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 chemistry mix and separate dry methods of separation process of separating components of a mixture method of separation

The freezing point of pure water is ____________.

  1. 100 degree centigrade

  2. 50 degree centigrade

  3. 20 degree centigrade

  4. zero degree centigrade
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Freezing point is the temperature where a liquid converts into solid. Water converts into ice at zero degrees centigrade. So degree zero centigrade is the freezing point of water. Hence option $D$ is correct.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

A block of ice of density $0.9\ gm\ cc^{-1}$ and of volume $100\ cc$ floats in water. Then volume of ice inside the water 

  1. $10\ cc$
  2. $90\ cc$
  3. $50\ cc$
  4. $9\ cc$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to the law of flotation, the weight of the floating object equals the weight of the displaced liquid. Therefore, Density_ice * Volume_ice = Density_water * Volume_submerged. Given 0.9 * 100 = 1 * Volume_submerged, the submerged volume is 90 cc.

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

150 g of ice is mixed with 100 g of water at temperature $80^oC$. The latent heat of ice is 80 ca/g and the specific heat of water is $1 cal/g-^oC$. Assuming no heat loss to the environment, the amount of ice which does not melt is

  1. 100 g

  2. 0 g

  3. 150 g

  4. 50 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat given by water $=100\times 1\times 800=8000 cal$
Heat taken by ice $=8000 cal=m\times 80$
$m=100 gm$
So amount of ice which does not melt $=150-100=50 gm$.

Multiple choice physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

A weight thermometer contains 52 gms of a liquid at 10$^{0}$C. When it is heated to 110$^{0}$C, 2 gm of the liquid is expelled. The coefficient of real expansion of the liquid is [$\alpha $ of glass is $9\times 10^{-6/0}C$]

  1. $27\times 10^{-6/0}C$
  2. $412\times 10^{-6/0}C$
  3. $373\times 10^{-6/0}C$

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$V _{expelled} = V _0(\gamma _l -\gamma _g)\Delta T$
$ \Rightarrow  2 = 52 \times ( \gamma _l -3 \times \alpha _g)\times(110-10)$
$ \Rightarrow 2 =52 \times (\gamma _l - 3 \times  9 \times 10^{-6} ) \times 100$
$ \Rightarrow \gamma _l = \frac{2}{52} \times 10^{-2} + 27 \times 10^{-6}=385 \times 10^{-6} + 27 \times 10^{-6}=412 \times 10^{-6}$

Multiple choice physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

When a liquid in a glass vessel is heated, its apparent expansion is $10.30\times 10^{-4}/^o$ C. Same liquid when heated in a metallic vessel, its apparent expansion is $10.06\times 10^{-4}/^o$ C. The coefficient of linear expansion of metal is :

($\alpha   _{glass}=9\times 10^{-6}/^o$ C)

  1. $51\times 10^{-6}/^oC$
  2. $43\times 10^{-6}/^oC$
  3. $25\times 10^{-6}/^oC$
  4. $17\times 10^{-6}/^oC$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${\gamma} _r={{\gamma} _{a1}} +{\gamma} _{glass}={{\gamma} _{a2}} +{\gamma} _{metal}$  ......(1)
where ${{\gamma} _{a1}}$ is the apparent  expansion coefficients when the the liquid is kept in the glass container and ${{\gamma} _{a2}}$is the apparent expansion coefficients when the the liquid is kept in the metal container
Given,
${{\gamma} _{a1}}=10.30 \times {10}^{-4}$
${{\gamma} _{a2}}=10.06 \times {10}^{-4}$
${\alpha} _{glass}=9 \times {10}^{-6}$
${\gamma} _{glass}=3{\alpha} _{glass}=27 \times {10}^{-6}$
${\gamma} _{metal}=3{\alpha} _{metal}$


Substituting the values in (1)
$10.30 \times {10}^{-4} + 27 \times {10}^{-6}=10.06 \times {10}^{-4} + 3{\alpha} _{metal}$
$ \Rightarrow 24 \times {10}^{-6}  + 27 \times{10}^{-6} = 3{\alpha} _{metal}$
$ \Rightarrow {\alpha} _{metal}=(\dfrac{51}{3}){10}^{-6}=17 \times {10}^{-6}$

Multiple choice chemistry element, compound and mixture different types of solutions various mixtures introduction to solutions

Which of the following is used as antifreeze on cars for cooling the engine?

  1. $10\%$ (V/V) ethanol solution in water
  2. $35\%$ (V/V) ethanol solution in glycol
  3. $45\%$ (V/V) ethanol solution in glycol
  4. $20\%$ (V/V) ethanol solution in water
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ethanol-water mixtures are commonly used as antifreeze in cooling systems because they lower the freezing point of the coolant.

Multiple choice geography precipitation elements of weather and climate :rainfall and air pressure forms of precipitation moisture in atmosphere

 As per Celsius scale, the freezing point of water is ________ degree and the boiling point is ___  degree Celsius.

  1. Zero, 100

  2. Zero, 200

  3. Zero, 300

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

Celsius is the scale which measures temperature. It was developed by Swedish astronomer Andres Celsius in 1742. Freezing point of the water is the temperature at which water starts converting into ice. As per the Celsius scale, zero degrees is the freezing point of the water. The boiling point is the temperature at which water starts converting into vapors. As per Celsius scale, 100 degrees is the boiling point of the water.

Multiple choice evs weather, climate and adaptation of animals to different climates climate and weather weather, climate and adaptations of animals to climate atmosphere and its components

If one litre of water at $\displaystyle { 25 }^{ \circ  }C$ is mixed with one litre of water at $\displaystyle { 45 }^{ \circ  }C$ the temperature of the water will be

  1. $\displaystyle { 70 }^{ \circ }C$
  2. $\displaystyle { 20 }^{ \circ }C$
  3. $\displaystyle { <25 }^{ \circ }C$
  4. $\displaystyle >{ 25 }^{ \circ }C\& <{ 45 }^{ \circ }C$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The temperature of the mixture would be $\displaystyle >{ 25 }^{ \circ  }C&amp; <{ 45 }^{ \circ  }C$, as there would be transfer of heat in between  water molecules from higher temperature to lower temperature.

Multiple choice physics measurements and units kinds of units fundamental quantities standardized measurement

The specific heat capacity of water in SI unit is :

  1. $4.2 Jg^{-1}K^{-1}$
  2. $42 Jg^{-1}K^{-1}$
  3. $420 J kg^{-1}K^{-1}$
  4. $4200 J kg^{-1}K^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The specific heat capacity  of a substance is the heat needed to raise

the temperature of 1 kg of the substance by 1K (or by ${1}^{0}C$).
The specific heat capacity of water is $4200 J{kg}^{-1}{K}^{-1}$.

Multiple choice physics energy and its forms idea of energy introduction to work and energy work and energy

Define the term calorie. How is it related to joule ?

  1. The amount of energy requires to raise the temperature of 1 gram of water by 1° Celsius, 1 Cal = 4.18 J
    <a href="http://scienceworld.wolfram.com/physics/Celsius.html&quot;&gt;&lt;/a>

  2. The amount of energy requires to raise the temperature of 1 gram of water by 1 Kelvin, 4.18 Cal = 1 J

  3. The amount of energy requires to raise the temperature of 1 gram of water by 1 Fahrenheit, 4.18 Cal = 1 J

  4. The amount of energy requires to raise the temperature of 1 gram of ether by 1° Celsius, 1 Cal = 4.18 J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The term calorie can be defined as the energy needed to raise the temperature of 1 gram of water by $ 1$ °C. 
$1 \ cal = 4.18 \ J$