Tag: heat transfer

Questions Related to heat transfer

The temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of :

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

  1. 2

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: D
Explanation:

We know that from Stefan's Boltzmann relation: $E\propto { T }^{ 4 }$
if the temperature of the sun will be doubled, then : $E\propto ({ 2T })^{ 4 }$
Hence, $E\  will\  increase \ by \ the \ factor \ of \  { 16}$.

A black body is at temperature $300K$. It emits energy at a rate, which is proportional to 

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

  1. ${(300)}^{4}$

  2. ${(300)}^{3}$

  3. ${(300)}^{2}$

  4. $300$

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Correct Option: A
Explanation:

For black body radiation
$E=\sigma{T}^{4}$ or $E\propto {T}^{4}$
Rate of emission of energy $\propto {(300)}^{4}$

If the absolute temperature of a blackbody is doubled, then the maximum energy density

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

  1. Increases to 16 times

  2. Increases to 32 times

  3. Decreases to 16 times

  4. Decreases to 32 times

Show answer
Correct Option: A
Explanation:

The power with which a body(in this case black body) radiates is directly proportional to the fourth power of absolute temperature:
$P = kT^{4}$
i.e.
$P _{1} = kT _{1}^{4}$
If the absolute temperature is doubled,
$P _{2} = k(2T _{1})^{4} = 16kT _{1}^{4}$
Now $\dfrac{P _{2}}{P _{1}} = \dfrac{16}{1}$ 

Hence energy density is increased 16 times.

Intensity of heat radiation emitted by body is believed to be proportional to fourth power of absolute temperature of the body. The proportionality constant also known as Boltzmann's constant may have possible value of :

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

  1. $5.67\times 10^{-8} watt/K^4 $

  2. $5.67\times 10^{-8} watt/m^2 K^4 $

  3. $5.67\times 10^{-8} J/K^4 $

  4. $5.67\times 10^{-8} Js/K^4 $

Show answer
Correct Option: B
Explanation:

From the given question,

$I\propto T^4$

$I=k T^4$

where $k=$ Stefan-Boltzmann's constant

$k=5.67\times10^{-8} W/m^2K^4 $

The correct option is B.

A black body at a temperature of $227^oC$ radiates heat energy at the rate 5 cal/cm$^{2}-s$. At a temperature of $727^oC$, the rate of heat radiated per unit area in cal/cm$^2$ will be

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

  1. 80

  2. 160

  3. 250

  4. 500

Show answer
Correct Option: A
Explanation:

According to Stefen's Law, the rate of heat radiation from body is proportional to the fourth power of body's temperature.

Thus $P\propto T^4$
$\implies \dfrac{P _2}{P _1}=\dfrac{T _2^4}{T _1^4}$
$=16$
$\implies P _2=80cal/cm^2-s$

For a block body temperature $727^{o}C,$ its rate of energy loss is $20\ watt$ and temperature of surrounding is $227^{o}C.$ If temperature of black body is changed to $1227^{o}C$ then its rate of energy loss will be:

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

  1. $320\ W$

  2. $\dfrac {304}{3}\ W$

  3. $240 W$

  4. $120 W$

Show answer
Correct Option: A
Explanation:

It is given that,

Temperature of surrounding

  $ {{T} _{0}}={{227}^{0}}C=500\ K $

 $ {{T} _{1}}={{727}^{0}}C=1000\ K $

 $ {{T} _{2}}={{1227}^{0}}C=1500\ K $

 $ {{E} _{1}}=20\ Watt\,\, $

 $ {{E} _{2}}=? $

According to Stefn boltzmann law:

$ E=\sigma {{T}^{4}} $

Or

 $ {{E} _{1}}=\sigma ({{T} _{1}}-{{T} _{0}})^4 $

 $ {{E} _{2}}=\sigma ({{T} _{2}}-{{T} _{0}})^4 $

Taking ratios of above equations:

For $ {{E} _{1}}=20\ Watt $

 $ \dfrac{20}{{{E} _{2}}}={{\left( \dfrac{500}{1000} \right)}^{4}} $

 $ \dfrac{20}{{{E} _{2}}}=\left( \dfrac{1}{16} \right) $

 $ {{E} _{2}}=320\ Watt $

The power received at distance $d$ from a small metallic sphere of radius $r(<<d)$ and at absolute temperature $T$ is $P$. If the temperature is doubled and distance reduced to half of the initial value, then the power received at that point will be:

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

  1. $4p$

  2. $8p$

  3. $32p$

  4. $64p$

Show answer
Correct Option: D
Explanation:
Energy received per second i.e., power $P\alpha \dfrac{T^4}{d^2}=k\dfrac{T^4}{d^2}$
if temperature is double than T become 2T and distance become half than d become $\dfrac{d}{2}$
than power $ p _{1}=k\dfrac{(2T)^4}{(\dfrac{d}{2})^2}=64k\dfrac{T^4}{d^2}=64P$
Hence D option is correct.

What is the value of solar constant if the energy received by $ 12$ m$^2$ area in $2$ minutes is $2016$ kJ?

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

  1. $1.4 \times 10^2 J s^{-1}  m^{-2}$

  2. $1400 J s^{-1}m^{-2}$

  3. $84 kJ s^{-1} m^{-2}$

  4. $84 J s^{-1} m^{-2}$

Show answer
Correct Option: B
Explanation:
Solar constant = Energy received by unit area in unit time from Sun
$=\cfrac{2016\times 10^3}{12\times (2\times 60)} \\=1400Js^{-1}m^2$

If a graph is plotted by taking spectral emissive power along $y-$axis and wavelength along x-axis is:

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

  1. Emissivity

  2. Total intensity of radiation

  3. Diffusivity

  4. Solar constant

Show answer
Correct Option: A
Explanation:

Emissive power varies according to Stefan-Boltzmann law as :

$E=\sigma {{T}^{4}}$

According to Planck’s distribution law:

${{E} _{\lambda }}(\lambda ,T)=\dfrac{{{C} _{1}}}{{{\lambda }^{5}}\left[ \exp (\dfrac{{{C} _{2}}}{\lambda T})-1 \right]}$

The graph shows that the emitted radiation varies with wavelength and also it shows Emissivity.

 

A spherical body of area A and emissivity $0.6$ is kept inside a perfectly black body. Total heat radiated by the body at temperature T is?

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

  1. $0.4\sigma AT^4$

  2. $0.8\sigma AT^4$

  3. $0.6\sigma AT^4$

  4. $1.0\sigma AT^4$

Show answer
Correct Option: D
Explanation:

When a non black body is placed inside a hollow enclosure the total radiation from the body is the sum of what it would emit in the open ( with e<1 ) and the part (1-a) of the incident radiation from the walls reflected by it.

The two add up to a black body radiation . Hence the total radiation emitted by the body is $1.0\sigma AT^4$
1.0σAT4.