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 measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

How much heat is required to raise the temperature of $150 g$ of iron from ${ 20 }^{ \circ  }C$ to ${ 25 }^{ \circ  }C$?

  1. $350 J$
  2. $345 J$
  3. $360 J$
  4. $330 J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given ,  $m=150g ,  \theta _{1}=20^{0}C , \theta _{2}=25^{0}C$

We have , specific heat of iron $c=0.46J/g-^{o}C$
Now , heat required to raise the temperature of iron is given by the definition of specific heat c ,
                    $Q=mc\Delta \theta=mc(\theta _{2}-\theta _{1})$
or                 $Q=150\times0.46\times(25-20)=345J$

Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

A copper ball of mass $100gm$ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100gm$, filled with $170gm$ of water at room temperature. Subsequently the temperature of the system is found to b4 ${75}^{o}$. $T$ is given by then (Given: room temperature $={30}^{o}C$, specific heat of copper $=0.1cal/gm _{  }^{ o }{ C }\quad $)

  1. ${ 825 }^{ o }C$
  2. ${ 800 }^{ o }C$
  3. ${ 885 }^{ o }C$
  4. ${ 1250 }^{ o }C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Final temperature of celomiter and its constant is given as
$To=75^o C$
$\Rightarrow \ 100\times 0.1\times (75-T)+100\times 0.1(75-30)+1.70 \times 1\times (75.32)$
$\therefore \ T=885^oC$
Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

400 g of vegetable oil of specific heat capacity $1.98 J g^{-1} {\;}^oC^{-1}$ is cooled from $100^oC$. Find the final temperature, if the heat energy given out by oil is 47376 J.

  1. $30.2^oC$
  2. $40.2^oC$
  3. $50.2^oC$
  4. $43.2^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$m=400 g, C=1.98 Jg^{-1} {\;}^oC^{-1}$
$T _1=100^oC, T _2=?$
Fall in temperature
$\Delta T=(100-x)$
Heat energy given out by oil
$=47376 J$
According to formula $Q=m.C.\Delta T$
$\Rightarrow 47376=400\times 1.98 (100-x)$
$\Rightarrow 100-x=\frac {47376}{400\times 1.98}=59.8$
$\Rightarrow x=100-59.8=40.2^oC$
$\therefore$ Final temperature of oil $=40.2^oC$.

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

What quantity of heat would be given out by 200 gm of copper in cooling from $80^oC$ to $20^oC$ (Specific heat of copper $=0.09 cal g^{-1} {\;}^oC^{-1})$?

  1. 1080 cal

  2. 1000 cal

  3. 1500 cal

  4. 1100 cal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m=200 g, C _{copper}=0.09 cal g^{-1} {\;}^oC^{-1}$
$\Delta T=T _1-T _2=80-20=60^oC$
$\therefore Q=mC \Delta T=200\times 0.09\times 60$
$=1080 cal$.

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

A man would feel iron and wooden balls equally cold or hot at

  1. $98.6^oC$
  2. $98.6^oF$
  3. $198.6^oF$
  4. $198.6^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A man would find them equally cold or hot only when the heat flowing in or out of them is equal.
But since there would be the same temperature difference between the object and the man.
The heat flow would only be the same, when there is no heat flow. i.e. the body's temperature should be equal to the body temperature of the man. Which is $98.6 ^{\circ} F$

Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

The specific heat for substance $A$ is twice the specific heat of substance $B$. The same mass of each substance is allowed to gain $50$ Joules of heat energy. As a result of the heating process:

  1. the temperature of $A$ rises twice as much as $B$
  2. the temperature of $A$ rises four times as much as $B$
  3. the temperature of $B$ rises twice as much as $A$
  4. the temperature of $B$ rises four times as much as $A$
  5. the temperature of both $B$ and $A$ rise the same amount
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let specific heat of substance $A$  is $2c$ and specific heat of substance $B$ is $c$ , 

we have ,  heat given  $Q=mc\Delta t$, where  $\Delta t $ denotes change in temperature , 
so , for substance $A$, $Q=m\times 2c\Delta t _{A}$ 
or $\Delta t _{A}=Q/2mc$ .........eq1
For substance $B$ ,   $Q=mc\Delta t _{B}$   
$\Delta t _{B}=Q/mc$ ...................eq2
by eq1 and eq2,
$2\Delta t _{A}=\Delta t _{B}$     

Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

An aluminium block of 2m mass and an iron block of m mass,each absorbs the same amount of heat, and both blocks remain solid. If the specific heat of aluminium is twice the specific heat of iron, then find out the correct statement?

