Tag: heat transfer

Questions Related to heat transfer

The temperature of a spherical planet is related to the distance from sun as :

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

  1. $T \propto 1/d^{2}$

  2. $T\propto \dfrac{1}{\sqrt{d}}$

  3. $T\propto d$

  4. $T\propto d^{2}$

Show answer
Correct Option: B
Explanation:
A planet reaches its equilibrium temperature when the amount of heat it absorbs from the sun is equal to the amount that it radiates back to space.
The energy absorbed is equal to the insolation at planet's distance (which is Sun's bolometric output divided by $4\pi$ times the distance squared) times the planet's profile area times the planet's Bond aldedo. The amount radiated (assuming that the thermal radiation is approximately black body radiation) is proportional to the planet's surface area times its emmisivity times the forth power of its temperature.

Now, throwing away 2, $\pi$ and other constants that do not vary with temperature or distance.
${ T }^{ 4 }\alpha \cfrac { 1 }{ { d }^{ 2 } } \\ \Rightarrow T\alpha \cfrac { 1 }{ \sqrt { d }  } $

Assertion (A): The radiation from the sun surface varies as the fourth power of its absolute temperature.
Reason (R): Sun is not a black body

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

  1. Both A and R are true, R is correct explanation of A

  2. Both A and R are true, R is not correct explanation of A

  3. A is true but R is false

  4. Both A and R are false

Show answer
Correct Option: B
Explanation:

According to Stefan's law, energy emitted is proportional to fourth power of absolute temperature.
So, the same will hold for the sun.
Also, Sun can absorb/emit radiations of all wavelengths - so it is a black body.
So, assertion and reason are both correct.
But reason is not a correct explanation of assertion.
Because all bodies emit energy which is proportional to fourth power of absolute temperature, not only sun.

Three bodies A, B, C are at $-27^{o}$C, $0^{o}$C, $100^{o}$C respectively. The body which does not radiate heat is:

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

  1. A

  2. B

  3. none as all the bodies radiate heat

  4. C

Show answer
Correct Option: C
Explanation:

All bodies radiate heat irrespective of temperature.
Heat radiated per unit time is given by Stefan's law.

A solid shpere and a hollow sphere of the same material and of equal radii are heated to the same temperature

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

  1. both will emit equal amount of radiation per unit time in the beginning.

  2. both will absorbs equal amount of radiation per second from the surrounding in the beginning.

  3. the initial rate of cooling will be the same for both the spheres

  4. the two spheres will have equal temperature at any instant

Show answer
Correct Option: A,B
Explanation:

According to many radiation laws like Stefan Boltzmann we know that radiation emission and absorption are a purely surface phenomenon. Since the two bodies are of same material, same radii, and same temperature they will at that instant radiate and absorb at the same rates.
But however since the hollow sphere has lesser mass, the rate at which it's temperature will rise will be different from that of the solid sphere. Hence their rates of cooling would be varied and they would have different temperatures at different times.

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 :

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

  1. $508^{o}C$

  2. $273^{o}C$

  3. $400^{o}C$

  4. $527^{o}C$

Show answer
Correct Option: D
Explanation:

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

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

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

  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. $\left ( 97\right )^{\dfrac{1}{4}}T$

Show answer
Correct Option: C

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

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

  1. $\displaystyle \frac{3}{8}E$

  2. $\displaystyle \frac{81}{16}E$

  3. $\displaystyle \frac{9}{16}E$

  4. $\displaystyle \frac{81}{64}E$

Show answer
Correct Option: D
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$\dfrac{l _2}{l _1}=\dfrac{1}{2}$       $\dfrac{b _2}{b _1}=\dfrac{1}{2}$       $\dfrac{T _2}{T _1}=\dfrac{3}{2}$

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

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

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

  1. 0.5% 

  2. 1.0%

  3. 1.5%

  4. 2.0%

Show answer
Correct Option: D
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 $=4\times0.5=2\%$

A black body is at a temperature of $500$K. It emits its energy at a rate which is proportional to :

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

  1. $500$

  2. $(500)^{2}$

  3. $(500)^{3}$

  4. $(500)^{4}$

Show answer
Correct Option: D
Explanation:

From stefan's boltzmann relation we know that: $E\propto { T }^{ 4 }$
So, $E\propto { 500}^{ 4 }$

The rate of emission of a black body at temperature $27$$^{o}$C is $E _{1}$. If its temperature is increased to $327$$^{o}$C, 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. $E _{2}=24E _{1}$

  2. $E _{2}=16E _{1}$

  3. $E _{2}=8E _{1}$

  4. $E _{2}=4E _{1}$

Show answer
Correct Option: B
Explanation:

We know the relation for emissive power of body: $\dfrac { { E } _{ 1 } }{ { E } _{ 2 } } =\dfrac { { T } _{ 1 }^{ 4 } }{ { T } _{ 2 }^{ 4 } } $
${T} _{1}=27+273=300\ K$
${T} _{2}=327+273=600\ K$
So, putting all these data in above formulas, we get
${E} _{2}=16{E} _{1}$