A body with area A and temperature T and emissivity e = 0.6 is kept inside a spherical black body. W

Q: Does the spherical black body affect the energy radiated by the inner body?

No, the surroundings only affect the *net* heat exchange (energy absorbed minus energy radiated). The energy *radiated* by the body itself depends solely on its own temperature, area, and emissivity.

Q: What is the difference between a black body and a regular body?

A perfect black body has an emissivity of $e = 1$, meaning it is a perfect emitter and absorber of radiation. A regular body has an emissivity between 0 and 1.

Q: Why is the answer not $1.00 \sigma AT^4$?

Because the body is not a perfect black body; its emissivity is given as $0.6$. Therefore, it radiates only $60\%$ of the energy a black body would radiate at the same temperature.

Q: What does the term 'maximum energy radiated' mean here?

It simply refers to the total thermal energy emitted by the body per unit time, which is given by the Stefan-Boltzmann law.

Q: What is the value of the Stefan-Boltzmann constant?

The Stefan-Boltzmann constant $\sigma$ is approximately $5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$.

Question

A body with area $A$ and temperature $T$ and emissivity $e = 0.6$ is kept inside a spherical black body. What will be the maximum energy radiated?

Options:

(A) $0.60 \sigma AT^4$

(B) $0.80 \sigma AT^4$

(C) $1.00 \sigma AT^4$

(D) $0.40 \sigma AT^4$

Sigma Percentile · Accepted Answer

### Understanding Thermal Radiation

When we talk about heat transfer, one of the most fascinating mechanisms is **thermal radiation**. Unlike conduction or convection, radiation doesn't need a medium to travel through; it can propagate through the vacuum of space. This is how the Sun warms the Earth!

### The Master Equation: Stefan-Boltzmann Law

The fundamental law governing thermal radiation is the **Stefan-Boltzmann Law**. It tells us exactly how much energy a body emits per unit time based on its temperature. The equation is beautifully simple yet powerful:

$$E = e \sigma A T^4$$

Let's break down what each term means:
- **$E$** is the energy radiated per second (Power).
- **$e$** is the **emissivity** of the body. It's a dimensionless number between $0$ and $1$ that tells us how good of an emitter the body is compared to a perfect black body.
- **$\sigma$** is the **Stefan-Boltzmann constant** ($\approx 5.67 	imes 10^{-8} 	ext{ W m}^{-2} 	ext{ K}^{-4}$).
- **$A$** is the surface area of the body.
- **$T$** is the absolute temperature in Kelvin.

### Analyzing the Setup

In our problem, we have a body with an area $A$, a temperature $T$, and an emissivity $e = 0.6$. It is placed inside a spherical black body.

A common trap here is to overthink the presence of the spherical black body. You might wonder, "Does the black body reflecting or emitting radiation back onto our object change how much our object radiates?"

The answer is a resounding **no**.

The energy *radiated* by a body is an intrinsic property. It depends **only** on its own temperature, its own surface area, and its own emissivity. The surroundings will certainly affect the *net* heat exchange (how much energy the body absorbs versus how much it emits), but they do not change the raw amount of energy the body pushes out into the universe.

### Final Calculation

Since the surroundings don't affect the emission rate, we can simply plug our given values directly into the Stefan-Boltzmann equation:

$$E = (0.6) \sigma A T^4$$

This gives us our final answer:

**$E = 0.60 \sigma A T^4$**

It's a straightforward application of the formula, but it tests a crucial conceptual understanding: the independence of emitted radiation from surrounding conditions.

Animated Solution for Physics - Properties of Solids and Liquids: A body with area $A$ and temperature $T$ and emissivity $e = 0.6$ is kept inside a spherical black body. What will be the maximum energy radiated?

Select Answer:

Visualized Solution

$E = e σ A T^{4}$

$E = 0.6 σ A T^{4}$

$E = 0.60 σ A T^{4}$

The Sigma Insight: Heat Transfer

Understanding Thermal Radiation

The Master Equation

Analyzing the Setup

Final Calculation

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