Problem 70 - Master Resource Book in JEE Main Physics

Question

The rectangular surface of area \(8 \mathrm{~cm} \times 4 \mathrm{~cm}\) of a black body at a temperature of \(127^{\circ} \mathrm{C}\) emits energy at the rate of \(E\) per second. If the length and breadth of the surface are each reduced to half of its initial value, and the temperature is raised to \(327^{\circ} \mathrm{C}\), the rate of emission of energy will become (a) \(\frac{3}{8} E_{1}\) (b) \(\frac{81}{16} E\) (c) \(\frac{9}{16} E\) (d) \(\frac{81}{64} E\)

Step-by-Step Solution

Answer

The rate of emission of energy will become \(\frac{81}{64} E\), option (d).

Step 1: Understanding the Problem

We are given a rectangular black body surface with dimensions \(8\text{ cm} \times 4\text{ cm}\) initially, and it's emitting energy \(E\) per second at a temperature of \(127^{\circ}\text{C}\). We need to find the new rate of emission \(E'\) when the dimensions are halved and the temperature is increased to \(327^{\circ}\text{C}\).

Step 2: Determine the New Dimensions

The original dimensions are \(8\text{ cm} \times 4\text{ cm}\). Halving each dimension gives us the new dimensions: \(4\text{ cm} \times 2\text{ cm}\).

Step 3: Calculate the New Area

The initial area \(A\) is \(8 \times 4 = 32 \text{ cm}^2\). After halving the dimensions, the new area \(A'\) becomes \(4 \times 2 = 8 \text{ cm}^2\).

Step 4: Applying the Stefan-Boltzmann Law

The total energy radiated per unit time (\(P\)) by a black body is given by Stefan-Boltzmann Law, \(P = \sigma A T^4\), where \(\sigma\) is the Stefan-Boltzmann constant. Let us denote the initial condition as \(P = E\) and the new condition as \(P' = E'\).

Step 5: Initial Energy Emission Rate Calculation

The initial emission rate, \(E\), is expressed as: \[ E = \sigma \times 32 \times (127 + 273)^4 \]

Step 6: Second Emission Rate Calculation

For the new conditions, \(E'\) is given by: \[ E' = \sigma \times 8 \times (327 + 273)^4 \]

Step 7: Express the Change in Emission Rate

Find the ratio \(\frac{E'}{E}\): \[ \frac{E'}{E} = \frac{8}{32} \times \left(\frac{600}{400}\right)^4 \]

Step 8: Simplify the Expression

After simplifying, we get: \[ \frac{E'}{E} = \frac{1}{4} \times \frac{81}{16} = \frac{81}{64} \]

Step 9: Compare the Results

Since \(\frac{E'}{E} = \frac{81}{64}\), option (d) is the correct answer.

Key Concepts

Black Body Radiation

Black body radiation refers to the phenomenon where an idealized object, called a black body, absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, and emits thermal radiation in a continuous spectrum. A black body is considered a perfect emitter and absorber of energy. This means that it can absorb all types of electromagnetic radiation without any reflection or transmission.

In physics, a black body emits radiation known as black body radiation. The energy emitted is a function of the black body's temperature. At a higher temperature, a black body emits radiation more intensely. This form of radiation plays an essential role in understanding thermal phenomena and forms the basis for the Stefan-Boltzmann Law, which quantifies the energy emission from such bodies.

Understanding black body radiation is fundamental when considering how objects emit energy in the form of thermal radiation, as it leads to deeper insights into heat and light interactions in various physical systems.

Thermal Radiation

Thermal radiation is the emission of electromagnetic waves from all matter that has a temperature above absolute zero. It is a mode of heat transfer where thermal energy is emitted by a body due to its temperature. Unlike conduction and convection, thermal radiation does not require any medium for transfer, allowing energy to emit and travel even through the vacuum of space.

The primary form of thermal radiation is infrared radiation, but visible light is also emitted by hotter objects. The amount and type of radiation depend intimately on the object's temperature and nature. The Stefan-Boltzmann Law helps in determining the energy emitted through thermal radiation by quantifying it with respect to temperature.

• Infants of cooler temperatures emit less thermal radiation, and vice versa for hotter objects.
• This process is why you feel the heat from the sun or a glowing coal in a fire.

In the context of the exercise, the change in thermal radiation due to changing dimensions and temperatures of a black body highlights how crucial the temperature and surface area are in determining the rate of thermal energy emission.

Rate of Energy Emission

The rate of energy emission refers to the amount of energy that a body emits in the form of radiation per unit of time. This is often expressed in watts (W) in the International System (SI) of units. For a black body, this rate can be precisely calculated using the Stefan-Boltzmann Law.

The law states that the power radiated per unit area of the black body is proportional to the fourth power of its temperature, expressed as:\[ P = \sigma A T^4 \]
Here:

• \( P \) is the power or rate of energy emission,
• \( \sigma \) is the Stefan-Boltzmann constant,
• \( A \) is the surface area of the body,
• \( T \) is the absolute temperature in Kelvin scale.

When black body's dimensions change, it impacts the surface area, thereby affecting the energy emission rate. In the step-by-step solution, halving the dimensions significantly impacted the area, reducing it to a quarter, which directly decreased the rate of energy emission. However, increasing the temperature causes a significant rise in the rate, as temperature in Kelvin raised to the fourth power far outweighs the area reduction effect.

Temperature and Energy Emission

Temperature is a critical factor in determining the rate of energy emission of a body. According to the Stefan-Boltzmann Law, the energy emitted by a body increases rapidly as its temperature rises. The fourth power relationship signifies that even a small increase in temperature can lead to a substantial rise in emitted energy.

For instance, in the provided exercise, the initial temperature is 127°C, which converts to 400 K. When the temperature is increased to 327°C, equivalent to 600 K, the effect on energy emission becomes magnified. The new temperature significantly boosts the rate of energy emission because the temperature term in the Stefan-Boltzmann Law has a larger influence compared to the surface area change.

Temperature impacts every level of energy emission, emphasizing the need to understand how thermal conditions affect physical interactions and energy states. Therefore, increasing the temperature implies a more substantial impact on energy emission than merely changing the object's physical dimensions, as it enhances radiative efficiency according to the law.