Tag: physics

Questions Related to physics

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction orconvection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic spectrum. Thermal radiations travel through vacuum with the speed oflight. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarization as light does.
The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengts. The total energy E emitted by a unit area of a black bodyper second is given by $E =\sigma T^{4}$ where T is the absolute temperature of the body and $\sigma $ is a constant known as Stefans constant. If the body is not a perfect black body, then $E =\varepsilon \sigma  T^{4}$where $\varepsilon $ is the emissivity of the body.

Which of the following devices is used to detect thermal radiations?

stefan's law black body radiation heat transfer thermal properties physics

  1. Constant volume air thermometer

  2. Platinum resistance thermometer

  3. Thermostat

  4. Thermopile

Show answer
Correct Option: D
Explanation:

Thermopile is a very sensitive device which converts thermal energy to electrical energy.
It is used to detect thermal radiations.

The rate of radiation from a black body at $0$$^{o}$C is $E$. The rate of radiation from this black body at $273$$^{o}$C is :

stefan's law black body radiation heat transfer thermal properties physics

  1. $2E$

  2. $E/2$

  3. $16E$

  4. $E/16$

Show answer
Correct Option: C
Explanation:

As we know that: ${E} \ {\propto} \ {T}^{4}$
So, $\dfrac{E}{{T} _{1}^{4}}=\dfrac{{E} _{2}}{{T} _{2}^{4}}$
${E} _{2}=\dfrac{{T} _{2}^{4}\times{E}}{{T} _{1}^{4}}=\dfrac{{546}^{4}}{{273}^{4}}$
${E} _{2}={16E}$

Two spherical black bodies of radii $r _{1} $ and $  r _{2}$ are with surface temperatures $T _{1} $ and $ T _{2}$ respectively radiate the same power. $r _{1} / r _{2}$ must be equal to

stefan's law black body radiation heat transfer thermal properties physics

  1. $(T _{1}/T _{2})^{2}$

  2. $(T _{2}/T _{1})^{2}$

  3. $(T _{1}/T _{2})^{4}$

  4. $(T _{2}/T _{1})^{4}$

Show answer
Correct Option: B
Explanation:

$E=\varepsilon \sigma A{ T }^{ 4 }$

Given that both spherical body radiate with same power.
So equating ${E} _{1}={E} _{2}$
${A}=4{\pi}{r}^{2}$
${ r } _{ 1 }^{ 2 }{ T } _{ 1 }^{ 2 }={ r } _{ 2 }^{ 2 }{ T } _{ 2 }^{ 4 }$

$\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } ={ (\dfrac { { T } _{ 2 } }{ { T } _{ 1 } } ) }^{ 2 }$

The temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of :

stefan's law black body radiation heat transfer thermal properties physics

  1. 2

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: D
Explanation:

We know that from Stefan's Boltzmann relation: $E\propto { T }^{ 4 }$
if the temperature of the sun will be doubled, then : $E\propto ({ 2T })^{ 4 }$
Hence, $E\  will\  increase \ by \ the \ factor \ of \  { 16}$.

A black body is at temperature $300K$. It emits energy at a rate, which is proportional to 

stefan's law black body radiation heat transfer thermal properties physics

  1. ${(300)}^{4}$

  2. ${(300)}^{3}$

  3. ${(300)}^{2}$

  4. $300$

Show answer
Correct Option: A
Explanation:

For black body radiation
$E=\sigma{T}^{4}$ or $E\propto {T}^{4}$
Rate of emission of energy $\propto {(300)}^{4}$

If the absolute temperature of a blackbody is doubled, then the maximum energy density

stefan's law black body radiation heat transfer thermal properties physics

  1. Increases to 16 times

  2. Increases to 32 times

  3. Decreases to 16 times

  4. Decreases to 32 times

Show answer
Correct Option: A
Explanation:

The power with which a body(in this case black body) radiates is directly proportional to the fourth power of absolute temperature:
$P = kT^{4}$
i.e.
$P _{1} = kT _{1}^{4}$
If the absolute temperature is doubled,
$P _{2} = k(2T _{1})^{4} = 16kT _{1}^{4}$
Now $\dfrac{P _{2}}{P _{1}} = \dfrac{16}{1}$ 

Hence energy density is increased 16 times.

