Question

Parallel rays of light of intensity I=912 Wm$^{-2}$ are incident on a spherical black body kept in surroundings of temperature 300 K . Take Stefan constant $\sigma =5.7 \times 1^{-8} Wm ^{-2} K^{-4}$ 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

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

Answer 1

The Correct Option is A

To find the final steady state temperature of a spherical black body, we need to compare the power absorbed from the incident light with the power radiated by the black body. Below is the step-by-step solution:

1. The power absorbed by the black body is given by the formula:

   P_{\text{abs}} = I \cdot A\_{proj},

   where I = 912 \, \text{W/m}^2 is the intensity of the incident light, and A\_{proj} = \pi r^2 is the projected area of the sphere.

2. The power radiated by the black body can be calculated using the Stefan-Boltzmann law:

   P_{\text{rad}} = \sigma \cdot A \cdot T^4,

   where \sigma = 5.7 \times 10^{-8} \, \text{Wm}^{-2}\text{K}^{-4} is the Stefan-Boltzmann constant, A = 4\pi r^2 is the surface area of the sphere, and T is the final steady state temperature.

3. In steady state conditions, the power absorbed equals the power radiated:

   I \cdot \pi r^2 = \sigma \cdot 4\pi r^2 \cdot T^4.

4. Simplifying the equation:

   \frac{912}{4 \times 5.7 \times 10^{-8}} = T^4.

5. Calculate T^4:

   T^4 = \frac{912}{4 \times 5.7 \times 10^{-8}}.

   T^4 \approx 4.004 \times 10^{10}.

6. Find T by taking the fourth root:

   T \approx \sqrt[4]{4.004 \times 10^{10}} \approx 330 \, \text{K}.

Therefore, the final steady state temperature of the black body is approximately 330 \, \text{K}. This confirms that the correct answer is 330 K.