  1. The increase in temperature of the aluminum block is twice the increase in temperature of the iron block

  2. The increase in temperature of the aluminum block is four times the increase in temperature of the iron block

  3. The increase in temperature of the aluminum block is the same as increase in temperature of the iron block

  4. The increase in temperature of the iron block is twice the increase in temperature of the aluminum block

  5. The increase in temperature of the iron block is four times the increase in temperature of the aluminum block
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The heat required to rise the temperature of body of mass $m$ of specific heat $s$ by $\Delta T=H=ms\Delta T$

Thus for same amount of heat, the rise in temperature ration of aluminium and iron is $\dfrac{\Delta T _{Al}}{\Delta T _{Fe}}=\dfrac{m _{iron}s _{iron}}{m _{aluminium}s _{aluminium}}=\dfrac{1}{4}$
Thus the rise in temperature of the iron block is four times the increase in temperature of the aluminum block

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

A mass of stainless steel spoon is 0.04 kg and specific heat is $0.50 kJ/kg \times ^oC$. Then calculate the heat which is required to raise the temperature $20^oC$ to $50^oC$ of the spoon.

  1. 200 J

  2. 400 J

  3. 600 J

  4. 800 J

  5. 1,000 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The required heat, $Q=ms\Delta T=0.04\times (0.50\times 10^3)\times(50-20)=600 J$

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

The original temperature of a black body is $727^\circ C$. Calculate temperature at which total radiant energy from this black body becomes double:

  1. $971K$
  2. $1189K$
  3. $2001K$
  4. $1458K$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Rediant Energy = \sigma T^2$

$Energy = \sigma (1000)^4$
$E _2 = 2 E _1$
$Then$
$\sigma T _2 ^{4} = 2 \times \sigma (1000)^4$
$T _2 = 2^\frac{1}{4} \times1000$
$T _2 = 1189 K$

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

Temp. of black body is $3000K$ when black body cools. Then change in wavelength $\Delta \lambda=9$ micron corresponding to maximum energy density. Now temp. of black body is:

  1. $300K$
  2. $2700K$
  3. $270K$
  4. $1800K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Wien's displacement law, lambda_max * T = constant. Initially, T1 = 3000K. If lambda_max changes by 9 microns, we need the initial lambda_max. Assuming lambda_max1 = 1 micron (typical for 3000K), then lambda_max2 = 10 microns. T2 = (1/10) * 3000 = 300K.

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

Initially a black body at absolute temperature $T$ is kept inside a closed chamber at absolute temperature $T _{o}$. Now the chamber is slightly opened to allow sun rays to enter. It is observed that temperatures $T$ and $T _{o}$ remains constant.Which of the following statement is/are true?

  1. The rate of emission of energy from the black body remains the same

  2. The rate of emission of energy from the black body increases

  3. The rate of absorption of energy by the black body increases.

  4. The energy radiated by the black body equals the energy absorbed by it
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

It is given that the absolute temperatures of both the black body and the surroundings are constant with time, even after sunlight(radiation) is incident on it.

  • When a body absorbs radiation, its temperature increases
  • When a body emits radiation, its temperature decreases
Also the sun, being a source of infinite radiation(very large source of radiation).
We infer from this that the incident radiation should be of constant magnitude.
And if the temperature of the black body is a constant, that means it's emission and absorption of radiation are matched and equal. The absorption is of constant magnitude, because the sun's radiation is of constant value. Hence the emission is of constant value also and is equal to the absorption. The options follow.

Multiple choice physics energy production perfectly black body black-body radiation black body radiation

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

  1. $ 0.8\ e\sigma AT^4$
  2. $ 0.4\ e\sigma AT^4$
  3. $ 0.6\ e\sigma AT^4$
  4. $ 1.0\ e\sigma AT^4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to Stefan's Boltzman law, the thermal energy radiated by a black body radiator per second per unit area is proportional to fourth power of the absolute temperature and is given by
$\dfrac{P}{A} = \sigma T^4$ ..............(1)
For the hot bodies other than black body radiator equation (1) becomes,
$\dfrac{P}{A} = e \sigma T^4$
$P = e \sigma A T^4$ .................(2)
where, $e$ is the emissivity of the body.
Now, when such hot body is kept inside a perfectly black body, the total thermal radiation is the sum of emitted radiations (in open) and the part of incident radiations reflected from the walls of the perfectly black body. This will give black body radiations, hence the total radiations emitted by the body will be,
$P = 1.0 e \sigma A T^4$.
Multiple choice physics energy production perfectly black body black-body radiation black body radiation

Emissivity of a perfect black body is

  1. always $0$.
  2. always $1$.
  3. between $0$ and $1$.
  4. always $>1$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Emissivity of a perfect black body is always 1.
The best absorber is defined as the object which can absorb all the electromagnetic radiations falling upon it. The black body is not only a perfect absorber but it is also the best in emitting radiation. Also, a black bosy in thermal equlibrium has emissivity, $\epsilon=1$
Multiple choice physics energy production perfectly black body black-body radiation black body radiation

The Wien's displacement law for a black body is
($T$ is the absolute temperature in $K$
$b$ is a constant of proportionality 
$e$ is the emissivity of the black body)

  1. $\lambda _{max} T = b$
  2. $\lambda _{max} T = e$
  3. $\lambda _{max} b = T$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
According to wein's displacement law there is inverse relation between $\lambda _{max}$ of radiation emitted by black body and its temperature (absolute)
$\lambda _{max}\; \alpha \; \cfrac{1}{T} \Rightarrow =b\cfrac{1}{T} \Rightarrow \lambda _{max} T=b$