Physics

Thermal Properties of Matter

274 Questions

Thermal properties of matter examine how substances react to heat, covering concepts like temperature, specific heat, and phase changes. This topic is regularly tested in the physics and general science sections of competitive exams. Use this collection to solve numericals on heat transfer, thermal equilibrium, and melting points.

Heat transfer and mixingMelting and freezing pointsThermal expansion and densityTriple point of waterCarnot cycle applications

Thermal Properties of Matter Questions

Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

 Water cannot be used as a liquid in a thermometer because it has:

  1. higher freezing point and lower boiling point than other thermometric liquids

  2. lower freezing point and higher boiling point than other thermometric liquids

  3. low specific heat capacity

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

Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids .  If water is used in a thermometer , it will start phase change at $0^{o}C$ and $100^{o}C$ and will not measure temperature , out of this range . This range is very small as compared  to other liquids as mercury , having freezing point about $-39^{o}C$ and boiling point $356^{o}C$ .

Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

One litre of water at $30^{o}C$ is mixed with one litre of water at $50^{o}C$. The temperature of the mixture will be

  1. 80C

  2. more than 50C but less than 80C

  3. 20C

  4. between 30C and 50C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the first volume of water be V1. Let the second volume of water be V2. Let the initial temperature of the first volume of water be θ1. Let the initial temperature of the second volume of water be θ2.

Let the energy required to heat a unit volume of water by unit temperature by C.

So, if the final temperature is θfinal, we have

heat gained by the first volume of water

$=V _1×C×(θ _{final}−θ _1)$

and

heat lost by the second volume of water

$=V _2×C×(θ _2−θ _{final}).$

By the law of conservation of energy,

heat lost = heat gained

∴$V _1×C×(θ _{final}−θ _1)=V _2×C×(θ _2−θ _{final})$

∴$V _1θ _{final}−V _1θ _1=V _2θ _2−V _2θ _{final}$

∴$θ _{final}=V _1θ _1+V _2θ _2V _1+V _2$

Substituting the given values,

$θ _{final}=\dfrac{30+50}{2}=40^∘C$.

The answer is option D
Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

If a thermometer reads freezing point of water as $20^{\circ}C$ and boiling point at $150^{\circ}C$ how much thermometer read when the actual temperature is $60^{\circ}C$.

  1. $98^{\circ}C$
  2. $110^{\circ}C$
  3. $40^{\circ}C$
  4. $60^{\circ}C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We let the unknown degree scale be called ${x}^{\circ}$
Now we know that the freezing point$={20}^{\circ}x$
The boiling point$={150}^{\circ}x$
This also means that ${20}^{\circ}x= {0}^{\circ}$ C($\because$ on a Celcius Scale, the freezing point is ${0}^{\circ}$)
And ${150}^{\circ}x={100}^{\circ}$C (On a  Celcius Scale, boiling point is ${100}^{\circ}$)
Also ${(150-20)}^{\circ}x= {100}^{\circ}$C ($\because$ on a Celcius Scale ${100}^{\circ}-{0}^{\circ}={100}^{\circ}$)
So, if ${100}^{\circ}$C$={130}^{\circ}x$
Then ${1}^{\circ}$C$={1.3}^{\circ}x$
To go from $C$ degree  to $x$, we need to use:
$x=1.3C+20$  
And $C=\dfrac{(x-20)}{1.3}$
So if the temperature given is ${60}^{\circ}$C, then we calculate $x$ as:
$C=\dfrac{(x-20)}{1.3}$  
$\Rightarrow 60=\dfrac{(X-20)}{1.3}$  
$\Rightarrow 60\times 1.3=x-20$  
$\Rightarrow 78=x-20$  
$\Rightarrow x= {98}^{\circ}$  
So ${60}^{\circ}$C will be equal to ${98}^{\circ}$ 
Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

The freezing point on a thermometer is marked as  $ 20^o$ and the boiling point as $150^o$. A temperature of $(60^o C)$on this thermometer will be read as:

  1. $40^o$
  2. $65^o$
  3. $98^o$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} By\, \, u\sin  g\, the\, \, formula \ \dfrac { { { T _{ t } }-{ T _{ fP } } } }{ { { T _{ BP } }-{ T _{ fp } } } } =\dfrac { { T-0 } }{ { 100-0 } }  \ \dfrac { { { T _{ t } }-20 } }{ { 150-20 } } =\dfrac { { 60 } }{ { 100 } }  \ { T _{ t } }-20=\dfrac { { 60\times 130 } }{ { 100 } }  \ { T _{ t } }=78+20 \ ={ 98^{ \circ  } } \ Hence,\, the\, option\, C\, \, is\, the\, correct\, answer. \end{array}\ $

Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

The boiling point of mercury is 367$^o$C. A mercury thermometer can be used to measure a temperature of 500$^o$C;

  1. by filling the space above mercury with oxygen at high pressure

  2. by filling the space above mercury with nitrogen at low pressure

  3. by filling the space above mercury with nitrogen at high pressure

  4. by keeping the space above mercury as vacuum
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mercury thermometer can be used above its boiling point by filling up the space above mercury with nitrogen at high pressure because as pressure increases, boiling point also increases.

Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

The steam point and the ice point of a mercury thermometer are marked $80^0$ and $20^0.$  what will be the temperature in centigrade mercury scale when this thermometer reads $32^0$

  1. $20 ^ { 0 } \mathrm { C }$
  2. $5 ^ { 0 } \mathrm { C }$
  3. $10 ^ { 0 } \mathrm { C }$
  4. $26 ^ { 0 } \mathrm { C }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(x-20)/(80-20) =(C-0)/(100-0)$
$(x-20)/60=(C-0)/100$

Now, given question$,$
$x=32 $
then$ ,$
$(32-20)/60=(C-0)/100$
$12/60=C/100$
$C=20^0$
hence in centigrade thermometer $20^0C$
Hence,
option $(A)$ is correct answer.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

Calculate the least amount of work that must be done to freeze one gram of water at $0^0C$ by means of the refrigerator.The temperature of the surrounding is $27^0C$.How much heat is passed on the surrounding in this process? Latent heat of fusion $L=80\ cal/g$.

  1. $87.91\ cal$
  2. $97.91\ cal$
  3. $88.95\ cal$
  4. $89.95\ cal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$L=80 \dfrac{cal}{g}$

$m=1g$
$T _1=27°=300K$
$T _2=0°=273K$
Least work done $W=L \times m \times \dfrac{T _1}{T _2}$
=$80 \times 1 \times \dfrac{300}{273}$
=$87.912 cal$

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

What would be the efficiency of a Carnot engine operating with boiling water as one reservoir and a freezing mixture of ice and water as the other reservoir?

  1. $27$ %
  2. $77$ %
  3. $20$ %
  4. $67$ %
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The boiling point of water is $100 ^oC$ and freezing point of water is $0^o C$

Thus, $T _h=100+273=373 K$ and $T _c=0+273=273 K$
The efficiency of a Carnot engine , $\eta=(1-\dfrac{T _c}{T _h})\times 100=[1-{373}/{273}]\times 100=26.8\sim 27$ %

Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

If, melting and boiling point of a liquid is $-{50}^{o}C$ and ${150}^{o}C$ respectively, and on the basis of this liquid a thermometer is formed with ${0}^{o}L$ representing the melting point as the minimum temperature on the scale and ${100}^{o}L$ representing the boiling point as the maximum temperature on the scale, find the melting and boiling point on this thermometer

  1. $0$ and ${100}^{o}L$ respectively
  2. ${30}^{o}L$ and ${80}^{o}L$ respectively
  3. ${20}^{o}L$ and ${70}^{o}L$ respectively
  4. ${25}^{o}L$ and ${75}^{o}L$ respectively
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

The water can be made to boil even at $0$ if the pressure of surrounding is:

  1. 56 cm of Hg

  2. 5 cm of Hg

  3. 0.1 cm of Hg

  4. 4.6 mm of Hg
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The water can be made to boil even at $0^\circ C$. 
As we know, the ideal equation for gases is $pV=nRT$. Now, if $V$ and $n$ are taken constants, $p=cT$, where $c$ is a constant.
So pressure is directly proportional to the temperature.
Water boils at $100^\circ C$ or $373K$ at $1atm$ pressure. So, to obtain boiling at $0^\circ C$ or $273K$, you need to reduce the pressure to $\dfrac {273\times 1}{373} atm = 0.732 atm=56\text{cm of Hg}$(approx).
Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

When 1 g of water at $100^\circ C$ gets converted into steam at the same temperature, the change in volume is approximately

  1. 1 cc

  2. 1000 cc

  3. 1500 cc

  4. 1670 cc
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

During change of state mass does not change.

Mass of water = 1gm
So mass of steam is also= 1gm
Density of steam =$0.0006gm/cm^3$
Initail volume of water =1$cm^3$(Density =$1gm/cm^3$) ($1=\dfrac { 1 }{ V } $) $,v=1cm^3$
Later volume of water converted to steam =$\dfrac { mass\quad of\quad steam }{ Density\quad of\quad steam } $
$=\dfrac { 1 }{ 0.0006 } $
$=\dfrac { 10000 }{ 6 } { cm }^{ 3 }$
$= 1670cm^3$
Change of volume = $1670cm^3$

Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

When 1 g of ice melts at $0^\circ C$

  1. 80 cal of heat is liberaed

  2. 80 cal of heat is absorbed

  3. no heat is required

  4. 160 cal heat is absorbed
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For melting of $1g$ of ice $80cal$ of heat is required.
As latent heat of fusion for water is the heat required or released when $1g$ of ice melts into water or water changes into ice.

Latent heat of fusion for water $=80cal/g$