Tag: thermal properties

Questions Related to thermal properties

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

  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.


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?

  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.


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.  

What is the process in which heat energy of both hot and cold body equalizes?

  1. Calorimetry

  2. Fermentation

  3. Latent heat

  4. Hidden heat


Correct Option: A
Explanation:

Principle of calorimetry states that heat lost by a hotter body = heat gained by a colder body, therefore calorimetry is a process in which heat energy of both hot and cold body equalizes.

The value of solar constant is approximately  :

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

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

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

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


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

  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 $2$


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

  1. size

  2. mass

  3. temperature

  4. density


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

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

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

  3. $ A \propto d$

  4. $ A \propto d^2$


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

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


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

  1. 13 times

  2. 2.27 times

  3. 1.9 times

  4. 3.9 times


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

  1. $300:1$

  2. $600:1$

  3. $900:1$

  4. $1200:1$


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