Problem 77 - Master Resource Book in JEE Main Physics

Question

The rate of emission of radiation of a black body at temperature \(27^{\circ} \mathrm{C}\) is \(E_{1}\). If its temperature is increased to \(327^{\circ} \mathrm{C}\), the rate of emission of radiation is \(E_{2} .\) The relation between \(E_{1}\) and \(E_{2}\) is (a) \(E_{1}=24 E_{1}\) (b) \(E_{2}=16 E_{1}\) (c) \(E_{2}=8 E_{1}\) (d) \(E_{2}=4 E_{1}\)

Step-by-Step Solution

Answer

The relation is \( E_2 = 16 E_1 \) (Option b).

Step 1: Understanding the Stefan-Boltzmann Law

The rate of emission of radiation by a black body is described by the Stefan-Boltzmann Law, which states that the power emitted per unit area by a perfect black body is proportional to the fourth power of the temperature in Kelvin. Mathematically, it is given by \( E = \sigma T^4 \), where \( E \) is the emissive power, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the temperature in Kelvin.

Step 2: Convert Temperatures to Kelvin

The given temperatures are \(27^{\circ} \mathrm{C}\) and \(327^{\circ} \mathrm{C}\). Convert these to Kelvin by adding 273.15 to each: - \(27^{\circ} \mathrm{C} = 27 + 273.15 = 300.15 \mathrm{K}\) - \(327^{\circ} \mathrm{C} = 327 + 273.15 = 600.15 \mathrm{K}\). For simplicity in calculations, we round these to \(300 \mathrm{K}\) and \(600 \mathrm{K}\), respectively.

Step 3: Apply Stefan-Boltzmann Law to Both Temperatures

For the initial temperature \(300 \mathrm{K}\), the rate of emission is \( E_1 = \sigma (300)^4 \). For the increased temperature \(600 \mathrm{K}\), the rate of emission is \( E_2 = \sigma (600)^4 \).

Step 4: Calculate the Ratio of Emissions

The ratio of the emissions \( \frac{E_2}{E_1} \) can be calculated using:\[ \frac{E_2}{E_1} = \frac{\sigma (600)^4}{\sigma (300)^4} = \left(\frac{600}{300}\right)^4 = 2^4 = 16 \]

Step 5: Conclude the Relation between Emission Rates

Based on the ratio calculated, \( E_2 = 16 E_1 \). This corresponds to option (b).

Key Concepts

Black Body Radiation

Black body radiation is a fundamental concept in physics that describes the way all matter emits and absorbs radiant energy. A black body is an idealized object that absorbs all incoming light without reflecting any. Because of this property, black bodies are perfect emitters of radiation. The radiation emitted by a black body is only dependent on its temperature.
The emitted radiation covers a broad spectrum of wavelengths; however, the intensity of radiation peaks at a particular wavelength depending on temperature. This concept helps us understand the behavior of real-world objects approximated as black bodies.
In practical terms, many objects can be treated like black bodies for simplicity because they emit radiation in a manner close to black body radiation. For instance:

• Stars, including our sun, are often modeled as black bodies because of their radiation emission characteristics.
• Studying black body radiation gives insights into the temperature and energy dynamics in physical systems such as planetary atmospheres and everyday technologies.

Understanding black body radiation allows physicists to predict and calculate how different bodies emit energy depending on their temperature.

Temperature Conversion to Kelvin

Temperature conversion to Kelvin is a key step when applying the Stefan-Boltzmann Law for calculating emissive power. Most physics problems involving temperatures require conversion to the Kelvin scale. This is because Kelvin is an absolute temperature scale, starting from absolute zero, where particle motion ceases.
Converting Celsius to Kelvin is straightforward. You add 273.15 to the temperature in Celsius. For instance:

• A temperature of \(27^{\circ} \mathrm{C}\) becomes \(27 + 273.15 = 300.15 \mathrm{K}\). For simplicity in calculations, this might be approximated to \(300 \mathrm{K}\).
• Similarly, \(327^{\circ} \mathrm{C}\) becomes \(327 + 273.15 = 600.15 \mathrm{K}\), which can be rounded to \(600 \mathrm{K}\).

Using the Kelvin scale ensures that calculations involving physical laws related to temperature, such as the Stefan-Boltzmann Law, are accurate and consistent. It’s crucial to use Kelvin when working with these laws because they depend on the principle of absolute temperatures, unlike Celsius, which is relative.

Emissive Power

Emissive power refers to the amount of thermal radiation emitted by a surface at a given temperature. It is an essential concept in the Stefan-Boltzmann Law, which connects temperature with the power emitted by a black body. This law is expressed as \( E = \sigma T^4 \), where \( E \) represents the emissive power, \( \sigma \) is a constant, and \( T \) is the temperature in Kelvin.
A few critical points to understand about emissive power include:

• Emissive power increases with the fourth power of temperature. Doubling the temperature of a black body results in a thermal radiation output increase by a factor of sixteen.
• This exponential relationship highlights the massive impact that temperature can have on energy emission as demonstrated in the example, when temperatures change from \(300 \mathrm{K}\) to \(600 \mathrm{K}\), the emissive power increases by a factor of 16.
• The Stefan-Boltzmann constant \(\sigma\) in the equation is a predefined value and is crucial for quantifying the amount of energy radiated.

Emissive power helps us understand thermal energy management in everyday devices and scientific equipment, impacting fields like astronomy, climatology, and engineering.