Tag: thermal properties

Questions Related to thermal properties

Match the physical quantities given in Column I with their SI units given in Cloumn II :

Column-I Column-II
(a) Thermal conductivity (p) Wm$^{-2}$K$^{-4}$
(b) Stefans constant (q) m-K
(c) Wiens constant (r) J kg$^{-1}$K$^{-1}$
(d) Specific heat (s)Wm$^{-1}$K$^{-1}$
stefan's law black body radiation heat transfer thermal properties physics

  1. a-s, b-p, c-q, d-r

  2. a-s, b-p, c-r, d-q

  3. a-s, b-r, c-p, d-q

  4. a-r, b-s, c-p, d-q

Show answer
Correct Option: A
Explanation:

Thermal conductivity =$\dfrac{Watt}{Metre.Kelvin}$

Stefan's constant $=5.64\times$ ${10}^{-8}$$\dfrac { w }{ { m }^{ 2 }{ K }^{4}}$

Wien's constant $={\lambda}{\theta}={constant}={Metre}  {Kelvin}$

Specific heat $=\dfrac{Joule} {{Kg}  {Kelvin}}$

Which of the following statements is true/correct?

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

  1. During clear nights, the temperature rises steadily upward near the ground level

  2. Newton's law of cooling, and approximate form of Stefan's law, is valid only for natural convection

  3. The total energy emitted by a black body per unit time per unit area is proportional to the square of its temperature in the Kelvin scale

  4. Two spheres of the same material have radii $1 m$ and $4 m$ and temperatures $4000 K$ and $2000 K$ respectively. The energy radiated per second by the first sphere is greater than that radiated per second by the second sphere

Show answer
Correct Option: B
Explanation:

During clear nights object on surface of earth radiate out heat and temperature falls. Hence option (a) is wrong.
The total energy radiated by a body per unit time per unit area $E \propto {T}^{4}$. Hence option (c) is wrong.
Energy radiated per second is given by
$\dfrac { Q }{ t } =PA\varepsilon \sigma { T }^{ 4 }$
$\Rightarrow \dfrac { { P } _{ 1 } }{ { P } _{ 2 } } =\dfrac { { A } _{ 1 } }{ { A } _{ 2 } } { \left( \dfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 4 }={ \left( \dfrac { { r } _{ 1 } }{ { r } _{ 2 } }  \right)  }^{ 2 }\cdot { \left( \dfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 2 }$
$={ \left( \dfrac { 1 }{ 4 }  \right)  }^{ 2 }\left( \dfrac { 4000 }{ 200 }  \right) =\dfrac { 1 }{ 1 } $
$\because    {P} _{1} = {P} _{2}$ hence option (d) is wrong.
Newton's law is an approximate from of Stefan's law of radiation and works well for natural convection. Hence option (b) is correct.

STATEMENT-1 : Animals curl into a ball, when they feel very cold.
STATEMENT-2 : Animals by curling their body reduces the surface area.

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

  1. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

  2. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

  3. STATEMENT-1 is True, STATEMENT-2 is False

  4. STATEMENT-1 is False, STATEMENT-2 is True

Show answer
Correct Option: A
Explanation:

Both statements are true as the animals curl their body in very cold environment so they can reduce their surface area , and reduce the heat leaving their body in form of radiation as heat emitted is directly proportional to surface area.

The dimensions of Stefan's constant are

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

  1. $\left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -3 }{ K }^{ -4 } \right] $

  2. $\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -3 }{ K }^{ -3 } \right] $

  3. $\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ K }^{ -4 } \right] $

  4. $\left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -3 }{ K }^{ -4 } \right] $

Show answer
Correct Option: D
Explanation:

Power radiated by a body $P = \sigma AeT^4$

where $\sigma$ is the Stefan's constant, $e$ is the emmissivity of the body, $A$ is the surface area of the body and $T$ is its temperature.
Dimensions of power $[P] = [ML^2T^{-3}]$
Dimensions of area $[A] = [L^2]$
Dimensions of temperature $[t] = [K]$
Emmissivity $e$ is a dimensionless quantity.
$\therefore$ Dimensions of Stefan's constant $[\sigma] = \dfrac{[ML^2T^{-3}]}{[L^2] [K^4]}$
$\implies$ $[\sigma] = [M^1 L^0 T^{-3} K^{-4}]$

