Question

A spherical black body with a radius of $12\;cm$ radiates $450\;W$ power at $500\;K$ . The radius was halved and the temperature doubled, the power radiated in watts would be:
(A) $225\;$
(B) $450\;$
(C) $900\;$
(D) $1800\;$

Answer 1

Here we will use Stefan’s law to obtain the relation between the area, temperature and the power radiated by a body. Since it is a black body, $e = 1$ is used. Then we find the ratios of the new temperature and are with the original values and substitute in the formula. Thus we will get the new value of the power radiation in terms of the original radiation.

Formula used:
$P = \sigma Ae{T^4}$ .

Complete step by step answer:
It is given in the question that the black body radiates a given amount of power. Any such radiation is given by Stefan’s law
$P = \sigma Ae{T^4}$
where $e$ is the emissivity of the body,
$\sigma$ is Stefan's constant, and
$A$ is the area of the body from which the radiation is taking place.
The Stefan’s constant has a value $\sigma = 5.67 \times {10^{ - 8}}W/{m^2}{K^4}$ .
Now for a perfect black body, the emissivity, $e = 1$ .
Thus we have our initial value of the power radiation as
${P_1} = \sigma {A_1}{T_1}^4$
where ${A_1}$ is the area of the black body and ${T_1}$ is the absolute temperature of the surface of the body given as ${T_1} = 500\;K$ .
It is given that the radius of the black body is ${r_1} = 12\;cm$ .
$\therefore {A_1} = 4\pi {r_1}^2$
Now when the radius of the black body is halved, the new radius becomes ${r_2} = 6\;cm$ .
$\therefore {A_2} = 4\pi {r_2}^2$
Upon substituting the values we get,
$\Rightarrow {A_2} = 4\pi {(\dfrac{{{r_1}}}{2})^2}$
$\Rightarrow {A_2} = \dfrac{{4\pi {r_1}^2}}{4}$
Thus we have,
$\Rightarrow {A_2} = \dfrac{{{A_1}}}{4}$
Also when the temperature of the surface of the black body is doubled, the new temperature ${T_2}$ becomes, ${T_2} = 1000\;K$ .
i.e., ${T_2} = 2{T_1}$
Now applying the Stefan’s law on the changed conditions, we have,
${P_2} = \sigma {A_2}{T_2}^4$
Substituting the values of the area and the temperature, we get,
$\Rightarrow {P_2} = \sigma \dfrac{{{A_1}}}{4}{(2{T_1})^4}$
$\Rightarrow {P_2} = \dfrac{{16}}{4}\sigma {A_1}{T_1}^4$
This is related to the initial power transmitted as,
$\Rightarrow {P_2} = 4{P_1}$
Substituting the value of ${P_1}$ in the above equation, we have the new power radiated as,
${P_2} = 4 \times 450\;W$
Thus ${P_2} = 1800\;W$ is the power radiated after the changes in temperature and the radius.
Therefore the correct answer is option (D).

Note:
The value of emissivity is a fraction which lies between $0$ and $1$ . This value is unity for a perfect black body. Also, the temperature used in the formula is the absolute value of the temperature in Kelvin. The area given here is assumed to be the total area of a sphere without any minimization due to contacts of placement.