What is the rate of energy radiation per unit area of a black body at (a)\({\bf{273}}\;{\bf{K}}\) and

(b) \({\bf{2730}}\;{\bf{K}}\)?

The temperature of the black body is\(T = 273\;{\rm{K}}\).

Emissivity of a perfect black body is \(e = 1\).

Area through which emissions are taking place is \(A = 1\;{{\rm{m}}^{\rm{2}}}\).

Stefan-Boltzmann constant: \(\sigma = 5.67 \times {10^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}\)

A perfect black body is the one which can absorb all the wavelengths of light, when it is incident on the black body. Also, on heating, the body should be able to emit all the wavelengths of light. The emissivity of such a body is 1. The rate heat radiation through a perfect black body, on heating is given as-

\(\frac{{{\bf{dQ}}}}{{{\bf{dt}}}}{\bf{ = A\sigma e}}{{\bf{T}}^{\bf{4}}}\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Using equation (1), the rate of heat radiation through a black body is given as-

\(\frac{{dQ}}{{dt}} = A\sigma e{T^4}\)

For the given values the above equation becomes-

\(\begin{array}{c}\frac{{dQ}}{{dt}} = 1\;{{\rm{m}}^2} \times 5.67 \times {10^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}} \times 1 \times {\left( {273\;{\rm{K}}} \right)^4}\\ = 314.9\;{\rm{W}}\end{array}\)

The rate of transfer of heat from the black body at \(273\;{\rm{K}}\) is \(314.9\;{\rm{W}}\).