Find the temperatue of an oven if it radiates 8.28 cal per second through an opening, whose area is `6.1 cm^(2)`. Assume that the radiation is close to that of a black body.

Explanation

The emittance of the oven (the energy radiated in one second by unit surface area) is
E = `(8.28 "cal//s")/(6.1 cm^(2))`
= `(8.28 xx 4.2 J //s )/(6.1 xx 10^(-4) m^(2)) = 5.7 xx 10^(4) "watt /"m^(2)`
From stefan - boltzmann formula E = `sigma - T^(4)`
Where `sigma = 5.67 x 10^(-8) W//m^(2) .K^(4)`
`T^(4) = (E)/(sigma ) = (5.7 xx 10^(4))/(5.67 xx 10^(-8)) = 1xx 10^(12) K^(4)`
`therefore T = 10^(3) K = 1000 K`.