The surface of a black body is at a temperature `727^(@)C` and its cross-section is `1 m^(2)`. Heat radiated from this surface in one minute in joules is (Stefan's constant = `5.7 xx 10^(-8) W//m^(2) // k^(4)`)

Options

A. `34.2 xx 10^(5)`
B. `2.5 xx 10^(5)`
C. `3.42 xx 10^(5)`
D. `2.5 xx 10^(4)`

Answer

a

Explanation

Given, Temperature of black body `(T) = 727^(@) C`
= 727 + 273 = 100 K
Cross sectional area of black body `(A) = 1m^(2)`
Stefan's canstant (a) `= 5.7 xx 10^(-8) W//m^(2) // K^(4)`
We know that,
Heat radiated second
`E sigma AT^(4)`
`= 5.7 xx 10^(-8) xx 1 xx (1000)^(4)`
or E = 57000 J
`:.` Heat radiated from this surface in one minute
`E = 57000 xx 60`
or, `E = 3.42 xx 10^(6)` or `34.2 xx 10^(5)`