Radiating power of a black body

$= {E_0} = \sigma \left( {{T^4} - T_0^4} \right)A$

where $\sigma$ is known as the Stefan-Boltzmann constant, A is the surface area of a black body, T is the temperature of the black body and T₀ is the temperature of the surrounding.

$\therefore$ 60 = $\sigma$(1000⁴ – 500⁴)    ...(i)

[T = 727^oC = 727 + 273 = 1000 K, T0 = 227^oC = 500 K].

In the second case, T = 1227^oC = 1500 K and let E' be the radiating power.

$\therefore$ E' = $\sigma$(1500⁴ – 500⁴)    ...(ii)

From (i) and (ii) we have

${{E'} \over {60}} = {{{{1500}^4} - {{500}^4}} \over {{{1000}^4} - {{500}^4}}} = {{{{15}^4} - {5^4}} \over {{{10}^4} - {5^4}}} = {{50000} \over {9375}}$

$\therefore$ $E' = {{50000} \over {9375}} \times 60 = 320\,W$