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An enclosure has an inside area of 50 m2, and its inside surface is black and is maintained at a constant temperature. A small opening in the enclosure has an area of 0.01 m2. The radiant power emitted from this opening is 48 W. What is the temperature of the interior enclosure wall, in K

Sagot :

Answer:

Explanation:

The enclosure will behave as black body . For black body , the formula for radiant energy is given by Stefan's law as follows

E = σ A T⁴ where σ = 5.67 x 10⁻⁸ W m⁻² T⁻⁴

Area A = .01 m²

E = 48 W

48 = .01 x 5.67 x 10⁻⁸ T⁴

T⁴ =  846.56 x 10⁸

T= 539 K

Following are the calculation to the temperature of the interior enclosure wall:

Given:

[tex]\to A_s=50\ m^2\\\\\to A_0=0.01\ m^2\\\\\to q_{rad}=48\ w\\\\\to \sigma = 5.67 \times 10^{-8}\ \frac{W}{m^{-2} T^{-4}} \\\\[/tex]

To find:

[tex]T_s=?[/tex]

Solution:

Using the value to calculate the surface is black radiates power:

[tex]\to q_{rad}=A_0 E_0 T_s\\\\\to q_{rad}=A_0 \sigma T_{s}^4\\\\\to 48=0.01 \times 5.67 \times 10^{-8} \times T_{s}^4\\\\[/tex]

[tex]\to T_{s}^4 =\frac{48}{0.01 \times 5.67 \times 10^{-8}} \\\\[/tex]

        [tex]=\frac{48\times 100}{ 5.67 \times 10^{-8}} \\\\=\frac{48\times 100\times 10^{8} }{ 5.67 } \\\\=\frac{48\times 10^{10} }{ 5.67 } \\\\=8.46\times 10^{10} \\\\[/tex]

[tex]\to \bold{T_s=\pm 539.31 \ K}[/tex]

Therefore the final answer is "[tex]\bold{\pm 539.31 \ K}[/tex]".

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