A metal ball of surface area `200 cm^(2)` and temperature `527^(@)C` is surrounded by a vessel at `27^(@)C` . If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is `(sigma = 5.67 xx10^(-8)J//m^(2)-s-k^(4))`

Options

A. 108 joule
B. 168 joule
C. 182 joule
D. 192 joule

Answer

C

Explanation

According to Stefan.s Law, the rate of loss of heat is
`(Q)/(t)=sigmaA(T_(1)^(4)-T_(2)^(4))xx e`
here `sigma=5.67 xx 10^(-8)J//m^(2) xx sec. K^(2),`
`T_(1)=527+273=800K,`
`T_(2)=27+273=300K & A=200xx10^(-4)m^(2)`
So. `(Q)/(t)=5.67xx10^(-8)xx2xx10^(-2)`
`[(800)^(4)-(300)^(4)] xx 0.4`
`~=182` joule