Question

The total emissive power at 0 °C is Wâ. So total emissive power at 546 °C is .... .

Answer 1

Answer:

The total emissive power at 546 °C is 80.88 W.

Explanation:

Total emissive power for a black body radiation is the total energy emitted by the black body per unit time. It is related with the temperature as

$P = \dfrac Et = \dfrac 1t \sigma T^4$

where,

$\sigma$ = Boltzmann constant.

$T$ = temperature.

Given that,

The total emissive power at 0 °C is W, such that,

$T_1 = 0^\circ C= 273.15\ K\\P_1 = \dfrac{E_1}{t} = W\\\\\\\therefore P_1=\dfrac 1t \sigma T_1^4\\W=\dfrac 1t \sigma (273.15)^4\\\sigma = \dfrac{Wt}{273.15^4}$

For the temperature, $T_2 = 546^\circ C= 819.15\ K.$

The total emissive power is given by

$P_2=\dfrac {E_2}{t} = \dfrac 1t \sigma T_2^4\\P_2=\dfrac 1t\sigma ( 819.15)^4\\\text{Putting the value of } \sigma ,\\P_2=\dfrac 1t \dfrac{Wt}{273.15^4} ( 819.15)^4 = 80.88\ W.$