Intensity of heat radiation emitted by body is believed to be proportional to fourth power of absolute temperature of the body. The proportionality constant also known as Boltzmann's constant may have possible value of :

stefan's law black body radiation heat transfer thermal properties physics

  1. $5.67\times 10^{-8} watt/K^4 $

  2. $5.67\times 10^{-8} watt/m^2 K^4 $

  3. $5.67\times 10^{-8} J/K^4 $

  4. $5.67\times 10^{-8} Js/K^4 $

Show answer
Correct Option: B
Explanation:

From the given question,

$I\propto T^4$

$I=k T^4$

where $k=$ Stefan-Boltzmann's constant

$k=5.67\times10^{-8} W/m^2K^4 $

The correct option is B.

A black body at a temperature of $227^oC$ radiates heat energy at the rate 5 cal/cm$^{2}-s$. At a temperature of $727^oC$, the rate of heat radiated per unit area in cal/cm$^2$ will be

stefan's law black body radiation heat transfer thermal properties physics

  1. 80

  2. 160

  3. 250

  4. 500

Show answer
Correct Option: A
Explanation:

According to Stefen's Law, the rate of heat radiation from body is proportional to the fourth power of body's temperature.

Thus $P\propto T^4$
$\implies \dfrac{P _2}{P _1}=\dfrac{T _2^4}{T _1^4}$
$=16$
$\implies P _2=80cal/cm^2-s$

For a block body temperature $727^{o}C,$ its rate of energy loss is $20\ watt$ and temperature of surrounding is $227^{o}C.$ If temperature of black body is changed to $1227^{o}C$ then its rate of energy loss will be:

stefan's law black body radiation heat transfer thermal properties physics

  1. $320\ W$

  2. $\dfrac {304}{3}\ W$

  3. $240 W$

  4. $120 W$

Show answer
Correct Option: A
Explanation:

It is given that,

Temperature of surrounding

  $ {{T} _{0}}={{227}^{0}}C=500\ K $

 $ {{T} _{1}}={{727}^{0}}C=1000\ K $

 $ {{T} _{2}}={{1227}^{0}}C=1500\ K $

 $ {{E} _{1}}=20\ Watt\,\, $

 $ {{E} _{2}}=? $

According to Stefn boltzmann law:

$ E=\sigma {{T}^{4}} $

Or

 $ {{E} _{1}}=\sigma ({{T} _{1}}-{{T} _{0}})^4 $

 $ {{E} _{2}}=\sigma ({{T} _{2}}-{{T} _{0}})^4 $

Taking ratios of above equations:

For $ {{E} _{1}}=20\ Watt $

 $ \dfrac{20}{{{E} _{2}}}={{\left( \dfrac{500}{1000} \right)}^{4}} $

 $ \dfrac{20}{{{E} _{2}}}=\left( \dfrac{1}{16} \right) $

 $ {{E} _{2}}=320\ Watt $

The power received at distance $d$ from a small metallic sphere of radius $r(<<d)$ and at absolute temperature $T$ is $P$. If the temperature is doubled and distance reduced to half of the initial value, then the power received at that point will be:

stefan's law black body radiation heat transfer thermal properties physics

  1. $4p$

  2. $8p$

  3. $32p$

  4. $64p$

Show answer
Correct Option: D
Explanation:
Energy received per second i.e., power $P\alpha \dfrac{T^4}{d^2}=k\dfrac{T^4}{d^2}$
if temperature is double than T become 2T and distance become half than d become $\dfrac{d}{2}$
than power $ p _{1}=k\dfrac{(2T)^4}{(\dfrac{d}{2})^2}=64k\dfrac{T^4}{d^2}=64P$
Hence D option is correct.