Physics

Thermal Properties and Thermodynamics

380 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 physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The heat generated in a circuit is dependent on the resistance, current and time of flow of electric current. If the percentage errors measured in the above physical quantities are 1%, 2% and 1% respectively, the maximum error in measuring the heat is :

  1. $2\%$
  2. $3\%$
  3. $6\%$
  4. $1\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


$ H={ I }^{ 2 }RT$ (by dimensional analysis)
$\displaystyle \frac { \triangle n }{ n } =(2\frac { \triangle I }{ I } +\frac { \triangle R }{ R } +\frac { \triangle T }{ T } )$%
          $=2(2)+1+1$
          $= 6 $ %

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The heat generated in a circuit is dependent upon the resistance, current and time for which the current is flown. If the error in measuring the above are 1%, 2% and 1% respectively. The maximum error in measuring the heat is :

  1. 8%

  2. 6%

  3. 18%

  4. 12%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ Q = I^2 R t$


$ \dfrac{\delta Q}{ Q }= 2\dfrac{\delta i}{i}+ \dfrac{\delta R}{R}+\dfrac{\delta t}{t}$

$ \dfrac{\delta Q}{ Q }= 2\times 2$ % $+ 1$%$+1$%

$ \dfrac{\delta Q}{ Q }=6$ % 

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The energy of a system as a function of time t is given as $E(t) = A^2 exp(- \alpha t)$, where $\alpha = 0.2 s^{-1}$. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E(t)$ at t = 5 s is:

  1. 2%

  2. 4%

  3. 3%

  4. 5%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$E(t) = A^2 e^{-\alpha t}$
Taking natural logarithm on both sides,
$ln(E) = 2ln(A) + (- \alpha t)$
Differentiating both sides
$\displaystyle \frac{dE}{E} = 2\frac{dA}{A} + (\alpha dt)$
Errors always add up for maximum error.
$\displaystyle \therefore \frac{dE}{E} = 2\frac{dA}{A} + \alpha \left( \frac{dt}{t} \right) \times t$
Here, $\displaystyle \frac{dA}{A} = 1.25$ %, $\displaystyle \frac{dt}{t} = 1.5$%, $t = 5s$, $\displaystyle \alpha = 0.2 s^{-1}$
$\therefore \displaystyle \frac{dE}{E} = (2 \times 1.25$%$\displaystyle ) + (0.2) \times (1.5$%$) \times 5 = 4$%

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

If temperature of body increases by 10%, then increase in radiated energy of the body is :

  1. 10 %

  2. 40 %

  3. 46 %

  4. 1000 %
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the Stefan-Boltzmann law, radiated energy E is proportional to T^4. If T increases by 10%, T_new = 1.1T. Then E_new = (1.1)^4 * E = 1.4641 * E. The increase is 46.41%.

Multiple choice rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

Three copper blocks of masses ${ M } _{ 1 }$, ${ M } _{ 2 }$, and ${ M } _{ 3 }$, kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at ${ T } _{ 1 }$,${ T } _{ 2 }$,${ T } _{ 3 }$ $\left( { T } _{ 1 }{ >T } _{ 2 }>{ T } _{ 3 } \right)$. Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)     

  1. $T=\dfrac { { T } _{ 1 }{ +T } _{ 2 }+{ T } _{ 3 } }{ 3 } $
  2. $T=\dfrac { { { { M } _{ 1 }T } _{ 1 }{ +{ M } _{ 2 }T } _{ 2 }+{ M } _{ 3 }{ T } _{ 3 } } }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } } $
  3. $T=\dfrac { { { M } _{ 1 }T } _{ 1 }{ +{ M } _{ 2 }T } _{ 2 }+{ M } _{ 3 }{ T } _{ 3 } }{ 3\left( { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } \right) } $
  4. $T=\dfrac { { { M } _{ 1 }T } _{ 1 }s{ +{ M } _{ 2 }T } _{ 2 }s+{ M } _{ 3 }{ T } _{ 3 }s }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let us assume that $T _1>T _2,T _3$ and $T _1>T>T _2,T _3$

Now heat loss by $M _1=$ Heat gained by $M _2$ and $M _3$

$M _1S(T _1-T)=M _2S(T-T _1)+M _3S(T-T _3)$

$\implies M _1T _1+M _2T _2+M _3T _3=(M _1+M _2+M _3)T$

$\implies T=\dfrac{M _1T _1+M _2T _2+M _3T _3}{M _1+M _2+M _3}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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 $2$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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$

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

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. $(97)^{\frac{1}{4}}T$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In steady state, the heat flux through each gap between the plates must be equal. Using the Stefan-Boltzmann law for radiative heat transfer between parallel plates, the heat flux q = sigma * (T1^4 - T2^4). Setting the flux between plate 1 and 2 equal to the flux between plate 2 and 3, we get (2T)^4 - T2^4 = T2^4 - (3T)^4. Solving for T2 gives T2 = ((2^4 + 3^4)/2)^(1/4) * T = (97/2)^(1/4) * T.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 ${E} _{1}={\sigma}{T} _{1}^{4}$
$\dfrac { E }{ { E } _{ 1 } } =\dfrac { { T }^{ 4 } }{ { T } _{ 1 }^{ 4 } }$
${ E } _{ 1 }=\dfrac { E\times { T } _{ 1 }^{ 4 } }{ { T }^{ 4 } } =\dfrac { E\times { 954 }^{ 4 } }{ { 727 }^{ 4 } } =2.96E$

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

In pyrometer , temperature measured is proportional to $\underline{\hspace{0.5in}}$ energy emitted by the body 

  1. light

  2. electric

  3. radiation

  4. All the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stefan- Boltzann law, $j^{ \star  }=\varepsilon \sigma T^{ 4 }$ connects temperature T with thermal radiation or irradiance  $j^{ \star  }$.
Thus measuring the irradiance with pyrometer yields the temperature of the body.

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

Two bodies of same shape and having emissivities 0.1 and 0.9 respectively radiate same energy per second. The ratio of their temperature is :

  1. $\sqrt{3}:1$
  2. $1:\sqrt{3}$
  3. $3:1$
  4. $1:3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{E}{t}=e \sigma A T^4$
From above equation which is Stefan's Law of radiation, it is clear that:
$\dfrac{E _1}{E _2} = \dfrac{{e} _{1}\sigma{T} _{1}^{4}}{{e} _{2}\sigma{T} _{2}^{4}}$

$1 = \dfrac{{0.1}{T} _{1}^{4}}{{0.9}{T} _{2}^{4}}$

$\dfrac{{T} _{1}}{{T} _{2}} = \dfrac{\sqrt{3}}{1} $
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

Two bodies A and B are kept in an evacuated chamber at $27^oC$. The temperature of A and B are $327^oC$ and $427^oC$ respectively. The ratio of rate of loss of heat from A and B will be

  1. 0.25

  2. 0.52

  3. 1.52

  4. 2.52
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The power radiated is directly proportional to fourth power of absolute temperature.
i.e.
$P \propto T^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{T _{1}}{T _{2}})^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{327+273}{427+273})^{4} = 0.53$
Hence the ratio of rate of heat loss = 0.53
Hence option B is correct.

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

The thermal radiation emitted by a body is proportional to $T^{n}$ where $T$ is its absolute temperature. The value of $n$ is exactly $4$ for

  1. a blackbody

  2. all bodies

  3. bodies painted balck only

  4. polished bodies only
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Stefan's Law, rate of thermal radiation is directly proportional to fourth power of temperature of the body.
$Q = \sigma {T}^{4}$

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 $\propto {(300)}^{4}$