Physics

Thermal Properties and Thermodynamics

431 Questions

Thermal properties and thermodynamics questions evaluate concepts of heat transfer, thermal efficiency, and temperature variations. Problems involve calculating heat content, conductivity, and the performance of heat engines. This subject is regularly tested in physics sections across multiple competitive platforms.

Heat transfer calculationsThermal efficiencyBlack body radiationTemperature variationsRefrigeration performance

Thermal Properties and Thermodynamics Questions

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

A black body radiates energy at the rate of $E\ watt/m$$^{2}$ at a high temperature $T^{o}K$ when the temperature is reduced to $\left [ \dfrac{T}{2} \right ]^{o}K$ Then radiant energy is

  1. $4E$
  2. $16E$
  3. $\dfrac{E}{4}$
  4. $\dfrac{E}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that from stefans-boltzman law: $E\propto { T }^{ 4 }$
if temperature will be reduces half form the initial value, then
${E} _{1}\propto ({ \dfrac { T }{ 2 } ) }^{ 4 }$
${E} _{1}\propto\dfrac{E}{16}$

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

Two spherical black bodies of radii $r _{1} $ and $  r _{2}$ are with surface temperatures $T _{1} $ and $ T _{2}$ respectively radiate the same power. $r _{1} / r _{2}$ must be equal to

  1. $(T _{1}/T _{2})^{2}$
  2. $(T _{2}/T _{1})^{2}$
  3. $(T _{1}/T _{2})^{4}$
  4. $(T _{2}/T _{1})^{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Given that both spherical body radiate with same power.
So equating ${E} _{1}={E} _{2}$
${A}=4{\pi}{r}^{2}$
${ r } _{ 1 }^{ 2 }{ T } _{ 1 }^{ 2 }={ r } _{ 2 }^{ 2 }{ T } _{ 2 }^{ 4 }$

$\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } ={ (\dfrac { { T } _{ 2 } }{ { T } _{ 1 } } ) }^{ 2 }$

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

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

  1. ${(300)}^{4}$
  2. ${(300)}^{3}$
  3. ${(300)}^{2}$
  4. $300$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. Increases to 16 times

  2. Increases to 32 times

  3. Decreases to 16 times

  4. Decreases to 32 times
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 :

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

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

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

  1. 80

  2. 160

  3. 250

  4. 500
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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:

  1. $320\ W$
  2. $\dfrac {304}{3}\ W$
  3. $240 W$
  4. $120 W$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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:

  1. $4p$
  2. $8p$
  3. $32p$
  4. $64p$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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?

  1. $0.4\sigma AT^4$
  2. $0.8\sigma AT^4$
  3. $0.6\sigma AT^4$
  4. $1.0\sigma AT^4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

The rate of emission of radiation of ablack body at temperature $27^oC $ is $ E _1 $ . If its temperature is increased to $ 327^oC $ the rate of emission of radiation is $ E _2 . $ The relation between $ E _1 $ and $ E _2 $ is:

  1. $ E _2 = 24 E _1 $
  2. $ E _2 =16 E _1 $
  3. $ E _2 = 8 E _1 $
  4. $ E _2 = 4 E _1 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In black body radiation 

$\dfrac{d\theta}{dt}=(4\pi r^{2})\sigma T^{4}$
If at $T=27^{o}C=300\ K, \dfrac{d\theta}{dt}=E _{1}$
Then, 
$E _{1}=(4\pi R^{2})\sigma(300)^{4}$
If at $T=327^{o}C=600\ k, \dfrac{d\theta}{dt}=E _{2}$
$E _{2}=(4\pi R^{2})\sigma (2)^{4}(300)^{4}$
So, $\dfrac{E _{2}}{E _{1}}=(2)^{4}\dfrac{(4\pi R^{2}\sigma (300)^{4}}{4\pi R^{2}\sigma(300)^{4}}$
$E _{2}=(2)^{4}E _{1}$
$\Rightarrow E _{2}10 E _{1}$
Option $B$ is correct






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

Two identical objects $A$ and $B$ are at temperatures $T _A$ and $T _B$. respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature $T$ ($T _A$ > $T$> $T _B$). The objects $A$ and $B$ attain the temperature $T$ eventually. Select the correct statements from the following

  1. $A$ only emits radiation, while $B$ only absorbs it until both attain the temperature $T$
  2. $A$ loses more heat by radiation than it absorbs, while $B$ absorbs more radiation than it emits until they attain the temperature $T$
  3. Both $A$ and $B$ only absorb radiation, but do not emit it, until they attain the temperature $T$
  4. Each object continuous to emit and absorb radiation even after attaining the temperature $T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the temperature of $A$ is higher than the temperature of the surrounding hence $A$ radiates heat much larger than it absorbs heat. Since the temperature of $B$ is lower than the temperature of the surrounding hence $B$ absorbs heat much larger than it radiates.
This process goes on until both $A$ and $B$ reach the temperature $T$.
Even after reaching thermal equilibrium, both bodies keep radiating and absorbing.
Hence options $B$, $D$. 

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

A planet is at an average distance $d$ from the sun and its average surface temperature is $T$. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If $T$ $\propto d^{-n}$, the value of $n$ is :

  1. $2$
  2. $1$
  3. $\displaystyle \frac{1}{2}$
  4. $\displaystyle \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let P=power radiated by Sun
R=Radius of planet
E=energy received by planet=$\cfrac{P}{4 \pi d^2}\times \pi R^2$
Energy radiated by planet=$(4 \pi R^2)\sigma T^4 $
For thermal equilibriums:
$\Rightarrow \cfrac { P }{ 4\pi d^{ 2 } } \pi R^{ 2 }=4\pi R^{ 2 }\sigma T^{ 4 }\\ \Rightarrow T^{ 4 }\alpha \cfrac { 1 }{ d^{ 2 } } \\ \Rightarrow T\alpha \cfrac { 1 }{ d^{ { 1 }/{ 2 } } } \\ \Rightarrow T\alpha { d }^{ -\cfrac { 1 }{ 2 }  }$
So n=$\cfrac{1}{2}$
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

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

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 $\dfrac{T _A}{T _B}=3$
$\therefore 2 (\lambda _A) _m = 10^{-6}$
$\therefore (\lambda _A) _m= 0.5\times 10^{-6}$
$\therefore (\lambda _B) _m = 0.5\times 10^{-6} + 1\times 10^{-6} =1.5 \times  10^{-6}=1.5 \mu m$