At what net rate does heat radiate from a 275−m2 black roof on a night when the roof’s temperature is 30.0ºc and the surrounding temperature is 15.0ºc?

Explanation:

We shall apply Stefan- Boltzmann  Law of black body radiation here which is applied to radiation made by black body in a particular surrounding.

E = σ A ( T⁴₀ - T⁴ )

Energy E is radiated by surface area A of a black body per second having absolute temperature of T₀ and having surrounding temperature of T.

Given T₀ = 30 degree = 303 K.

T = 288 K

A = 275 m²

σ = constant =5.67 x 10⁻⁸ Wm⁻²s⁻¹

Putting these values in the relation written above

E = 5.67 X 10⁻⁸  X 275  ( 303⁴ - 288⁴ )

= 5.67 X 10⁻⁸ X 275 (84.28 - 68.79) 10⁵

24.15 J A⁻¹ S⁻¹

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i think the answer is tomato juice? not entirely sure though