Steady State Temperature of Spherical Black Body in Radiation

To find the final steady-state temperature $T$ of the spherical black body, we equate the total rate of energy absorbed by the sphere to the total rate of energy radiated by it.

1. Rate of Energy Absorbed

Let the radius of the spherical black body be $R$.

• The projected area of the sphere perpendicular to the parallel incident light rays is: $A_{\text{proj}} = \pi R^2$ Since it is a black body (absorptivity $a = 1$), the energy absorbed per second from the incident parallel light beam is: $P_{\text{incident}} = I \cdot A_{\text{proj}} = I \pi R^2$

• The total surface area of the sphere through which it exchanges radiation with the surroundings is: $A_{\text{total}} = 4\pi R^2$ The thermal radiation absorbed from the surrounding environment at temperature $T_0 = 300\text{ K}$ is: $P_{\text{surr}} = \sigma T_0^4 \cdot A_{\text{total}} = 4\pi R^2 \sigma T_0^4$

Therefore, the total power absorbed by the body is: $P_{\text{in}} = P_{\text{incident}} + P_{\text{surr}} = I \pi R^2 + 4\pi R^2 \sigma T_0^4$

2. Rate of Energy Radiated

At the steady-state temperature $T$, the rate of energy emitted by the black body (with emissivity $e = 1$) from its entire surface is given by the Stefan-Boltzmann law: $P_{\text{out}} = \sigma T^4 \cdot A_{\text{total}} = 4\pi R^2 \sigma T^4$

3. Steady-State Equilibrium

At steady state, the net rate of heat transfer is zero, which means: $P_{\text{in}} = P_{\text{out}}$ $I \pi R^2 + 4\pi R^2 \sigma T_0^4 = 4\pi R^2 \sigma T^4$

Dividing both sides by $\pi R^2$: $I + 4\sigma T_0^4 = 4\sigma T^4$

Rearranging the expression gives: $T^4 - T_0^4 = \frac{I}{4\sigma}$

4. Numerical Calculation

Substitute the given values:

• $I = 912\text{ W m}^{-2}$
• $\sigma = 5.7 \times 10^{-8}\text{ W m}^{-2}\text{K}^{-4}$
• $T_0 = 300\text{ K}$

$\frac{I}{4\sigma} = \frac{912}{4 \times 5.7 \times 10^{-8}} = \frac{912}{22.8 \times 10^{-8}} = 40 \times 10^8\text{ K}^4$

The term $T_0^4$ is: $T_0^4 = (300)^4 = 81 \times 10^8\text{ K}^4$

Thus: $T^4 = T_0^4 + \frac{I}{4\sigma} = 81 \times 10^8 + 40 \times 10^8 = 121 \times 10^8\text{ K}^4$

Taking the fourth root: $T = \left(121 \times 10^8\right)^{1/4} = \sqrt{\sqrt{121}} \times 100 = \sqrt{11} \times 100\text{ K}$

Since $\sqrt{11} \approx 3.317$: $T \approx 3.317 \times 100 \approx 331.7\text{ K} \approx 330\text{ K}$

Therefore, the final steady-state temperature attained by the black body is close to $330\text{ K}$.

Correct Answer: (A)