Question

Two spheres of the same material have radii $1m$ and $4m$ and temperature $4000K$ and $2000K$ respectively. Which of the following is the energy radiated per second by the first sphere?
A. Greater than of the second sphere
B. Less than of the second sphere
C. Equal to that of the second sphere
D. The information is incomplete to draw any conclusion.

Answer 1

Hint : Calculate the radiated energy by using Stefan’s law of radiation. The formula for total energy radiated from a body with surface area $A$ at absolute temperature $T$ is,
$E = e\sigma {T^4}A$ , Where $e$ is the emissivity of that body and $\sigma$ is the Stefan’s constant.

Complete Step By Step Answer:
We know that Stefan's law of blackbody radiation states that the energy radiated by a black body from per unit area per second is proportional to the fourth power of it’s absolute temperature.
That means, energy radiated per second per unit area from a blackbody, $E\alpha {T^4}$
$E$ is energy radiated per unit area per second from the blackbody,
$T$ is the absolute temperature of the blackbody.
Now, Equating this equation we know,
$E = \sigma {T^4}$
where, $\sigma$ is a constant and is called Stefan’s constant. It’s value is $\sigma = 5.672 \times {10^{ - 5}}W \cdot {m^{ - 2}} \cdot {K^{ - 4}}$
Now, If the body is a normal one rather than a blackbody then, energy radiated from the body per second per unit area becomes,
$E = e\sigma {T^4}$ Where $e$ is the emissivity of that body.
It is defined as, the ratio of the energy radiated from a material's surface to that radiated from a perfect emitter or known as a blackbody, at the same temperature and wavelength and under the same conditions. That means the geometry of the blackbody and the material must be the same.
So that means, emissivity $e = \dfrac{E}{{{E_B}}}$ where, $E$ is the radiated energy of a body with surface area $A$ and ${E_B}$ is the radiated energy per second from a blackbody of same surface area $A$ at absolute temperature $T$ .
$\therefore$ Total energy radiated from a body with surface area $A$ at absolute temperature $T$ is,
$E = e\sigma {T^4}A$
Here, we have two spheres built from same material of radius ${r_1} = 1m$ and ${r_2} = 4m$
Also, we have the two spheres are made of the same material, hence emissivity $e$ is the same for both the bodies.
$\therefore$ Energy radiated by the first sphere ${E_1} = e\sigma {T_1}^4{A_1}$ where, ${A_1} = 4\pi {r_1}^2 = 4\pi \cdot {1^2}$ , ${T_1} = 4000K$
$\therefore$ Energy radiated by the second sphere ${E_2} = e\sigma T_2^4{A_2}$ Where, ${A_2} = 4\pi {r_2}^2 = 4\pi \cdot {4^2}$ and ${T_2} = 2000K$
$\therefore$ $\therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{e\sigma T_1^4{A_1}}}{{e\sigma T_2^4{A_2}}} = \dfrac{{T_1^4{A_1}}}{{T_2^4{A_2}}}$
That becomes,
$= \dfrac{{T_1^44\pi r_1^2}}{{T_2^44\pi r_2^2}} = \dfrac{{T_1^4r_1^2}}{{T_2^4r_2^2}}$
Putting the values, ${r_1} = 1$ , ${T_1} = 4000K$ , ${r_2} = 4$ and ${T_2} = 2000K$ we get,
$= \dfrac{{{{\left( {40000} \right)}^4} \times {1^2}}}{{{{\left( {2000} \right)}^4} \times {4^2}}} = \dfrac{1}{1}$
$= 1$
Hence, energy radiated by the spheres is equal
Hence, correct option is option (C).

Note :
$\bullet$ Stefan’s law of radiation states that the energy radiated per second per unit area from a blackbody is which is basically the power radiated by that body per unit area.
$\bullet$ Ratio of the energy radiated from two bodies with the same material does not depend on the emissivity.