Tag: heat transfer

Questions Related to heat transfer

The rate of emission of radiation of ablack body at temperature $27^oC $ is $ E _1 $ . If its temperature is increased to $ 327^oC $ the rate of emission of radiation is $ E _2 . $ The relation between $ E _1 $ and $ E _2 $ is:

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

  1. $ E _2 = 24 E _1 $

  2. $ E _2 =16 E _1 $

  3. $ E _2 = 8 E _1 $

  4. $ E _2 = 4 E _1 $

Show answer
Correct Option: B
Explanation:

In black body radiation 

$\dfrac{d\theta}{dt}=(4\pi r^{2})\sigma T^{4}$
If at $T=27^{o}C=300\ K, \dfrac{d\theta}{dt}=E _{1}$
Then, 
$E _{1}=(4\pi R^{2})\sigma(300)^{4}$
If at $T=327^{o}C=600\ k, \dfrac{d\theta}{dt}=E _{2}$
$E _{2}=(4\pi R^{2})\sigma (2)^{4}(300)^{4}$
So, $\dfrac{E _{2}}{E _{1}}=(2)^{4}\dfrac{(4\pi R^{2}\sigma (300)^{4}}{4\pi R^{2}\sigma(300)^{4}}$
$E _{2}=(2)^{4}E _{1}$
$\Rightarrow E _{2}10 E _{1}$
Option $B$ is correct






Two identical objects $A$ and $B$ are at temperatures $T _A$ and $T _B$. respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature $T$ ($T _A$ > $T$> $T _B$). The objects $A$ and $B$ attain the temperature $T$ eventually. Select the correct statements from the following

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

  1. $A$ only emits radiation, while $B$ only absorbs it until both attain the temperature $T$

  2. $A$ loses more heat by radiation than it absorbs, while $B$ absorbs more radiation than it emits until they attain the temperature $T$

  3. Both $A$ and $B$ only absorb radiation, but do not emit it, until they attain the temperature $T$

  4. Each object continuous to emit and absorb radiation even after attaining the temperature $T$

Show answer
Correct Option: B
Explanation:

Since the temperature of $A$ is higher than the temperature of the surrounding hence $A$ radiates heat much larger than it absorbs heat. Since the temperature of $B$ is lower than the temperature of the surrounding hence $B$ absorbs heat much larger than it radiates.
This process goes on until both $A$ and $B$ reach the temperature $T$.
Even after reaching thermal equilibrium, both bodies keep radiating and absorbing.
Hence options $B$, $D$. 

A planet is at an average distance $d$ from the sun and its average surface temperature is $T$. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If $T$ $\propto d^{-n}$, the value of $n$ is :

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

  1. $2$

  2. $1$

  3. $\displaystyle \frac{1}{2}$

  4. $\displaystyle \frac{1}{4}$

Show answer
Correct Option: C
Explanation:
Let P=power radiated by Sun
R=Radius of planet
E=energy received by planet=$\cfrac{P}{4 \pi d^2}\times \pi R^2$
Energy radiated by planet=$(4 \pi R^2)\sigma T^4 $
For thermal equilibriums:
$\Rightarrow \cfrac { P }{ 4\pi d^{ 2 } } \pi R^{ 2 }=4\pi R^{ 2 }\sigma T^{ 4 }\\ \Rightarrow T^{ 4 }\alpha \cfrac { 1 }{ d^{ 2 } } \\ \Rightarrow T\alpha \cfrac { 1 }{ d^{ { 1 }/{ 2 } } } \\ \Rightarrow T\alpha { d }^{ -\cfrac { 1 }{ 2 }  }$
So n=$\cfrac{1}{2}$

A planet radiates heat at a rate proportional to the fourth power of its surface temperature $T$. If such a steady temperature of the planet is due to an exactly equal amount of heat received from the sun then which of the following statements is true?

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

  1. The planet's surface temperature varies inversely as the distance of the sun

  2. The planet's surface temperature varies directly as the square of its distance from the sun

  3. The planet's surface temperature varies inversely as the square root of its distance from the sun

  4. The planet's surface temperature is proportional to the fourth power of distance from the sun

Show answer
Correct Option: C
Explanation:
Planet's surface temperature varies inversely as square root of its distance from the Sun.
${ T }^{ 4 }\alpha \cfrac { 1 }{ { d }^{ 2 } } \Rightarrow T\alpha \cfrac { 1 }{ \sqrt { d }  } $

The radiation emitted by a star $A$ is $1000$ times that of the sun. If the surface temperatures of the sun and star $A$ are $6000 K$ and $2000 K$, respectively, the ratio of the radii of the star $A$ and the Sun is:

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

  1. 300:1

  2. 600:1

  3. 900:1

  4. 1200:1

Show answer
Correct Option: C
Explanation:

$E\propto A{ T }^{ 4 }$


$\displaystyle \frac { { E } _{ 1 } }{ { E } _{ 2 } } =\frac { 1000 }{ 1 } =\frac { \pi { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ \pi { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { 2000 }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times 6000^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 } }{ { { r } _{ 2 } }^{ 2 }\times 81 } $

