At what net rate does heat radiate from a 300-m^2 black roof on a night when the roof's temperature is 33.0°C and the surrounding temperature is 18.0°C? The emissivity of the roof is 0.900. (Enter the magnitude.)

Respuesta :

Answer:

24445.85 J/s

Explanation:

Area, A = 300 m^2

T = 33° C = 33 + 273 = 306 k

To = 18° C = 18 + 273 = 291 k

emissivity, e = 0.9

Use the Stefan's Boltzman law

[tex]E = \sigma  \times e \times A\times\left ( T^4 -T_{0}^{4}\right )[/tex]

Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.

The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4

By substituting the values

[tex]E = 5.64 \times 10^{-8}\times 0.9 \times 300 \times  (306^{4}-291^{4})[/tex]

E = 24445.85 J/s

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