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The surface area of a black body is 5 ti...

The surface area of a black body is `5 times 10^(-4)m^(2)` and its temperature is `727^(@)C`. The energy radiated by it per minute is :
`" "(sigma=5.67 times 10^(-8)J//m^(2)-s-k^(4))`

A

`1.7 times 10^(3)J`

B

`2.5 times 10^(2)J`

C

`8 times 10^(3)J`

D

`3 times 10^(4)J`

Text Solution

Verified by Experts

The correct Answer is:
A
From Stefan.s law
`" "E=sigmaT^(4)A`
Given, `T=727^(@)C=(727+273)=1000K, A=55 times 10^(-4)m^(2)`
`therefore "Energy"=(5*67 times 10^(-8)) (1000)^(4) (5 times 10^(-4))60`
`therefore " "E=1.7 times 10^(-3)J`
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