$\displaystyle \frac { { r } _{ 1 } }{ { r } _{ 2 } } =\sqrt { \frac { 1000 }{ 1 } \times \frac { 81 }{ 1 }  } =284.6:1$

The number of oxygen molecules in a cylinder of volume $1 \mathrm { m } ^ { 3 }$ at a temperature of $27 ^ { \circ } C$ and pressure $13.8 Pa$ is
 (Boltzmaan's constant $k = 1.38 \times 10 ^ { - 23 } \mathrm { JK } ^ { - 1 }$)

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

  1. $6.23 \times 10 ^ { 26 }$

  2. $0.33 \times 10 ^ { 28 }$

  3. $3.3 \times 10 ^ { 21 }$

  4. none of these

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Correct Option: A

A solid sphere of mass m and radius $R$ is painted black and placed inside a vacuum chamber. The walls of the chamber are maintained at temperature $T 0$ the initial temperature of the sphere is $3T _0$. The specific heat capacity of the sphere material varies with its temperature $T$ as $\alpha T^3$ where $\alpha$ is a constant. Then the sphere will cool down to temperature $2T _0$ in time ________ ($\sigma$ = Stefan Boltzmann constant)

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

  1. $\dfrac{m\alpha}{16\pi R^2\sigma}\ell n\left(\dfrac{16}{3}\right)$

  2. $\dfrac{m\alpha}{8\pi R^2\sigma}\ell n\left(\dfrac{4}{3}\right)$

  3. $\dfrac{m\alpha}{8\pi R^2\sigma}\ell n\left(\dfrac{3}{2}\right)$

  4. $\dfrac{m\alpha}{4\pi R^2\sigma}\ell n\left(\dfrac{8}{3}\right)$

Show answer
Correct Option: A

In the nuclear fusion, $ _{1}^{2}{H}+ _{1}^{3}{H}\rightarrow _{2}^{4}{He}+n$ given that the repulsive potential energy between the two nuclie is $7.7\times 10^{-14}J$, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant $k=1.38\times 10^{-23}J/K$]-

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

  1. $10^{7}K$

  2. $10^{5}K$

  3. $10^{3}K$

  4. $10^{9}K$

Show answer
Correct Option: D
Explanation:

Energy    $E \approx kT$

So,     $7.7\times 10^{-14} \approx 1.38\times 10^{-23}\times T$
$\implies \ T\approx  5.6\times 10^9 \ K$
Correct answer is option D.

Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surface area of the two bodies are the same. The two bodies radiate energy at the same rate. The wavelength $\lambda _{B}$, corresponding to the maximum spectral radiancy in the radiation from $B$, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from $A$ by $1.00 :\mu m$. If the temperature of $A$ is $5802 :K$, then:

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

  1. the temperature of $B$ is $1934:K$

  2. $\lambda _{B}=1.5:\mu m$

  3. the temperature of $B$ is $11604:K$

  4. the temperature of $B$ is $2901:K$

Show answer
Correct Option: A,B
Explanation:

From Stefan's Law:
$\sigma A\epsilon _AT _A^4=\sigma A\epsilon _BT _B^4$  ....(1)
where, $T _A=5802:K$ is temp of A and $T _B$ is temp of B,

$\epsilon _A=0.01$ is emissivity of A,
$\epsilon _B=0.81$ is emissivity of B,
$\sigma$ is Stefan's constant,
$A$ is the surface area of the bodies A and B

Substituting the values in (1)

$0.01 \times 5802^4 = 0.81 T _B^4$

or, $\left (\dfrac{T _B}{5802}\right )^4 = \dfrac{0.01}{0.81}=\left ( \dfrac{1}{3} \right )^4$

$\therefore T _B= \dfrac{5802}{3}=1934:K$

From Wien's displacement Law
$(\lambda _A) _mT _A=(\lambda _B) _mT _B$    ......(2)

Given, $(\lambda _B) _m = (\lambda _A) _m + 1\times 10^{-6}$  ....(3)

Substituting $(\lambda _B) _m$ from (3) in (2)
$(\lambda _A) _mT _A=( (\lambda _A) _m + 1\times 10^{-6}) T _B$
$ \therefore (\lambda _A) _m \times  3 = (\lambda _A) _m + 1\times 10^{-6}$ since $\dfrac{T _A}{T _B}=3$
$\therefore 2 (\lambda _A) _m = 10^{-6}$
$\therefore (\lambda _A) _m= 0.5\times 10^{-6}$
$\therefore (\lambda _B) _m = 0.5\times 10^{-6} + 1\times 10^{-6} =1.5 \times  10^{-6}=1.5 \mu m$

Energy associated with each molecule per degree of freedom o a system at room temperature $(27^{\circ}C)$ will be ($k$ is Boltzmann's constant)

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

  1. $150\;k$

  2. $(27/2)\;k$

  3. $1/2\;k$

  4. None of these

Show answer
Correct Option: C