Tag: stefan's law

Questions Related to stefan's law

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

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?

  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

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

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:

  1. 300:1

  2. 600:1

  3. 900:1

  4. 1200:1

Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $

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

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

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 $k = 1.38 \times 10 ^ { - 23 } \mathrm { JK } ^ { - 1 }$)

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

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

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

  4. none of these

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

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)

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

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of heat loss is dQ/dt = -sigma * A * (T^4 - T0^4). Since Q = mcT and c = alpha * T^3, dQ = m * alpha * T^3 * dT. Equating these, m * alpha * T^3 * dT/dt = -sigma * (4 * pi * R^2) * (T^4 - T0^4). Integrating from 3T0 to 2T0 gives the time.

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

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 $k=1.38\times 10^{-23}J/K$]-

  1. $10^{7}K$

  2. $10^{5}K$

  3. $10^{3}K$

  4. $10^{9}K$

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

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

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:

  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 $2901:K$

Reveal answer Fill a bubble to check yourself
A,B Correct answer
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 $\dfrac{T _A}{T _B}=3$
$\therefore 2 (\lambda _A) _m = 10^{-6}$
$\therefore (\lambda _A) _m= 0.5\times 10^{-6}$
$\therefore (\lambda _B) _m = 0.5\times 10^{-6} + 1\times 10^{-6} =1.5 \times  10^{-6}=1.5 \mu m$

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

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)

  1. $150\;k$

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

  3. $1/2\;k$

  4. None of these

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the equipartition theorem, the energy associated with each degree of freedom per molecule is (1/2)kT, regardless of the temperature.

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

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

  1. 1.36 times

  2. 2 times

  3. 4 times

  4. 8 times

Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(1.08)^{4}$ = $1.36$

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

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:

  1. 40

  2. 160

  3. 200

  4. 400

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

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

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

  1. 5.67 J

  2. 56.7 J

  3. 567 J

  4. 5670 J

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