1.

An aluminiumum foil of relative emittance 0.1is placed in between two concentric spheres at temperatures 300K and 200k respectively.Calculate the temperature of the foil after the steady state is reached .Assume that the spheres are perfect black body radiators. Also calculate the rater of energy transfer between one of the spheres and the foil. [sigma=5.67xx10^(-8)S.I. Unit]

Answer»

Solution :Let T be the temperature of the foil .When the steady STATE is rerached,
`esigma(T_(1)^(4)-T^(4)=esigma(T^(4)-T_(2)^(4))`
`300^(4)-T^(4)=T^(4)-(200)^(4)`
`T^(4)=4.85xx10^(9)`
T = 263.8 K
Rate of energy transfer= E =`esigma(T_(1)^(4)-T^(4))`
`0.1xx5.67xx10^(-8) [300^(4)-(263.8)^(4)]`
`18.5 W// m^(2)`


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