Problem 128 - Master Resource Book in JEE Main Physics

Question

Two spheres made of same material have radii in the ratio \(1: 2 .\) Both are at same temperature. Ratio of heat radiation energy emitted per second by them is (a) \(1: 2\) (b) \(1: 4\) (c) \(1: 8\) (d) \(1: 16\)

Step-by-Step Solution

Answer

The heat radiation energy emitted per second ratio is 1:4 (answer b).

Step 1: Understanding the Problem

We are given two spheres made of the same material and at the same temperature, with radii in the ratio 1:2. We need to find the ratio of heat radiation energy emitted per second by both spheres.

Step 2: Applying the Formula for Radiated Heat Energy

The heat radiation energy emitted per second by a body is given by the formula for power: \ P = \epsilon \sigma A T^4 \. Here, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the temperature. Since both spheres are at the same temperature and made of the same material, \(\epsilon\), \(\sigma\), and \(T\) remain constant and do not affect the ratio.

Step 3: Calculating Surface Area Ratio

Since only the surface area \(A\) affects the ratio and the sphere's surface area is \(4\pi r^2\), the surface areas of the spheres are in relation to their radii. If the radii ratio is 1:2, the surface area ratio is \((4\pi(1)^2):(4\pi(2)^2) = 1:4\).

Step 4: Determining the Ratio of Radiated Energy

Given the surface area ratio is 1:4 and the formula for power includes \(A\), the ratio of the heat radiation energy emitted per second will follow the ratio of the area: 1:4, making the radiated power ratio also 1:4.

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is fundamental in understanding the radiation of energy from objects due to their temperature. This law states that the power radiated by an object, or the energy emitted per second, is proportional to the fourth power of the object's absolute temperature. This relationship is expressed mathematically as \[ P = \ \ \epsilon \ \sigma A T^4 \], where:

• \( P \) stands for the power radiated by the object.
• \( \epsilon \) is the emissivity, which determines how efficiently an object emits thermal radiation compared to an ideal black body.
• \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \ W \, m^{-2} \, K^{-4} \).
• \( A \) represents the surface area of the object.
• \( T \) is the absolute temperature in Kelvin.

This law helps us understand why objects at higher temperatures emit more thermal radiation, emphasizing the effect of temperature raised to the fourth power.

Surface Area of Spheres

To determine the amount of heat radiated, understanding the surface area of spheres is essential. A sphere's surface area can be calculated using the formula:\[ A = 4\pi r^2 \]where \( A \) is the surface area and \( r \) is the radius of the sphere. This relationship shows that the surface area of a sphere increases with the square of its radius. In the context of the problem involving two spheres with radii in the ratio of 1:2, their surface areas will be in a ratio calculated as follows:

• Spherical surface with radius \( r_1 \): \( A_1 = 4\pi (1)^2 = 4\pi \)
• Spherical surface with radius \( r_2 \): \( A_2 = 4\pi (2)^2 = 16\pi \)

Thus, the surface area ratio of the two spheres is \( 1:4 \). This calculation is crucial as it directly influences the ratio of heat energy they radiate, given all other factors equal.

Radiation Emissivity

Radiation emissivity is a measure of a material's efficiency in emitting thermal radiation compared to an ideal black body. The value of emissivity, denoted by \( \epsilon\ \), ranges between 0 and 1.

• A value of \( 1 \) means the material is a perfect emitter, also known as a black body.
• A value of \( 0 \) describes a material that does not emit thermal radiation.

In practical terms, emissivity affects how much heat energy an object can emit. For the spheres described in the original problem, because they are made of the same material and at the same temperature, their emissivity values are equal and thus cancel out in the ratio calculation of emitted heat energy. This scenario allows us to directly compare their surface areas to determine the final energy ratio, where emissivity plays an important background role in the overall emission capacity.