Determine the net heat radiated by the black body disc of mass 6g and radius 4cm at 400C, initially kept at 300C.
Solution;
By Stefan-Boltzmann's law of radiation,the net rate of heat transfer by radiation is;
Qnett=σAe(T24−T14)\frac {Q_{net}}{t}=\sigma Ae(T_2^4-T_1^4)tQnet=σAe(T24−T14)
For a black body,e=1
A=πr2=π×0.042=A=πr^2=π×0.04^2=A=πr2=π×0.042=5.026×10-3m2
Qnett=5.67×10−8×5.026×10−3×(3134−3034)\frac{Q_{net}}{t}=5.67×10^{-8}×5.026×10^{-3}×(313^4-303^4)tQnet=5.67×10−8×5.026×10−3×(3134−3034)
Qnett=0.33331J/s\frac{Q_{net}}{t}=0.33331J/stQnet=0.33331J/s
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