Question

For a black body at temperature 727°C, its radiating power is 20 watt and the temperature of the surrounding is 227°C. If the temperature of the black body is changed to 1227°C, then its rate of energy loss will be?

Answer 1

As we all know that a black body absorbs all the radiation falling on it and hence it is called the perfect absorber. We also know that a perfect absorber is also a perfect emitter. For a perfectly black body, the absorptivity is equal to the emissivity.

Complete step by step solution:
Given: The initial temperature of the black body ${T_1} = 727^\circ C$ .
The final temperature of the black body ${T_2} = 1227^\circ C$.
The temperature of the surrounding ${T_0} = 227^\circ C$.
The radiating power for body at $727^\circ C$ is ${P_1} = 20\,W$
We have studied Stefan’s Boltzmann law and according to that, the radiating power of a black body is defined as:

$P = \sigma A({T^4} - T_0^4)$ …… (I)

Here, P is the radiating power, $\sigma$ is the Stefan's-Boltzmann constant, $T$ is the temperature of the black body and ${T_0}$ is the temperature of the surrounding and A is the surface area.
For initial condition 1, the temperature in kelvin is,

$\Rightarrow \,{T_1} = \left( {727 + 273} \right)K$ ,

$\Rightarrow {T_1} = 1000\,K$

For final condition 2, the temperature in kelvin is,

$\Rightarrow {T_2} = (1227 + 273)K$
$\therefore {T_2} = 1500K$,

We can find the temperature of the surrounding in Kelvin as,
$\Rightarrow {T_0} = \left( {227 + 273} \right)K$
$\therefore {T_0} = 500K$

Therefore, equation (I) can be written for condition 1 as,
$\Rightarrow {P_1} = \sigma A(T_1^4 - T_0^4)$ …… (II)

Therefore, equation (I) can be written for condition 2 as,
$\Rightarrow {P_2} = \sigma A(T_2^4 - T_0^4)$ …… (III)

Therefore, we will now divide equation(II) by equation (III), the result will become as,
$\Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \left( {\dfrac{{T_1^4 - T_0^4}}{{T_2^4 - T_0^4}}} \right)$ …… (IV)

We will now substitute ${T_2} = 1500K$ , ${T_1} = 1000\,K$ , ${P_1} = 20\,W$ , ${T_0} = 227^\circ C$ and the result would simplify as,
$\Rightarrow \dfrac{{20W}}{{{P_2}}} = \left( {\dfrac{{{{1000}^4} - {{500}^4}}}{{{{1500}^4} - {{500}^4}}}} \right)$
$\Rightarrow \dfrac{{20W}}{{{P_2}}} = \dfrac{{{{500}^4}}}{{{{500}^4}}}\left( {\dfrac{{{2^4} - {1^4}}}{{{3^4} - {1^4}}}} \right)$
$\Rightarrow \dfrac{{20W}}{{{P_2}}} = \left( {\dfrac{{16 - 1}}{{81 - 1}}} \right)$
$\Rightarrow \dfrac{{20W}}{{{P_2}}} = \left( {\dfrac{{15}}{{80}}} \right)$
$\Rightarrow {P_2} = \left( {\dfrac{{80 \times 20W}}{{15}}} \right)$

$\therefore {P_2} = 106.67W$

$\therefore$ The rate of energy loss when the temperature of the black body is changed to $1227^\circ C$ is $106.67W$.

Note:
We can notice that the perfect example of a practical black body is a box painted black from inside and having a small hole in it. Practically talking, no radiation could reflect from the black surface after striking into it. All the radiation would be absorbed eventually.