An ideal black body at room temperature is thrown into a furnace. It is observed that
(a) initially

Q: Why doesn't the black body reflect any light initially?

By definition, an ideal black body is a perfect absorber. It absorbs 100% of the electromagnetic radiation that strikes it, regardless of wavelength or angle of incidence, meaning zero reflection.

Q: If it absorbs all light, shouldn't it always be dark?

It absorbs all *incident* light, but it also *emits* its own thermal radiation. At room temperature, this emission is mostly in the invisible infrared spectrum. But at high temperatures, it emits intensely in the visible spectrum.

Q: What is Kirchhoff's Law of radiation?

Kirchhoff's Law states that for an arbitrary body in thermal equilibrium with its environment, its emissivity is equal to its absorptivity. A perfect absorber is therefore a perfect emitter.

Q: Why does it become the *brightest* body and not just as bright as the furnace?

The furnace walls and other objects might not be ideal black bodies (their emissivity is less than 1). Since the ideal black body has an emissivity of exactly 1, it will emit more radiation per unit area than any non-ideal body at the same temperature.

Q: Does the black body ever stop absorbing heat?

Once it reaches thermal equilibrium with the furnace, it continues to absorb radiation, but it emits radiation at the exact same rate. The *net* heat exchange becomes zero, keeping its temperature constant.

Question

An ideal black body at room temperature is thrown into a furnace. It is observed that

Options:

(A) initially it is the darkest body and at later times the brightest

(B) it is the darkest body at all times

(C) it cannot be distinguished at all times

(D) initially it is the darkest body and at later times it cannot be distinguished

Sigma Percentile · Accepted Answer

The behavior of an ideal black body when subjected to extreme temperature changes is one of the most fascinating thought experiments in thermodynamics. Let's embark on a journey to understand exactly what happens when we throw a room-temperature black body into a blazing furnace.

### Analyzing the Initial State

Imagine a perfectly black body, initially at room temperature, being tossed into a hot furnace. By definition, an ideal black body is a perfect absorber of electromagnetic radiation. It absorbs $100\%$ of the light and heat that falls upon it, reflecting absolutely nothing.

Since the black body is initially at room temperature, its own thermal emission is negligible, especially in the visible spectrum. Because it absorbs all the intense, glowing radiation from the furnace walls without reflecting any of it back to our eyes, it will visually appear as a completely dark void.

Therefore, initially, it is the **darkest body** in the furnace.

### The Transition to Thermal Equilibrium

But the black body won't stay cold forever. As it continuously absorbs the massive amount of heat energy from the furnace, its internal temperature begins to rise steadily.

This heating process continues until the black body reaches thermal equilibrium with its surroundings. At this point, the temperature of the black body becomes exactly equal to the temperature of the furnace, $T_f$.

### Kirchhoff's Law and the Final State

This brings us to a beautiful principle known as Kirchhoff's Law of thermal radiation. The law states that for any body in thermal equilibrium, its emissivity is equal to its absorptivity.

$$e_{\lambda} = a_{\lambda}$$

For an ideal black body, the absorptivity $a_{\lambda}$ is exactly $1$. Consequently, its emissivity $e_{\lambda}$ must also be exactly $1$. This means that a perfect absorber is also a perfect emitter.

Once the black body reaches the high temperature of the furnace, it emits thermal radiation at the maximum possible rate for that temperature. If there are other non-ideal bodies in the furnace (which have an emissivity less than $1$), the ideal black body will outshine them all.

Because it emits the most energy per unit area, it will visually appear as the **brightest body** in the furnace.

### Conclusion

Our thought experiment leads us to a clear and elegant conclusion. When an ideal black body is thrown into a furnace, it starts off by absorbing everything and reflecting nothing, making it the darkest object. However, once it heats up to the furnace's temperature, its perfect emissivity takes over, making it the brightest object.

Thus, initially it is the darkest body, and at later times, it becomes the brightest.

Animated Solution for Physics - Properties of Solids and Liquids: An ideal black body at room temperature is thrown into a furnace. It is observed that

Select Answer:

Visualized Solution

$The Setup$

$Initial State: Absorption$

$Visual Appearance Initially$

$Reaching Thermal Equilibrium$

$Kirchhoff’s Law of Radiation$

$Final State: Emission$

$Visual Appearance Later$

$Conclusion$

The Sigma Insight: Heat Transfer

Analyzing the Initial State

The Transition to Thermal Equilibrium

Kirchhoff's Law and the Final State

Conclusion

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