Answer:
259.274 kW
Explanation:
Given:
Area of the lava, A = 1.00 m²
Temperature of the surrounding, T₁ = 30.0° C = 303 k
Temperature of the lava, T₂ = 1190° C = 1463 K
emissivity, e = 1
Now,
from the Stefan-Boltzmann law of radiation the rate of heat loss is given as,
u = σeA(T₂⁴ - T₁⁴)
where,
u = rate of heat loss
σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²∙K⁴
on substituting the respective values, we get
u = 5.67 × 10⁻⁸ × 1 × 1 × (1463⁴ - 303⁴)
or
u = 259274.957 W
or
u = 259.274 kW
