Tag: heat transfer

Questions Related to heat transfer

Statement-1:Good reflectors are poor emitters of thermal radiation.
Statement-2:The ratio of the emissive power (e) and absorptive power(a) is constant for all substances at any given temperture and for radiation of the same wavelength

kirchoff's laws black body radiation heat transfer thermal properties physics

  1. Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement- 1

  2. Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Staement-1

  3. Statement-1 is true, Statement-2 is false.

  4. Statement-1 is false, Statement-2 is true.

Show answer
Correct Option: A
Explanation:

From the krichoff's law we know that absorptivity is equal to emissivity,so the body which emitt more wil absorbe more nor reflect more. So good reflector are poor emittor.
Now, by taking there ratio it validate krichoff's law because at any fixed or constant temperature both are equal.

In a dark room with ambient temperature $T _o$, a black body is kept at a temperature $T$. Keeping the temperature of the black body constant (at $T$), sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct?

kirchoff's laws black body radiation heat transfer thermal properties physics

  1. The quantity of radiation absorbed by the black body in unit time will increase

  2. Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will

    increase

  3. Black body radiates more energy in unit time in the visible spectrum.

  4. The reflected energy in unit time by the black body remains same.

Show answer
Correct Option: A,B,C,D
Explanation:

Since the radiation is continuously falling on the black body, the quantity radiation absorbed per second will increase.

The reason given in question is self explanatory.
With an increase in temperature, the entire Plank's curve shifts upwards and hence radiation in any spectrum will increase.
Reflected energy per unit time will be zero since black body has 0 reflectivity and hence it will remain constant.  

Good conductors of heat is

physics heat transfer applications of heat conduction applications of insulation clothes - our necessity

  1. Copper

  2. Water

  3. Wood

  4. Glass

Show answer
Correct Option: A
Explanation:

Substances become bad conductors of heat when they contain a large number of free electrons. This is because the electrons contained in atom of these elements are not closely bond to nucleus.

The value of solar constant is approximately  :

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

  1. $ 1340\ watt/m^{2}$

  2. $ 430\ watt/m^{2}$

  3. $ 340\ watt/m^{2}$

  4. $ 1388\ watt/m^{2}$

Show answer
Correct Option: D
Explanation:

The solar constant is defined as the amount of heat energy received per second per unit area by a perfect black body placed at the surface of the Earth with its surface being held perpendicular to the direction of the sun's rays.

The value of solar constant is $1388$($\dfrac{watt}{{meter}^{2}}$) or $2$($\dfrac{cal}{{cm}{\times} {min}}$)

A heated body emits radiation which has maximum intensity at frequency $v _m$. If the temperature of the body is doubled

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

  1. the maximum intensity radiation will be at frequency $2v _m$

  2. the maximum intensity radiation will be at frequency $\displaystyle\dfrac{1}{2}v _m$

  3. the total emitted energy will increase by a factor of $16$

  4. the total emitted energy will increase by a factor of $2$

Show answer
Correct Option: A,C
Explanation:

Wien's displacement law states maximum intensity wavength $ \lambda _{m}\propto \dfrac{1}{T}$
Also for any photon,$ \lambda \propto \dfrac{1}{\nu}$
Hence, frequency $\nu _m \propto T$
Doubling of temperature leads to doubling of frequency from $\nu _m$ to $ 2\nu _m$
From Stefan's law, power is directly proportional to $T^4$
Hence $ T \rightarrow 2T \Rightarrow E \rightarrow (\dfrac {2T}{T})^4E=16E$

The amount of radiations emitted by a black body depends on its

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

  1. size

  2. mass

  3. temperature

  4. density

Show answer
Correct Option: C
Explanation:

The radiations emitted by the body only depend on the type of surface emitting and the temperature difference between the body and the surroundings.

The amplitudes of radiations from a cylindrical heat source is related to the distance are

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

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

  2. $\displaystyle A \propto \frac{1}{ d} $

  3. $ A \propto d$

  4. $ A \propto d^2$

Show answer
Correct Option: B
Explanation:

The intensity is inversely proportional to  square of distance and intensity is directly proportional to square of amplitude. So, amplitude is inversely proportional to distance. 

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 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is

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

  1. $(\cfrac{65}{2})^{\frac{1}{4}}T$

  2. $(\cfrac{97}{4})^{\frac{1}{4}}T$

  3. $(\cfrac{97}{2})^{\frac{1}{4}}T$

  4. $(97)^{\frac{1}{4}}T$

Show answer
Correct Option: C

The energy emitted by a black body at $727^oC$ is E. If the temperature of the body is increased by $227^oC$, the emitted energy will become

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

  1. 13 times

  2. 2.27 times

  3. 1.9 times

  4. 3.9 times

Show answer
Correct Option: B
Explanation:

Here we know that energy emitted by any body is given by ${E}=\sigma{T^{4}}$

So, at temperature ${T}={727}^{o}C$ energy emitted will be ${E}$
at temperature ${T} _{1}={727+227}={954}^{o}C$ energy emitted will be ${E} _{1}={\sigma}{T} _{1}^{4}$
$\dfrac { E }{ { E } _{ 1 } } =\dfrac { { T }^{ 4 } }{ { T } _{ 1 }^{ 4 } }$
${ E } _{ 1 }=\dfrac { E\times { T } _{ 1 }^{ 4 } }{ { T }^{ 4 } } =\dfrac { E\times { 954 }^{ 4 } }{ { 727 }^{ 4 } } =2.96E$

The radiation emitted by a star $A$ is $10000$ times that of the sun. If the surface temperature 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:
Stars can be approximated as black bodies.
Hence by stefan's law, power emitted=P=$\sigma AT^4$
Thus,$ \dfrac{P _A}{P _{sun}}=\dfrac{r _A^2T _A^4}{r _{sun}^2T _{sun}^4}$
$\Rightarrow (\dfrac{r _A}{r _{sun}})^2=10000\times (\dfrac{6000}{2000})^4 \rightarrow \dfrac{r _A}{r _{sun}}=900$
So required ratio is 900:1