- 0.5
- 0.4
- 0.2
- 0.1

Option 2 : 0.4

Free

ST 1: Logical reasoning

5333

20 Questions
20 Marks
20 Mins

__Concept:__

The rate of heat transfer between two gray, diffuse and opaque surfaces which form an enclosure is given by

\({Q_{1 - 2}} = \frac{{\sigma {A_1}\left( {T_1^4 - T_2^4} \right)}}{{\left( {\frac{1}{{{ϵ _1}}} - 1} \right) + \frac{1}{{{F_{12}}}} + \frac{{{A_1}}}{{{A_2}}}\left( {\frac{1}{{{ϵ _2}}} - 1} \right)}}\)

Since the small body is enclosed in a large enclosure,

F_{12} = 1 and A_{1}/A_{2} is approximately zero

\( \Rightarrow {Q_{1 - 2}} = {ϵ_1}\sigma {A_1}\left( {T_1^4 - T_2^4} \right)\)

Now the equivalent emissivity is ϵ_{1}

__Calculation:__

Given ϵ_{1} = 0.4, ϵ_{2} = 0.5;

Therefore equivalent emissivity is 0.4

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