A black body is heated from $27^oC  $ to $927^oC  $. The ratio of radiation emitted will be:

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

  1. $1 : 4$

  2. $1 : 8$

  3. $1 : 16$

  4. $1 : 256$

Show answer
Correct Option: D
Explanation:

Energy radiated depends on the temperature of the body.
Stefan's law states that the total amount of energy radiated per second per unit area of a perfect black body is directly proportional to the fourth power of the absolute temperature of the surface of the body,ie,
$E\propto { T }^{ 4 }$
or $E=\sigma { T }^{ 4 }$
where $\sigma $ is Stefan's constant. It's value is $5.67\times { 10 }^{ -8 }W{ m }^{ -2 }{ K }^{ -4 }$
Here, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

Two bodies A and B of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelengths $\lambda _A$, and $\lambda _B$ respectively. Difference in these two wavelengths is 1 $\mu$. If the temperature of the body A is $5802\  K$, then value of $\lambda _B$ is :

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

  1. $\dfrac{3}{2}\mu m$

  2. $1\mu m$

  3. $2 \mu m$

  4. $\dfrac{3}{4} \mu m$

Show answer
Correct Option: A
Explanation:

We know that as per stephan's boltzman radiation law, $P\ \alpha\  \sigma A{ T }^{ 4 }$.
Since surface area of two bodies is same,

Therefore, ${ \sigma  } _{ A }{ T } _{ A }^{ 4 }={ \sigma  } _{ B }{ T } _{ B }^{ 4 }$. Hence, ${ T } _{ A }=3{ T } _{ B }$.

Now, as per Wein's displacement law, $\lambda T=constant=k$, $\lambda =\dfrac { k }{ T } $.

${ \lambda  } _{ B }-{ \lambda  } _{ A }=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ { T } _{ A } } )=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ 3{ T } _{ B } } )=\dfrac { 2k }{ 3{ T } _{ B } } =1\mu $


${ T } _{ B }=\dfrac { { T } _{ A } }{ 3 } =1934\ K$

Putting in the above equation to calculate $k$ and the solving for ${ \lambda  } _{ B }=\dfrac { k }{ { T } _{ B } } $, we get ${ \lambda  } _{ B }=1.5\mu m$.


A black body at a high temperature $T$ radiates energy at the rate of $U\left( in\quad W/{ m }^{ 2 } \right) $. When the temperature falls to half (i.e $T/2$), the radiated energy $\left( in\quad W/{ m }^{ 2 } \right) $ will be

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

  1. $U/8$

  2. $U/16$

  3. $U/4$

  4. $U/2$

Show answer
Correct Option: B
Explanation:

According to Stefan's law, rate of energy radiated by a black per unit area (inW/m2)(inW/m2) at temperature TT is given by
U=σT4...(i)U=σT4...(i)
when the temperature falls to half (i.e., T/2T/2) 
radiated energy (inW/m2)...(ii)(inW/m2)...(ii)
From (i) and (ii) we get
U′U=(12)4=116U′U=(12)4=116
or U′=U16

 If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is 0? (a stands for Stefan's constant.)

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

  1. $

    \left(\frac{4 \pi R^{2} Q}{\sigma}\right)^{1 / 4}

    $

  2. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{1 / 4}

    $

  3. $

    \frac{Q}{4 \pi R^{2} \sigma}

    $

  4. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{-1 / 2}

    $

Show answer
Correct Option: C

$\dfrac {watt} {kelvin}$ is the unit of 

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

  1. Stefan's constant

  2. Wien's constant

  3. Cooling's constant

  4. Thermal constant

Show answer
Correct Option: A

Assuming the Sun to be a spherical body of radius $R$ at a temperature of $T\ K$. Evaluate the intensity of radiant power, incident on Earth, at a distance $r$ from the Sun where $r _{0}$ is the radius of the Earth and $\sigma$ is Stefan's constant :

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

  1. $\dfrac{R^{2}\sigma T^{4} }{r^{2}}$

  2. $\dfrac{4\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$

  3. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$

  4. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{4\pi r^{2}}$

Show answer
Correct Option: A
Explanation:
Total power radiated by the sun

 $=\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }$

The intensity of power at earth surface

$=\cfrac{\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }}{4\pi { r }^{ 2 }} \\=\cfrac{\sigma { T }^{ 4 } { R }^{ 2 }}{{ r }^{ 2 }}$