Physics

Thermal Properties and Thermodynamics

431 Questions

Thermal properties and thermodynamics questions evaluate concepts of heat transfer, thermal efficiency, and temperature variations. Problems involve calculating heat content, conductivity, and the performance of heat engines. This subject is regularly tested in physics sections across multiple competitive platforms.

Heat transfer calculationsThermal efficiencyBlack body radiationTemperature variationsRefrigeration performance

Thermal Properties and Thermodynamics Questions

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

A black body at 127$^{o}$C emits the energy at the rate of 10$^{6}$ J/m$^{2}$ s. The temperature of a black body at which the rate of energy emission is 16x10$^{6}$ J/m$^{2}$ s is :

  1. $508^{o}C$
  2. $273^{o}C$
  3. $400^{o}C$
  4. $527^{o}C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Stefan's Law:
$\dfrac { { E } _{ 2 } }{ { E } _{ 1 } } ={ \left( \dfrac { { T } _{ 2 } }{ { T } _{ 1 } }  \right)  }^{ 4 }\ { T } _{ 2 }={ T } _{ 1 }\sqrt [ 4 ]{ \dfrac { { E } _{ 2 } }{ { E } _{ 1 } }  } \quad \ { T } _{ 2 }={ (127+273) }\sqrt [ 4 ]{ \dfrac { 16\times { 10 }^{ 6 } }{ { 10 }^{ 6 } }  } =\quad 800K\ { T } _{ 2 }={ 527 }^{ \circ  }C$

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

Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures of 2T and 3T respectively. The temperatures of the middle (i.e. second) plate under steady state condition is then

  1. $\left ( \dfrac{64}{2} \right )^{\dfrac{1}{4}}T$
  2. $\left ( \dfrac{97}{4} \right )^{\dfrac{1}{4}}T$
  3. $\left ( \dfrac{97}{2} \right )^{\dfrac{1}{4}}T$
  4. $\left ( 97\right )^{\dfrac{1}{4}}T$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is identical to question 534600. The steady state temperature of the middle plate is ( (T1^4 + T3^4) / 2 )^(1/4). With T1=2T and T3=3T, T2 = ((16T^4 + 81T^4)/2)^(1/4) = (97/2)^(1/4) * T.

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

The rectangular surface of area $8cm$ $\times$ $4 cm$ of a black body at temperature $127^{\circ}C$ emits energy $E$ per second. If the length and breadth are reduced to half of the initial value and the temperature is raised to $327^{\circ}C$, the rate of emission of energy becomes

  1. $\displaystyle \frac{3}{8}E$
  2. $\displaystyle \frac{81}{16}E$
  3. $\displaystyle \frac{9}{16}E$
  4. $\displaystyle \frac{81}{64}E$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By $Stefan's$ Law

$Power=\sigma A T^4$
$A=length\times breadth$
$\dfrac{P _2}{E}=\dfrac{l _2b _2T _2^4}{l _1b _1T _1^4}$
Here$\dfrac{l _2}{l _1}=\dfrac{1}{2}$       $\dfrac{b _2}{b _1}=\dfrac{1}{2}$       $\dfrac{T _2}{T _1}=\dfrac{3}{2}$

$\implies P _2=\dfrac{81}{64}E$

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

If the temperature of a hot body is raised by $0.5\%$, then the heat energy radiated would increase by :

  1. 0.5%

  2. 1.0%

  3. 1.5%

  4. 2.0%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The rate of heat energy radiated by black body is given as:
$Q=\sigma T^4A$ where $\sigma$ is the stefen-boltzmann constant, $A$ is the area of the radiating body.
So $Q\propto T^4=kT^4$, where $k$ is the propotionality constant that we have assumed here.
Taking log of above equation, $logQ=log(kT^4)=logk+4logT$
Taking differential of above equation, $\dfrac{dQ}{Q}=0+\dfrac{4dT}{T}$ (because logk is constant).
Here $dQ$ and $dT$ indicate very small change in the $Q$ and $T$ respectively.
Multiplying by $100$,
$\dfrac{dQ}{Q}\times100=4\dfrac{dT}{T}\times100$
$\Rightarrow$ Percentage change in Heat rate $=4\times $ percentage change in temperature
So here, percentage change in rate of heat energy radiated $=4\times0.5=2\%$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The temperature of a black body is increased by $50\%$ . Then the percentage of increase of radiation is approximately

