Tag: thermal properties

Questions Related to thermal properties

The temperature of a piece of metal is raised from $27^oC$ to $51.2^oC$. The rate at which the metal radiates energy increases nearly

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

  1. 1.36 times

  2. 2 times

  3. 4 times

  4. 8 times

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Correct Option: A
Explanation:

The rate at which a substance radiates is directly proportional to the fourth power of the absolute temperature.
The temperature increases from $300K$ to $324.2K$ which is an increase by $1.080$
Hence the rate at which the metal radiates would increase by $(1.08)^{4}$ = $1.36$

A black body at a temperature $77^oC$ radiates heat at a rate of $10 calcm^{-2}s^{-1}$. The rate at which this body would radiate heat in units of $cal \ cm^{-2} \ s^{-1}$ at $427^oC$ is closest to:

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

  1. 40

  2. 160

  3. 200

  4. 400

Show answer
Correct Option: B
Explanation:

Energy radiated $P=\sigma AT^4$
$ \displaystyle \frac{P _1}{P _2} = \cfrac{T _1^4}{T _2^4}= { \bigg ( \frac {350}{700} \bigg ) }^4 = \frac{10}{P _2} \space or P _2 = 160$

The amount of thermal radiations emitted from one square centimeter area of a black body in a second when at a temperature of 1000K

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

  1. 5.67 J

  2. 56.7 J

  3. 567 J

  4. 5670 J

Show answer
Correct Option: A
Explanation:
Stefan's law $\Delta Q = \sigma ST^4\Delta t$
$\sigma=5.67E- 8W/m^2K^4;\, S=1E-4m^2;\, T=1000K; \, \Delta t=1s;$
subsutituting value in Stefan's law $\Delta Q=5.67J$
The value can be directly calculated by the Stefan's equation. After substituting the parameters sigma $= 5.67 E-8  W/m^2/K^4, A=10^-4 m^4,$ $T=1000K$ the value comes $5.67$ J
Thus, A is correct answer.

Find the radiation pressure of solar radiation on the surface of earth. Solar constant is $1.4kW{{m}^{-2}}$

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

  1. $4.7\times { 10 }^{ -5 }Pa$

  2. $4.7\times { 10 }^{ -6 }Pa$

  3. $2.37\times { 10 }^{ -6 }Pa$

  4. $9.4\times { 10 }^{ -6 }Pa$

Show answer
Correct Option: B
Explanation:
METHOD-1:
${Pressure} _{absorbed}=\dfrac{E _{f}}{c}$
${E} _{f}=$ energy flux, $C=$ Speed of light

${ Pressure } _{ absorbed }=\dfrac { 1.4\times 1000 }{ 3\times { 10 }^{ 8 } } =4.66\times { 10 }^{ -6 }Pa$

METHOD-2: (checking the unit,if formula is not remembered)

We know that:
${Power}={force}\times{velocity}$-----(1)
${Pressure}=\dfrac{force}{area}$------(2)

${Force}=\dfrac{power}{velocity}=\dfrac { 1.4\times 1000 }{ 3\times { 10 }^{ 8 } } =4.66\times { 10 }^{ -6 } newton$

Put in equation (2)
${ Pressure } _{ absorbed }=\dfrac { 4.66\times { 10 }^{ -6 }N }{ { m }^{ 2 } } =4.66\times { 10 }^{ -6 }Pa$

The temperature of a black body corresponding to which it will emit energy at the rate of $1 watt/cm^2$ will be

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

  1. 650K

  2. 450K

  3. 350K

  4. 250K

Show answer
Correct Option: A
Explanation:

$E\propto { T }^{ 4 }\quad \Longrightarrow \quad E=\sigma { T }^{ 4 }$
$\sigma =5.67*{ 10 }^{ 8 }W{ m }^{ 2 }{ k }^{ 4 }$
$1*{ 10 }^{ -4 }=5.67*{ 10 }^{ 8 }*{ T }^{ 4 }$
$T=648k\cong 650k$



The solar constant for the earth is $\Sigma$. The surface temperature of the sun is $T$ K. The sun subtends an angle $\theta$ at the earth

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

  1. $\Sigma \space \propto \space T^4$

  2. $\Sigma \space \propto \space T^2$

  3. $\Sigma \space \propto \space \theta^4$

  4. $\Sigma \space \propto \space \theta$

Show answer
Correct Option: A
Explanation:

By Stephan Boltzman law,


$P=\sigma (4\pi { R }^{ 2 }){ T }^{ 4 }$

$\theta =\dfrac { 2r }{ R } $ where R is distance between earth and sun and r is radius of earth.

Hence, $\sum {  } =\dfrac { P }{ 4\pi { r }^{ 2 } } =C{ T }^{ 4 }{ \left( \dfrac { R }{ r }  \right)  }^{ 2 }=K{ T }^{ 4 }{ \theta  }^{ 2 }$

Hence $\sum {  } \alpha { T }^{ 4 }$ and $\sum {  } \alpha { \theta }^{ 2 }$

Answer is option A.

In the Orion stellar system the shining of a star is $17\space \times 10^3$ times that of the sun. If the temperature of the surface of the sun $6 \times 10^3 K$ then the temperature of this star will be

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

  1. 273 K

  2. 652 K

  3. 6520 K

  4. 68520 K

Show answer
Correct Option: D
Explanation:

The power radiated is directly proportional to the fourth power of the absolute temperature.
$P \propto T^{4}$
$P = kT^{4}$ 

Let P = Power radiated by the sun
Power radiated by star $ = 17 \times  10^{3}P $
$\dfrac{T _{star}^{4}}{T _{sun}^{4}} = 17 \times  10^{3}$

$\dfrac{T _{star}}{T _{sun}} = 11.418$

$T _{star} = 6000 \times  11.418 = 68,511K$

Hence, the answer is option D.

There are two planets $A$ and $B$ at a large distance Planet $A$ is bigger and hotter than planet $B$. The angular diameter of planet $A$ is $40$ minute of arc as seen from planet $B$. The energy received by planet $B$ is $3cal-cm^{-2}$ per minute. Assuming the radiation to be black body in character. Given that stefan costant is $5.67\times 10^{-8}\ Wm^{-2}\ K^{-4}$. The temperature of planet $A$ is

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

  1. $(10.93\times 10^{14})^{1/4}\ K$

  2. $(53.21\times 10^{14})^{1/4}\ K$

  3. $(63.63\times 10^{14})^{1/4}\ K$

  4. $(63.21\times 10^{14})^{1/4}\ K$

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Correct Option: C

A blackened steel plate is put in a dark room after being heated up to a high temperature. A white spot on the plate appears. 

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

  1. brighter than the plate

  2. as bright as the plate

  3. dull as compared to the plate

  4. appears to be yellow

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Correct Option: A

Boltzmann's constant$ K = 1.38 \times 10^{-23} J/k $ The energy associated with helium atom the surface of sun, where surface temperature is 6000 K is

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

  1. $ 1.242 \times 10^{-19} J $

  2. $ 2.484 \times 10^{-19} J $

  3. $ 207 \times 10^{-19} J $

  4. $ 0.621 \times 10^{-19} J $

Show answer
Correct Option: A