Tag: energy production

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Questions Related to energy production

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

Stefan-Boltzmanns Law for a perfect black body is represented by

  1. $\dfrac{dQ}{dt} = \sigma AT^2 $
  2. $\dfrac{dQ}{dt} = \sigma AT^3 $
  3. $\dfrac{dQ}{dt} = \sigma AT^4 $
  4. $ Q = \sigma AT^4 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Stefan boltzmann's law, $\Rightarrow \cfrac { dQ }{ dt } =\sigma A\varepsilon { T }^{ 4 }$
( for black body, $\varepsilon=1$)
So, for black body stefan's boltzmann's law is $\Rightarrow \cfrac { dQ }{ dt } =\sigma A { T }^{ 4 }$
Stefan's boltzmann's law states that total energy radiated from a surface is proportional to the forth power of its absolute temperature.
Multiple choice physics energy production perfectly black body black-body radiation black body radiation

Ferry's black body is accurately represented by 

  1. A fine hole in a double walled spherical cavity.

  2. A fine hole in a double walled spherical cavity, evacuated and painted black.

  3. A fine hole in a spherical cavity, evacuated and painted black.

  4. A fine hole in a black cavity.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ferry's black body is accurately represented by a fine hole in a double walled spherical cavity, evacuated and painted black.

$\rightarrow$ Ferry designed the simplest black body. It is a double walled evacuated spherical cavity whose inner wall is blackened. The space between wall is evacuated to prevent heat loss by conduction and radiation. There is a fine hole in it. All the radiations incident upon this hole are absorbed by this black body. 

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

The original temperature of a black body is $727^\circ C$. Calculate temperature at which total radiant energy from this black body becomes double:

  1. $971K$
  2. $1189K$
  3. $2001K$
  4. $1458K$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Rediant Energy = \sigma T^2$

$Energy = \sigma (1000)^4$
$E _2 = 2 E _1$
$Then$
$\sigma T _2 ^{4} = 2 \times \sigma (1000)^4$
$T _2 = 2^\frac{1}{4} \times1000$
$T _2 = 1189 K$

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

Temp. of black body is $3000K$ when black body cools. Then change in wavelength $\Delta \lambda=9$ micron corresponding to maximum energy density. Now temp. of black body is:

  1. $300K$
  2. $2700K$
  3. $270K$
  4. $1800K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Wien's displacement law, lambda_max * T = constant. Initially, T1 = 3000K. If lambda_max changes by 9 microns, we need the initial lambda_max. Assuming lambda_max1 = 1 micron (typical for 3000K), then lambda_max2 = 10 microns. T2 = (1/10) * 3000 = 300K.

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

The rate of emission of radiation of a black body at 273$^{ \circ  }{ C }$ is E, then the rate of emission of radiation  of this body at 0$^{ \circ  }{ C }$ will be :-

  1. $\dfrac { E }{ 16 } $
  2. $\dfrac { E }{ 4 } $
  3. $\dfrac { E }{ 8 } $

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Stefan-Boltzmann law states E is proportional to T^4. T1 = 273 + 273 = 546K. T2 = 0 + 273 = 273K. Ratio E2/E1 = (273/546)^4 = (1/2)^4 = 1/16. Thus E2 = E/16.

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

A blackbody does not

  1. emit radiation

  2. absorb radiation

  3. reflect radiation

  4. refract radiation
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An ideal blackbody is defined as an object that absorbs all electromagnetic radiation that falls on it, regardless of frequency or angle of incidence. Because it absorbs everything, it does not reflect any radiation.