Problem 58 - Master Resource Book in JEE Main Physics

Question

The temperature of a black body is increased by \(50 \%\), then the percentage of increase of radiation is approximately (a) \(100 \%\) (b) \(25 \%\) (c) \(400 \%\) (d) \(500 \%\)

Step-by-Step Solution

Answer

The radiation increases by approximately 406.25%, which is closest to option (d) 500%.

Step 1: Understanding the Problem

The problem asks us to find the percentage increase in radiation emitted by a black body when its temperature increases by 50%. We should use the Stefan-Boltzmann Law, which states that the power radiated by a black body is proportional to the fourth power of its temperature.

Step 2: Applying Stefan-Boltzmann Law

The Stefan-Boltzmann Law is given by the formula \( E = \sigma T^4 \), where \( E \) is the energy radiated per unit area, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the temperature in Kelvin. Our goal is to find the new energy output after the temperature change.

Step 3: Calculating New Temperature

Let the original temperature be \( T \). The new temperature after a 50% increase will be \( T_{new} = 1.5T \).

Step 4: Calculating Old Energy Output

The original energy output is \( E_{old} = \sigma T^4 \).

Step 5: Calculating New Energy Output

With the new temperature, the energy output is \( E_{new} = \sigma (1.5T)^4 \). Expanding this gives \( E_{new} = \sigma (1.5^4) T^4 = \sigma \times 5.0625 T^4 \).

Step 6: Find Percentage Increase

The increase in radiation is given by \( E_{new} - E_{old} \). Substitute the values to get: \( 5.0625\sigma T^4 - \sigma T^4 = 4.0625 \sigma T^4 \). To find the percentage increase, divide by the original energy and multiply by 100: \( \frac{4.0625 \sigma T^4 }{\sigma T^4} \times 100\% = 406.25\% \).

Key Concepts

Black Body Radiation

The concept of black body radiation is central to the understanding of how objects emit energy based on their temperature. A black body is an idealized object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.

It does not reflect or transmit any light, making it the perfect absorber and emitter of radiation. This property also means that a black body emits radiation with the maximum possible intensity for a given temperature.

• Perfect Absorber: Absorbs all electromagnetic radiation.
• Perfect Emitter: Radiates maximum energy for a given temperature.
• Ideal Model: Black bodies are theoretical constructs but are useful for understanding real-world objects.

The emissions from a black body are purely a function of its temperature. This relationship is essential for understanding how changes in temperature affect the radiation emitted.

Temperature Increase Effect

When the temperature of a black body increases, the amount of radiation it emits also changes significantly. The temperature increase effect is governed by the Stefan-Boltzmann Law.

This law states that the power emitted by a black body is proportional to its temperature raised to the fourth power. Mathematically, it can be expressed as:\[ E = \sigma T^4 \]where:

• \( E \): Energy radiated per unit area.
• \( \sigma \): Stefan-Boltzmann constant.
• \( T \): Temperature in Kelvin.

A small increase in temperature results in a much larger increase in energy output due to the fourth power relationship described by the law. Thus, when the temperature increases by 50%, this effect is very pronounced, leading to a dramatic increase in radiation.

Radiation Percentage Increase

The radiation percentage increase as temperature rises can be calculated by comparing the initial and final radiated energies. When a black body's temperature increases by 50%, we calculate the new temperature as \( 1.5T \) if the original temperature is \( T \).

To determine the new radiation, we apply the Stefan-Boltzmann Law again:\[ E_{new} = \sigma (1.5T)^4 \]Simplifying,

• Name: Equation simplifies to \( 5.0625 \sigma T^4 \).
• Initial Energy Output: \( E_{old} = \sigma T^4 \).

The increase in radiation is given by the difference between \( E_{new} \) and \( E_{old} \), which is \( 4.0625 \sigma T^4 \).

Finally, the percentage increase is determined by:\[ \frac{4.0625 \sigma T^4}{\sigma T^4} \times 100\% = 406.25\% \]Thus, the radiation emitted increases by approximately 406.25% when the temperature of a black body is increased by 50%.