A block having some emissivity is maintained at 500 K temperature in a surrounding of 300 K temperature. It is observed that, to maintain the temperature of the block, 210 W external power is required to be supplied to it. If instead of this block a black body of same geometry and size is used, 700 W external power is needed for the same. Find the emissivity of the material of the block.
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answer is 0.3.
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Detailed Solution
If emissivity of body is e, then Power required = Radiation rateP=σeAT4−T04 …….(i)For a block body PB=σAT4−T04 …………(ii) (i) (ii) gives PPB=e⇒ e=210700=0.3