Tag: thermal properties

Quizzes (8) Questions (112)

Questions Related to thermal properties

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

In pyrometer , temperature measured is proportional to $\underline{\hspace{0.5in}}$ energy emitted by the body 

  1. light

  2. electric

  3. radiation

  4. All the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stefan- Boltzann law, $j^{ \star  }=\varepsilon \sigma T^{ 4 }$ connects temperature T with thermal radiation or irradiance  $j^{ \star  }$.
Thus measuring the irradiance with pyrometer yields the temperature of the body.

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

Two bodies of same shape and having emissivities 0.1 and 0.9 respectively radiate same energy per second. The ratio of their temperature is :

  1. $\sqrt{3}:1$

  2. $1:\sqrt{3}$

  3. $3:1$

  4. $1:3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{E}{t}=e \sigma A T^4$
From above equation which is Stefan's Law of radiation, it is clear that:
$\dfrac{E _1}{E _2} = \dfrac{{e} _{1}\sigma{T} _{1}^{4}}{{e} _{2}\sigma{T} _{2}^{4}}$

$1 = \dfrac{{0.1}{T} _{1}^{4}}{{0.9}{T} _{2}^{4}}$

$\dfrac{{T} _{1}}{{T} _{2}} = \dfrac{\sqrt{3}}{1} $
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

Two bodies A and B are kept in an evacuated chamber at $27^oC$. The temperature of A and B are $327^oC$ and $427^oC$ respectively. The ratio of rate of loss of heat from A and B will be

  1. 0.25

  2. 0.52

  3. 1.52

  4. 2.52
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The power radiated is directly proportional to fourth power of absolute temperature.
i.e.
$P \propto T^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{T _{1}}{T _{2}})^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{327+273}{427+273})^{4} = 0.53$
Hence the ratio of rate of heat loss = 0.53
Hence option B is correct.

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

The radiation emitted by a perfectly black body is proportional to 

  1. temperature on ideal gas scale

  2. fourth root of temperature on ideal gas scale

  3. fourth power of temperature on ideal gas scale

  4. square of temperature on ideal gas scale
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stefan-Boltzmann law states that total power radiated by a perfectly black body is
$P=A\sigma { T }^{ 4 }$
so the radiation emitted by a perfectly black body is proportional to fourth power of temperature on ideal gas scale.
option (C) is the correct answer.

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

The amount of heat energy radiated per second by a surface depends upon:

  1. Area of the surface

  2. Difference of temperature between the surface and its surroundings

  3. Nature of the surface

  4. All the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Refer Stefan's Law of Radiation: $Q = \eta \sigma A \delta{T}^{4}$
where, $\sigma = conductivity\ A$  = area $T$ = difference of the temperature 

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

The thermal radiation emitted by a body is proportional to $T^{n}$ where $T$ is its absolute temperature. The value of $n$ is exactly $4$ for

  1. a blackbody

  2. all bodies

  3. bodies painted balck only

  4. polished bodies only
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Stefan's Law, rate of thermal radiation is directly proportional to fourth power of temperature of the body.
$Q = \sigma {T}^{4}$

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

A black body radiates energy at the rate of $E\ watt/m$$^{2}$ at a high temperature $T^{o}K$ when the temperature is reduced to $\left [ \dfrac{T}{2} \right ]^{o}K$ Then radiant energy is

  1. $4E$

  2. $16E$

  3. $\dfrac{E}{4}$

  4. $\dfrac{E}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that from stefans-boltzman law: $E\propto { T }^{ 4 }$
if temperature will be reduces half form the initial value, then
${E} _{1}\propto ({ \dfrac { T }{ 2 } ) }^{ 4 }$
${E} _{1}\propto\dfrac{E}{16}$

Multiple choice stefan's law black body radiation heat transfer thermal properties 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?

  1. Constant volume air thermometer

  2. Platinum resistance thermometer

  3. Thermostat

  4. Thermopile
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 :

  1. $2E$

  2. $E/2$

  3. $16E$

  4. $E/16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

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

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

  4. $(T _{2}/T _{1})^{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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