Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respecti

Q: Why did we assume the emissivities are equal?

The problem explicitly states that both spheres are made of the 'same material'. Emissivity is a fundamental property of a material's surface, so it remains constant for both spheres regardless of their size or temperature.

Q: Why is the area $4\pi r^2$ and not $\pi r^2$?

The energy is radiated outward from the entire 3D outer boundary of the sphere, not just a flat 2D cross-section. The total surface area of a 3D sphere is mathematically $4\pi r^2$.

Q: Does the distance between the spheres matter for this calculation?

No, the question asks for the energy *radiated* by each sphere individually. This emission depends only on their own intrinsic properties (size, temperature, material), not on their surroundings or how far apart they are.

Q: What if the temperatures were given in Celsius instead of Kelvin?

You MUST convert them to Kelvin before using Stefan's Law. The $T$ in the $T^4$ formula strictly represents absolute thermodynamic temperature. Using Celsius will yield completely incorrect ratios.

Q: Why does temperature have such a huge impact compared to the radius?

In Stefan's Law, radiated energy depends on the *fourth power* of temperature ($T^4$), but only on the *square* of the radius ($r^2$). This means doubling the temperature increases radiation by $16$ times, whereas doubling the radius only increases it by $4$ times.

Question

Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

Options:

(A) 1 : 1

(B) 16 : 1

(C) 4 : 1

(D) 1 : 9

Sigma Percentile · Accepted Answer

### The Battle of the Spheres: Size vs. Temperature in Thermal Radiation

Imagine a cosmic showdown between two glowing spheres. In one corner, we have a tiny sphere, just $1	ext{ m}$ in radius, but blazing with an intense heat of $4000	ext{ K}$. In the other corner stands a massive sphere, $4	ext{ m}$ in radius, but glowing with a cooler, more subdued temperature of $2000	ext{ K}$.

The question is: which of these two spheres radiates more energy into the universe every second?

To answer this, we must call upon one of the most elegant laws in thermodynamics.

### The Master Equation: Stefan-Boltzmann Law

The rate at which a black body (or any thermal body) radiates energy is governed by the **Stefan-Boltzmann Law**. It states that the energy radiated per second ($E$) is directly proportional to the surface area ($A$) and the *fourth power* of its absolute temperature ($T$).

Mathematically, it is written as:

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

Here, $e$ is the emissivity of the material (a measure of how well it radiates compared to a perfect black body), $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the temperature in Kelvin.

### The Mathematical Duel

We need to find the ratio of the energy radiated by the first sphere ($E_1$) to the second sphere ($E_2$). Let's set up the ratio:

$$\frac{E_1}{E_2} = \frac{e_1 \sigma A_1 T_1^4}{e_2 \sigma A_2 T_2^4}$$

Now, let's look at the constraints given in the problem. We are told that both spheres are made of the **same material**. This is a crucial piece of information! It means their emissivities are identical ($e_1 = e_2$), allowing us to cancel them out. The constant $\sigma$ also cancels out naturally.

Furthermore, the surface area of a sphere is given by $A = 4\pi r^2$. Substituting this into our ratio, the $4\pi$ terms will also vanish, leaving us with a beautifully simplified equation:

$$\frac{E_1}{E_2} = \left(\frac{r_1}{r_2}ight)^2 \left(\frac{T_1}{T_2}ight)^4$$

### The Final Showdown

Now, we bring in our contenders' stats. For the first sphere, $r_1 = 1$ and $T_1 = 4000$. For the second sphere, $r_2 = 4$ and $T_2 = 2000$. Let's plug these numbers into our refined equation:

$$\frac{E_1}{E_2} = \left(\frac{1}{4}ight)^2 \left(\frac{4000}{2000}ight)^4$$

Let's break down the math. The radius ratio squared gives us $\left(\frac{1}{4}ight)^2 = \frac{1}{16}$. This tells us that based purely on size, the smaller sphere has only $\frac{1}{16}	ext{th}$ the radiating surface area of the larger one.

But wait! Look at the temperature ratio. $\frac{4000}{2000} = 2$. Because temperature is raised to the fourth power, this factor becomes $2^4 = 16$. The smaller sphere's higher temperature gives it a massive $16	imes$ multiplier in radiation power!

Multiplying these two competing factors together:

$$\frac{E_1}{E_2} = \frac{1}{16} 	imes 16 = 1$$

### The Physical Insight

The result is a perfect $1:1$ tie. The massive size advantage of the second sphere was completely neutralized by the intense heat of the first sphere. This problem beautifully illustrates the sheer power of the $T^4$ dependence in thermal radiation. A body doesn't need to be huge to radiate massive amounts of energy; it just needs to be hot!

Animated Solution for Physics - Properties of Solids and Liquids: Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

Select Answer:

Visualized Solution

Visualizing the Spheres

Stefan-Boltzmann Law

Setting up the Ratio

Simplifying the Expression

Substituting Values

Calculating the Powers

Final Answer

The Way Forward

The Sigma Insight: Heat Transfer

The Battle of the Spheres

The Master Equation

The Mathematical Duel

The Final Showdown

The Physical Insight

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