Question

Parallel rays of light of intensity $I = 912W{m^{ - 2}}$ are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan-Boltzmann constant $\sigma = 5.7 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$ and assume that the energy exchange with the surrounding is only through radiation. The final steady state temperature of the black body is close to-

A.)330K
B.)660K
C.)990K
D.)1550K

Answer 1

Hint: The question has been asked about the steady state temperature. We achieve steady state temperature only when the energy radiated is equal to the energy lost. Energy radiated is in the black body while the energy lost is in the intensity of light.

Complete answer:
Formula used: Energies in both the forms: $e\sigma A{\left( {\Delta T} \right)^4}$, $I\pi {R^2}$

The sphere given to us is a black body which means that it will absorb light from every point over its surface area. The area of the emitting body will be circular in shape and its area will be equal to the area of the black body where the light falls which will also be circular in shape.

Hence, the energy radiated will be (according to Stefan’s law)-

$ \Rightarrow e\sigma A{\left( {\Delta T} \right)^4}$

Where,

E is the emissivity.
$\sigma $ is the Stefan-Boltzmann constant.
A is the area of the sphere
And, $\Delta T$ is the change in the temperature.
As we know that the emissivity of a black body is always considered as 1.
Let the radius of the sphere be R.

Substituting the value, we have-

$
   \Rightarrow e\sigma A{\left( {\Delta T} \right)^4} \\
    \\
   \Rightarrow 1 \times \sigma \times 4\pi {R^2} \times \left( {{T^4} - {T_0}^4} \right) \\
$

The value of energy lost will be-

$ \Rightarrow I\pi {R^2}$, (since the radius of the black body is equal to the radius of the circle from where the light is being emitted.)

Equating both the equation, we have-

$
   \Rightarrow 1 \times \sigma \times 4\pi {R^2} \times \left( {{T^4} - {T_0}^4} \right) = I\pi {R^2} \\
    \\
   \Rightarrow I = \sigma \times 4 \times \left( {{T^4} - {T_0}^4} \right) \\
$

We have to find out the value of T. The value of surrounding temperature ${T_0}$ is given to us by the question already which is 300K.

$
   \Rightarrow I = \sigma \times 4 \times \left( {{T^4} - {T_0}^4} \right) \\
    \\
   \Rightarrow {T^4} - {T_0}^4 = \dfrac{I}{{4\sigma }} \\
    \\
   \Rightarrow {T^4} = {\left( {\dfrac{I}{{4\sigma }} + {T_0}^4} \right)^{\dfrac{1}{4}}} \\
$

Substituting all the given values, we will have-

$
   \Rightarrow {T^4} = {\left( {\dfrac{I}{{4\sigma }} + {T_0}^4} \right)^{\dfrac{1}{4}}} \\
    \\
   \Rightarrow T = {\left( {\dfrac{{912}}{{5.7 \times {{10}^{ - 8}} \times 4}} + {{\left( {300} \right)}^4}} \right)^{\dfrac{1}{4}}} \\
    \\
   \Rightarrow T = {\left( {\dfrac{{912}}{{288 \times {{10}^{ - 9}}}} + {{\left( {300} \right)}^4}} \right)^{\dfrac{1}{4}}} \\
    \\
   \Rightarrow T = {\left( {3.16 \times {{10}^9} + 81 \times {{10}^8}} \right)^{\dfrac{1}{4}}} \\
    \\
   \Rightarrow T = {\left( {31.6 \times {{10}^8} + 81 \times {{10}^8}} \right)^{\dfrac{1}{4}}} \\
    \\
   \Rightarrow T = {\left( {112.6 \times {{10}^8}} \right)^{\dfrac{1}{4}}} \\
    \\
   \Rightarrow T = 331.66K \approx 330K \\
$

Hence, it is clear that option A is the correct option.

Note: A black body is an idealized physical structure which, irrespective of frequency or angle of occurrence, absorbs all electromagnetic incident radiation. The term 'black body' derives from consuming radiation at all wavelengths, and not just by processing it. The black body is able to radiate from the human eye. On the opposite, a white body has a 'raw surface which totally and uniformly reflects all incident rays in every direction.'