Physics

Thermal Properties of Matter

302 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

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
A Correct answer
Explanation

By definition, the new thermometer scale is constructed by setting the melting point of the liquid (-50 deg C) as 0 deg L and the boiling point (150 deg C) as 100 deg L. Therefore, by construction, the melting and boiling points on this new scale are 0 and 100 deg L respectively.

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 $=80cal/g$
Multiple choice physics the kinetic model of matter force and kinetic theory change of states phase change

Water in a lake is changing into ice at $0^{o}C$. When the atmospheric temperature is $10^{o}C$. If the time taken for $1\ cm$ thick ice layer to be formed is $7$ hour, the time required for the thickness of ice to increase from $1\ cm $ to $2\ cm$ is

  1. $7\ hour$
  2. $14\ hour$
  3. $<7\ hour$
  4. $>14\ hour$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The rate of ice formation is proportional to the thickness already formed because the heat must conduct through the existing ice layer. If it takes 7 hours to form the first 1 cm, the next 1 cm (from 1 cm to 2 cm) takes three times the duration of the first interval in some models, but standard physics problems of this type often follow the t proportional to x^2 rule. Specifically, t1 = k(x1^2), t2 = k(x2^2 - x1^2). Here, 7 = k(1^2), so k = 7. The time for the next cm is 7(2^2 - 1^2) = 7(3) = 21 hours. Given the options, 14 hours is often cited in simplified textbooks, but mathematically it should be 21.

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

How much heat energy is gained when 5 kg of water at  $20 ^ { \circ } \mathrm { C }$ is brought to its boiling point?

  1. 1680 KJ

  2. 1700 KJ

  3. 1720 KJ

  4. 1740 KJ
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Q = mc\Delta \theta $

$ = 5 \times \left( {1000 \times 4.2} \right) \times \left( {100 - 20} \right)$
$ = 1680 \times {10^3}J$
$ = 1680\,KJ$
Hence,
option $(A)$ is correct answer.

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

Given: When 1g of water at $100^∘C$ gets converted into steam at the same temperature, 

To find the change in volume
Solution:
We know, 
Density of water, $\rho=1000kg/m^3=1 g/cm^3$
Density of steam, $\rho=0.6kg/m^3=0.0006 g/cm^3$
Mass of the water and mass of water, $m=1g$
Volume of water is
$m=\rho V\\implies V=\dfrac m\rho=\dfrac 11\\implies V=1cc$
Volume of steam, 
$V'=\dfrac m\rho=\dfrac 1{0.0006}=1666.7cc\approx1667cc$
The change in volume, 
$\Delta V=V'-V\\implies \Delta V=1667-1=1666cc$