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

Step 1: Identify the given parameters.

The given values for the system are the incident intensity, the surrounding temperature, and the Stefan-Boltzmann constant.

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

Step 2: Determine the power absorbed by the spherical black body.

Since the incident light consists of parallel rays, the beam is intercepted only by the projected cross-sectional area of the sphere.

$$P_{\text{abs}} = I \pi R^{2}$$

Step 3: Determine the net power emitted by the spherical black body.

The black body radiates energy into the surroundings from its entire surface area.

$$P_{\text{emit}} = \sigma (4\pi R^{2}) (T^{4} - T_0^{4})$$

Step 4: Apply the steady-state condition.

At steady state, the rate of energy absorption equals the rate of net energy emission.

$$P_{\text{abs}} = P_{\text{emit}}$$ $$I \pi R^{2} = 4\pi R^{2} \sigma (T^{4} - T_0^{4})$$

Step 5: Solve for the steady-state temperature $T$.

Simplify the equation to express $T^{4}$ in terms of the known variables.

$$T^{4} = T_0^{4} + \frac{I}{4\sigma}$$

Substitute the given values into the equation.

$$T^{4} = (300)^{4} + \frac{912}{4(5.7 \times 10^{-8})}$$ $$T^{4} = 8.1 \times 10^{9} + 4.0 \times 10^{9}$$ $$T^{4} = 1.21 \times 10^{10}$$ $$T = (1.21 \times 10^{10})^{1/4} \approx 331\,\text{K}$$

Answer: The final steady-state temperature of the black body is close to $330\,\text{K}$ (or more precisely, $331\,\text{K}$).