Question

At what net rate (in kW) does heat radiate from a 295 m² black roof on a night when the roof's temperature is 28.0°C and the surrounding temperature is 16.0°C? The emissivity of the roof is 0.900. (Enter the magnitude.)

  kW

Please explain, I am very confused

Answer 1

Solution) A = 295 m^2

T1 = 28°C = 28+273 = 301 K

T2 = 16°C = 16+273 = 289 K

e = 0.9

Power , P = ? ( net rate of heat radiated is power )

P = Ae(sigma)((T2)^(4) - (T1)^(4))

sigma = 5.67×10^(-8) W/(m^2)(K^4)

P = 295×0.9×5.67×10^(-8)((289^(4)) - (301^(4)))

P = - 18558.14 W

P = - 18.558 KW