calculate radiative heat

Question

Caclculate the radiative heat gain in watts by a loaf of bread at surface temperature of 100C. The syrrounding oven surface temperature is 1000C. The total surface area of the bread is 0.15m^2 and the emissivity of the bread surface is 0.80. Assume the oven is a blackbody radiator.

Answer 1

Use the Stefan-Boltzmann law: heat gained =

 

q =A σ (ε T₁⁴ - φT₂⁴)

 

where σ=5.669E-8 W/m²/K⁴ is Boltzmann's constant, A is the surface area of the absorbing surface and φ and T₂ its absorptivity and temperature, and ε and T₁ are the emissivity and temperature of the radiating surface. In this case, the radiating surface is a blackbody, so ε=1. The temperatures are T₁=1000 °C = 1273 K and T₂=100 °C = 373 K. You get

 

q = 0.15 m² (5.669E-8 W/m²/K⁴) (1*1273⁴ - 0.8*373⁴)K⁴ = 22,200 W.