}_1 \mathrm{~T}_1^4=\mathrm{e}_2 \mathrm{~T}_2^4 \
	herefore \quad & \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}ight)^4=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{3}{1} \
& \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sqrt[4]{3}}{1}=\frac{3^{\frac{1}{4}}}{1}
\end{aligned}$$

Answer

$$1: 3$$

Answer

$$3^{1 / 4}: 1$$

Answer

$$9^{1 / 4}: 1$$

Answer

$$81: 1$$

Thermodynamics & KTG

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