Animated Solution for Physics - Properties of Solids and Liquids: Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength ${\lambda }_B$ corresponding to maximum spectral radiancy in the radiation from $B$ shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from $A$, by $1.00\mu \text{m}$. If the temperature of $A$ is $5802\text{ K}$

The beauty of thermal radiation lies in its universal laws. Every object around us is constantly emitting and absorbing electromagnetic radiation. In this thrilling problem, we are presented with two spherical bodies, $A$ and $B$, that are locked in a fascinating thermodynamic dance. They have the exact same surface area, yet their surfaces are fundamentally different. Body $A$ is highly reflective with an emissivity of $e_A=0.01$, while Body $B$ is much darker, boasting an emissivity of $e_B=0.81$. Despite this stark contrast, they are radiating total power at the exact same rate!

To unravel this mystery, we must first turn to the Stefan-Boltzmann Law. This fundamental principle states that the total power $P$ radiated by a black body is proportional to the fourth power of its absolute temperature $T$. For a non-ideal body, we introduce the emissivity $e$, giving us the master equation:

Here, $\sigma$ is the Stefan-Boltzmann constant, and $A$ is the surface area. We are given that the power radiated by both bodies is identical, so we can confidently set their equations equal to each other:

Notice the elegance of the physics here! The Stefan-Boltzmann constant $\sigma$ and the surface area $A$ appear on both sides of the equation. They perfectly cancel out, leaving us with a pure relationship between emissivity and temperature:

We know the temperature of Body $A$ is a blazing $T_A=5802\text{ K}$. Our goal is to find the temperature of Body $B$. Let's rearrange our simplified equation to isolate $T_B$:

Taking the fourth root of both sides, we get:

Now, we substitute the given values into our derived formula. The ratio of the emissivities is $\frac{0.01}{0.81}$, which simplifies beautifully to $\frac{1}{81}$.

Since $3^4=81$, the fourth root of $\frac{1}{81}$ is exactly $\frac{1}{3}$.

This result makes profound physical sense. Body $B$ is a much more efficient radiator (higher emissivity). Therefore, it doesn't need to be nearly as hot as Body $A$ to emit the exact same amount of total power.

Having conquered the temperature, we now turn our attention to the color of the radiation. Wien's Displacement Law tells us that the wavelength at which a body emits the maximum spectral radiancy (its peak color) is inversely proportional to its absolute temperature:

Because this product is a constant, we can relate the peak wavelengths and temperatures of our two bodies:

We want to find the relationship between the wavelengths, so we rearrange the equation:

We already know that $T_A$ is exactly three times $T_B$ ($5802/1934=3$). Therefore:

The problem provides one final piece of the puzzle: the peak wavelength of Body $B$ is shifted by $1.00\mu \text{m}$ compared to Body $A$. Since Body $B$ is cooler, Wien's Law dictates that its peak wavelength must be longer. Thus, the shift is positive:

We can substitute our previous finding, ${\lambda }_A=\frac{{\lambda }_B}{3}$, into this equation:

Solving for ${\lambda }_B$, we arrive at our final destination:

In conclusion, the temperature of Body $B$ is $1934\text{ K}$, and its peak emission wavelength is $1.5\mu \text{m}$. This makes options (a) and (b) the correct choices for this brilliantly crafted problem!