  1. 100%

  2. 25%

  3. 400%

  4. 500%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By $Stefan's$ $Law$,

Power = ${\sigma A {T}^{4}}$ for a black body
$\sigma$ is known as Stefan's constant
$\dfrac{P _1}{P _2}$ = $\dfrac{ T _1^4}{ T _2^4}$
${T _2}$ = $\dfrac{3 T _1}{2}$
$\dfrac{T _1^4}{T _2^4}$= $\dfrac{16}{81}$
By the above equations we get
$P _2$=$\dfrac{81 P _1}{16}$
Percentage increase in radiation = $\dfrac{P _2 - P _1}{P _1}$ x $100$ = $406.25$%

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

The rays of sun are focussed on a piece of ice through a lens of diameter $5$ cm, as a result of which $10$ grams of ice melts in $10$ min. The amount of heat received from Sun is (per unit area per min)

  1. 4 $cal/cm^{2} \: min$
  2. 40 $cal/cm^{2} \: min$
  3. 4 $J/cm^{2} \: min$
  4. 400 $J/cm^{2} \: min$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The latent heat of fusion is $80cal/g$.

So heat received per unit times is $10g\times 80cal g^{-1}/ 10min=80cal/min$
The are is $\pi r^2=\pi\times(2.5)^2cm^2\approx 20cm^2$
So, amountof heat per unit area per unit time is $80/20=4cal/cm^2\ min$

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

Which of the following statements is true/correct?

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $1 : 4$
  2. $1 : 8$
  3. $1 : 16$
  4. $1 : 256$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

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

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 :

  1. $\dfrac{3}{2}\mu m$
  2. $1\mu m$
  3. $2 \mu m$
  4. $\dfrac{3}{4} \mu m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${ \lambda  } _{ B }=1.5\mu m$.


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

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

  1. $U/8$
  2. $U/16$
  3. $U/4$
  4. $U/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

 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.)

  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}

    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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 :

  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. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{4\pi r^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Total power radiated by the sun

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

The intensity of power at earth surface

$=\cfrac{\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }}{4\pi { r }^{ 2 }} \\=\cfrac{\sigma { T }^{ 4 } { R }^{ 2 }}{{ r }^{ 2 }}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rectangular surface of area $8 cm \times 4 cm$ of a black body at a temperature of $127^0C$ emits energy at the rate of $E$ per second. If both length and breadth of the surface are reduced to half of its initial value, and the temperature is raised to $327^0C$, then the rate of emission of energy will become :

  1. $\dfrac{3}{8}E$
  2. $\dfrac{81}{16}E$
  3. $\dfrac{9}{16}E$
  4. $\dfrac{81}{64}E$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $A _1=32$ as given.
Let $A _2$ be the area when length and breadth are reduced by half. Thus the area will be $\dfrac {1}{4}$th of $A _1$
$ \therefore A _2=\dfrac {1}{4} *32=8$
Given $T _1={127}^0C={400}^0K$
Given $T _2={327}^0C={600}^0K$
From Stefan's law $E=\sigma AT^4$
$ \therefore\dfrac{E _1}{E _2}= \dfrac {A _1{T _1}^4}{A _2{T _2}^4}=\dfrac {32*(400)^4}{8*(600)^4}=\dfrac{64}{81}$
$ \therefore \dfrac{E _2}{E _1}= \dfrac{81}{64}$
Multiple choice thermodynamic principles of metallurgy general principles and processes of isolation of elements metallurgy chemistry

Hardened steel on heating in the range of $220^{\circ} C$ to $330^{\circ}C$ and on slow cooling gives :

  1. hard steel

  2. Brittle steel

  3. hard and brittle steel

  4. hard and tough steel
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hardened steel on heating in the range of $220^{\circ} C$ to $330^{\circ}C$ and on slow cooling gives Hard and tough steel.