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The rate of radiation of a black body at \(0^{\circ} \mathrm{C}\) is \(E\) watt. The rate of radiation of this body at \(273^{\circ} \mathrm{C}\) will be (a) \(16 \bar{E}\) (b) \(8 E\) (c) \(4 \underline{E}\) (d) \(E\)

Short Answer

Expert verified
The rate of radiation at \(273^{\circ} \mathrm{C}\) is \(16E\).

Step by step solution

01

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. Mathematically, this can be expressed as \( P = \sigma A T^4 \), where \( P \) is the power radiated, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the area of the black body, and \( T \) is the absolute temperature in kelvin.
02

Convert Temperatures to Kelvin

We need to convert the given temperatures from Celsius to Kelvin. The absolute temperature in Kelvin is obtained by adding 273.15 to the Celsius temperature.- For \(0^{\circ} \mathrm{C}\): \[ T_1 = 0 + 273.15 = 273.15 \text{ K} \]- For \(273^{\circ} \mathrm{C}\): \[ T_2 = 273 + 273.15 = 546.15 \text{ K} \]
03

Apply the Stefan-Boltzmann Law to Both Cases

From the Stefan-Boltzmann Law, the radiation rates at these temperatures are:1. At \(0^{\circ} \mathrm{C}\): \[ E = \sigma A (273.15)^4 \]2. At \(273^{\circ} \mathrm{C}\): \[ E_2 = \sigma A (546.15)^4 \]
04

Calculate the Ratio of the Radiation Powers

Find the ratio of the radiation rates at the two temperatures:\[ \frac{E_2}{E} = \frac{\sigma A (546.15)^4}{\sigma A (273.15)^4} = \left( \frac{546.15}{273.15} \right)^4 \]Calculating the ratio:\[ \frac{546.15}{273.15} = 2 \]Thus,\[ \frac{E_2}{E} = 2^4 = 16 \]
05

Determine the New Rate of Radiation

The radiation rate at \(273^{\circ} \mathrm{C}\) is 16 times the rate at \(0^{\circ} \mathrm{C}\). Therefore, \( E_2 = 16E \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Black Body Radiation
A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It neither reflects nor transmits any light, making it a perfect absorber.
In actuality, a true black body does not exist, but plenty of materials come close, allowing us to apply the concept in real-world physics problems and theoretical discussions.
Black body radiation is the thermal electromagnetic radiation emitted from a black body. This radiation has a characteristic spectrum that solely depends on the temperature of the body.
  • The radiation emanates from the body in a continuous spectrum.
  • At higher temperatures, the peak of the emission spectrum shifts to shorter wavelengths.
  • This behaviour is in line with Wien's Displacement Law.
Understanding black body radiation allows scientists and people like you and me to dive deeper into complex topics such as stellar optics and the study of cosmic microwave background radiation.
Absolute Temperature
Absolute temperature is a temperature measured on a scale starting from absolute zero, the point where all thermal motion ceases. The Kelvin scale is the most commonly used scale for absolute temperature in scientific measurements.
Absolute temperature plays a crucial role in physics, especially when dealing with thermodynamic processes. It is distinguished from Celsius and Fahrenheit by its absolute nature, where no negative values exist. Thus, zero on the Kelvin scale is the coldest possible temperature, any material can theoretically reach known as absolute zero.
  • Absolute zero is 0 Kelvin, equivalent to -273.15 °C.
  • To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
  • Kelvin is commonly used in scientific calculations, such as those involving the gas laws and thermodynamics.
In the context of the Stefan-Boltzmann Law, the power radiated is closely tied to the fourth power of the absolute temperature, demonstrating how pivotal it is in achieving precise scientific comprehension.
Power Radiated
The power radiated by an object is the total energy emitted per unit time. In the context of the Stefan-Boltzmann Law, the power radiated by a black body is defined as proportional to the fourth power of its absolute temperature. This fundamental relationship provides insight into how temperature influences energy emission.
The equation used to determine the power radiated by a black body is:
\[ P = \sigma A T^4 \]
Where:
  • \( P \) is the power radiated.
  • \( \sigma \) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \, \text{W m}^{-2} \text{K}^{-4})\).
  • \( A \) is the surface area of the black body.
  • \( T \) is the absolute temperature in Kelvin.
By analyzing the power radiated at different temperatures, researchers can infer the temperature of distant objects, like stars, by measuring the radiation they emit. This is pivotal in astronomical observations and helps us understand the universe better.

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Most popular questions from this chapter

A wheel is \(80.3 \mathrm{~cm}\) in circumference. An iron tyre measures \(80.0 \mathrm{~cm}\) around its inner face. If the coefficient of linear expansion for iron is \(1.2 \times 0^{-5}{ }^{\circ} \mathrm{C}^{-1}\), the temperature of the tyre must be raised by (a) \(105^{\circ} \mathrm{C}\) (b) \(417^{\circ} \mathrm{C}\) (c) \(312^{+} \mathrm{C}\) (d) \(223^{\circ} \mathrm{C}\)

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of \(0.1 \mathrm{~g}\) of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of the material of the second rod is \(1 / 4\) that of the first, the rate at which ice melts in \(\mathrm{gs}^{-1}\) will be (a) \(3.2\) (b) \(1.6\) (c) \(0.2\) (d) \(0.1\)

Two rods of same length and material transfer a given amount of heat in \(12 \mathrm{~s}\), when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in (a) 245 (b) \(3 \underline{5}\) (c) \(48 \mathrm{~s}\) (d) \(1.5 \mathrm{~s}\)

An oxygen cylinder of volume \(30 \mathrm{~L}\) has an initial gauge pressure of \(15 \mathrm{~atm}\) and a temperature of \(27^{\circ} \mathrm{C}\). After some oxygen is withdrawn from the cylinder the gauge pressure drops to 11 atm and its temperature drops to \(17^{\circ} \mathrm{C} .\) The mass of oxygen taken out of the cylinder \(\left(R=8.31 . \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\). molecular mas of \(\mathrm{O}_{2}=32 \mathrm{u}\) ) is [NCERT] (a) \(0.14 \mathrm{~g}\) (b) \(0.02 \mathrm{~g}\) (c) \(0.14 \mathrm{~kg}\) (d) \(0.014 \mathrm{~kg}\)

Assuming the sun to have a spherical outer surface of radius \(r\) radiating like a black body at temperature \(t^{\circ} \mathrm{C}\), the power received by a unit surface (normal to the incident rays) at a distance \(R\) from the centre of the sun is ( \(\sigma\) is stefan's constant) (a) \(4 \pi r^{2} \sigma t^{4}\) (b) \(\frac{r^{2} \sigma(t+273)^{4}}{4 \pi R^{2}}\) (c) \(\frac{16 \pi^{2} r^{2} \sigma t^{4}}{R^{2}}\) (d) \(\frac{r^{2} \sigma(1+273)^{4}}{R^{2}}\)

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