Black Body Radiation and Power Emission

This problem involves calculating the ratio of the total power emitted by two perfect black bodies, A and B, based on their temperatures and surface areas. Perfect black bodies are ideal objects that absorb all incoming radiation and emit thermal radiation according to the Stefan-Boltzmann Law.

Stefan-Boltzmann Law

The total energy radiated per unit surface area of a black body per unit time (also known as the emissive power or intensity) is directly proportional to the fourth power of its absolute temperature. The total power emitted by a black body with surface area $A$ at absolute temperature $T$ is given by the Stefan-Boltzmann Law:

$$P = \sigma A T^4$$

where:

• $P$ is the total power emitted.
• $\sigma$ is the Stefan-Boltzmann constant, a universal physical constant.
• $A$ is the surface area of the black body.
• $T$ is the absolute temperature of the black body in Kelvin.

For two black bodies A and B, the power emitted will be:

• Power emitted by A: $P_A = \sigma A_A T_A^4$
• Power emitted by B: $P_B = \sigma A_B T_B^4$

Calculating the Power Ratio

We are asked to find the ratio of the total power emitted by A to that by B, which is $\frac{P_A}{P_B}$.

Using the formulas above, the ratio is:

$$\frac{P_A}{P_B} = \frac{\sigma A_A T_A^4}{\sigma A_B T_B^4}$$

The Stefan-Boltzmann constant $\sigma$ is the same for both bodies and cancels out:

$$\frac{P_A}{P_B} = \frac{A_A T_A^4}{A_B T_B^4} = \left(\frac{A_A}{A_B}\right) \left(\frac{T_A}{T_B}\right)^4$$

Substituting Given Values

From the question, we are given:

• Temperature of A, $T_A = 400$ K
• Temperature of B, $T_B = 200$ K
• Surface area of A is twice that of B, $A_A = 2 A_B$. This means $\frac{A_A}{A_B} = 2$.

Now, substitute these values into the ratio equation:

$$\frac{P_A}{P_B} = \left(\frac{A_A}{A_B}\right) \left(\frac{T_A}{T_B}\right)^4 = (2) \left(\frac{400 \text{ K}}{200 \text{ K}}\right)^4$$

Simplify the temperature ratio:

$$\frac{T_A}{T_B} = \frac{400}{200} = 2$$

Now, substitute this back into the power ratio equation:

$$\frac{P_A}{P_B} = (2) (2)^4$$

Calculate the power of 2:

$$(2)^4 = 2 \times 2 \times 2 \times 2 = 16$$

Finally, calculate the ratio:

$$\frac{P_A}{P_B} = 2 \times 16 = 32$$

Thus, the ratio of the total power emitted by A to that by B is 32.

Conclusion

The ratio of total power emitted by black body A to that by black body B is 32. This result shows the strong dependence of emitted power on both temperature (to the fourth power) and surface area.