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. The emissivity of the material of the block is

Question

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. The emissivity of the material of the block is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

If emissivity of body is e, we use Power required = Radiation rate P=σ e A(T4-T54).....(i) For a block body PB=σ A(T4-T54) .....(ii) ( text (i) / text (ii) ) text gives (P/PB)=e ⇒ e=(210/700)=0.3

If emissivity of body is e, we use Power required = Radiation rate P=σeA(T4−T54​).....(i) For a block body PB​=σA(T4−T54​) .....(ii) (ii) (i) ​ gives PB​P​=e ⇒e=700210​=0.3

If emissivity of body is $e$ , we use
Power required $=$ Radiation rate
$P = σ e A (T^{4} - T_{5}^{4})$ .....(i)
For a block body $P_{B} = σ A (T^{4} - T_{5}^{4})$ .....(ii)
$\frac{(i)}{(ii)} gives \frac{P}{P _{B}} = e$
$\Rightarrow e = \frac{210}{700} = 0.3$