The net heat transfer by radiation from a body at temperature (T₁) to another body or surrounding at temperature (T₂) is given by

(where, σ = Radiation constant for a perfect black body and ε = Emissivity of a body at a particular temperature)

Question

Question

This question was previously asked in

BPSC AE Paper IV General Engineering 19 Dec 2024 Official Paper

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\(\mathrm{Q}=\sigma \varepsilon_2\left(\mathrm{T}_1^2-\mathrm{T}_2^2\right) \mathrm{W} / \mathrm{m}^2\)
\( \mathrm{Q}=\sigma \varepsilon_1\left(\mathrm{T}_1^4+\mathrm{T}_2^4\right) \mathrm{W} / \mathrm{m}^2 \)
\( \mathrm{Q}=\sigma \varepsilon_2\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right) \mathrm{W} / \mathrm{m}^2\)
\( \mathrm{Q}=\sigma \varepsilon_1\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right) \mathrm{W} / \mathrm{m}^2\)

Answer 1

Concept:

Net heat transfer by radiation occurs when a body at higher temperature radiates energy toward a cooler body or surroundings.

According to Stefan–Boltzmann Law, a perfect black body radiates heat as,

\( Q = σ T^4 \)

For real bodies, emissivity \( \varepsilon \) is introduced:

\( Q = σ \varepsilon T^4 \)

Here,

σ = 5.67 × 10^-8 W/m²-K⁴ (Stefan–Boltzmann constant), and ε is the emissivity (0 ≤ ε ≤ 1)

Calculation:

Let Body 1 be at temperature T₁ and surroundings (or Body 2) at temperature T₂ .

The net radiative heat transfer is given by:

\( Q = σ \varepsilon_1 (T_1^4 - T_2^4) \)