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 dynamics - explaining motion system of unit summary of si units system of units

On a hypothetical scale X, the ice point in ${ 40 }^{ \circ  }$ and then steam point is ${ 120 }^{ \circ  }$. For another scale Y the ice point and stem points are ${ -30 }^{ \circ  }$ and ${ 130 }^{ \circ  }$ respectively. If X-reads ${ 50 }^{ \circ  }$ The read of Y is 

  1. $-{ 5 }^{ \circ }$
  2. $-{ 8 }^{ \circ }$
  3. $-{ 10 }^{ \circ }$
  4. $-{ 12 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Scale X: (50 - 40) / (120 - 40) = 10 / 80 = 1/8. Scale Y: (T_y - (-30)) / (130 - (-30)) = (T_y + 30) / 160. Setting them equal: 1/8 = (T_y + 30) / 160. 20 = T_y + 30. T_y = -10.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

By cooling two liquids of equal volume from temperature $60 ^ { \circ } \mathrm { C }$  to $50 ^ { \circ } \mathrm { C }$ in same conditions time required are 324 and 810 sec respectively. If ratio of specific heat of both are 3:4. Then ratio of their. densities (water equivalent of calorimeter is negligible ):

  1. 3 /4

  2. 4 /9

  3. 8 /15

  4. 9 /20
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Newton's law of cooling states that the rate of heat loss is proportional to temperature difference, or time t is proportional to (mc / A). By setting up the ratio of times for identical volume, specific heat, and surface area conditions, the density ratio works out to 8/15.

Multiple choice physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

The resistance of a platinum wire of a platinum resistance thermometer at the ice point is $5 \Omega$ and at steam point is $5.4 \Omega$. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $6.2 \Omega$. Find the temperature of the hot bath.

  1. $300^\circ C$
  2. $30^\circ C$
  3. $3000^\circ C$
  4. $300 \ K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

Resistance of platinum wire at ice point, $R _0=5\,\Omega$

Resistance of platinum wire at hot bath $R _H=6.2\,\Omega$

Temperature at hot bath $T _H=?$

We have,

$R _T=R _0[1+\alpha(T-T _0)]$

$\implies R _{100}=R _0[1+\alpha(T _{100}-T _0)]$

$\implies 5.4=5[1+\alpha(100-0)]$

$\implies \dfrac{5.4}{5}-1=100\alpha$

$\implies \alpha=\dfrac{1}{1250}  \, ^0 C^{-1}$

Also,

$R _H=R _0[1+\alpha (T _H-T _0)]$

That is,

$6.2=5[1+\dfrac{1}{1250}(T _H-0)]$

$\dfrac{6.2}{5}-1=\dfrac{1}{1250}\times T _H$

$\implies T _H=300^0 C$


Multiple choice physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

45 gm of alcohol are needed to completely fill up a weight thermometer at $15^{\circ}C$. Find the weight of alcohol which will overflow when the weight thermometer is heated to $33^{circ}C$.
(Given ${ \gamma  } _{ a }=121\times { 10 }^{ -5 }{ { \circ  } _{ C } }^{ -1 }$

  1. 0.96 gm

  2. 0.9 gm

  3. 1 gm

  4. 2 gm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The overflow volume is given by V * deltaT * (gamma_liquid - gamma_glass). Assuming the weight thermometer expansion is negligible or included in the effective coefficient, the overflow mass is m * deltaT * gamma_alcohol. 45 * (33-15) * 121 * 10^-5 = 45 * 18 * 0.00121 = 0.9801 gm, which rounds to 1 gm.

Multiple choice chemistry materials liquid crystals liquid state the liquid state

The latent heat of vaporisation of water is:

  1. $2.25\times 10^6 J/kg$
  2. $2.25\times 10^8 J/kg$
  3. $2.25\times 10^4 J/kg$
  4. $2.25\times 10^{-6} J/kg$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,250 joules (533 calories) per gram or $2.25 *10^6$ J/kg.

Multiple choice chemistry materials around us and different types of houses materials around us matter and its states different states of matter

At which temperature the density of water will be highest?

  1. $0^o C$
  2. $4^o C$
  3. $80^o C$
  4. $-4^o C$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In ice, the water molecules are in a crystal structure that has a lot of empty space.

When the ice melts to liquid water, the structure collapses and the density of the liquid increases.

At temperatures well above freezing, the molecules move faster and get further apart. The density decreases as temperature increases.

Now, let's cool the water.

As the temperature of warm water decreases, the water molecules slow down and the density increases.

At 4 °C, the clusters start forming.

Cluster formation is the bigger effect, so the density starts to decrease. 
Thus, the density of water is a maximum at 4 °C.

Option (B) is correct here.
Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

An athelete is given 100 g of glucose $(C _6H _{12}O _6)$ of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, Determine the weight of water he would need to perspire. (The enthalpy of evaporation of water is 44 kJ/mole.)

  1. 319 gm

  2. 323 gm

  3. 342 gm

  4. 312
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 Net amount of energy given to athlete $= 1560 kJ$

$1560 × 50/100$ Energy lost in an event $= 780 kJ$
 Energy left out $= 1560 – 780 kJ = 780 kJ$
 Now, consider the evaporation of water $H _2O(l) \rightarrow H _2O(g)$; $\Delta H = 44 kJ mole^{–1}$
 Thus, for consumption of $44 kJ$ of energy the amount of water evaporated $=1mole= 18 g$ 
For consumption of $780 kJ$ of energy the amount of water to be evaporated $ \frac {18 \times  780}{44} = 319·09 g$

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

A piston exerting a pressure of 1.0 atmosphere rests on the surface of water at $100^{\circ}C$. The pressure is reduced to smaller extent and as a result 10 g of water evaporates and absorbs 22.2 kJ of heat. The change in internal energy is:

  1. 18.24 kJ

  2. 20.477 kJ

  3. 22.05 kJ

  4. 23.923 kJ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relationship between the enthalpy change and the change in the internal energy is
 $\Delta H=\Delta U+P\Delta V=\Delta U+\Delta n _gRT$
$\Delta U=\Delta H-\Delta n _gRT$
Substitute values in the above expression.
$\Delta U=\displaystyle 22.2-\frac{10}{18}\times 8.314\times 10^{-3}\times 373$
                       $=20.477 :kJ$
The change in internal energy is 20.477 kJ

Multiple choice evs water in our life where do we get water from? why conserve water uses of water

Water occurs in frozen state over the land is

  1. 0.75%

  2. 1.97%

  3. 2.1%

  4. 2.5%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • 70% of the Earth's surface is water. Of this 70%, 98% is salt water, leaving 2% as fresh water. Of the 2% that is fresh, about 90% is frozen(i.e 1.97%of the total water). This frozen water is locked up in the Antarctic ice sheets and glaciers on the Alps, etc.
  • Hence Water occurs in the frozen state over the land is 1.97%.
  • So, the correct answer is '1.97%'.
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.