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The surface of a black body is at a temp...

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)`)

A

`34.2xx10^(5)`

B

`2.5xx10^(5)`

C

`3.42xx10^(5)`

D

`2.5xx10^(6)`

Text Solution

Verified by Experts

The correct Answer is:
A
Temperature , `T=727^@c`
Cross section Area, `A=1 m^2`
`sigma = 5.7 xx 10^(-8) w//m^(2)//k^(4)`
Heat radiated per second,
`E= sigma A T^4`
`=5.7 xx10^(-8) xx 1 xx (1000)^4`
`=5.7 xx 10^(-8) xx(10^3)^4`
`=5.7 xx 10^(-8) xx 10^(12)`
`=5.7 xx 10^(-4)`
Heat radiated in one minute,
`E= 5.7 xx 10^(4) xx 60 `
`=342 xx 10^(4)`
`=34.2 xx 10^(5) J`.
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