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 energy transfers heat and heat transfer heat energy transfer transfer of heat

A geyser heats water flowing at the rate of 4 litre per minutes from 30 C to 85 C. If the geyser operates on gas burner then the amount of heat used per minute is :

  1. $9.24\times { 10 }^{ 5 }$J
  2. $6.24\times { 10 }^{ 7 }$J
  3. $9.24\times { 10 }^{ 7 }$J
  4. $6.24\times { 10 }^{ 5 }$J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For one minute

$H=mC\theta$
$=4kg\times 4200\times55\=924000J\=9.24\times 10^5j$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

If 10 g of a sample of water contains 16.2 g of  $ Ca(HCO _{3}) _{2} $, then the hardness of water is:

  1. 200 ppm

  2. 300 ppm

  3. 60 ppm

  4. 100 ppm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If density of water $=1\text{ }g/ml$

$\therefore 10\text{ }ml$ water contains $16.2\text{ }mg$ $Ca(HCO _3) _2$.
$\therefore 1L$ water contains $16.2\times 10^2\text{ }mg$ $Ca(HCO _3) _2$.
Equivalent of $CaCO _3=16.2\times 10^2\times \cfrac{100}{162}=100$
$\therefore $ Hardness in $ppm=100ppm$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

The hardness of water due to $ HCO^{2-} _{3} $ is 122pm. Select the correct statement(s).

  1. The hardness of water in terms of $ CaCO _{3} $ is 200 ppm
  2. The hardness of water in terms of $ CaCO _{3} $ is 100 ppm
  3. The hardness of water in terms of $ CaCO _{3} $ is 222 ppm
  4. The hardness of water in terms of $ CaCO _{3} $ is 95 ppm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The hardness of water due to ${ HCO } _{ 3 }^{ 21 }$ is $122ppm$. Because the hardness of water in terms of ${ CaCO } _{ 3 }$ is $200$ $ppm$.

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

The density of water is not same at all temperatures because of its anomalous expansion. The density is maximum at:

  1. $0^oC$
  2. $4^oC$
  3. $40^oC$
  4. $100^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Due to anomalous expansion of water the density of water is not same at all temperatures. When water is cooled from room temperature it first contracts in volume and becomes increasingly dense as do other liquids but at $4^oC$ water reaches maximum density.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A mass of $50g$ of water in a closed vessel with surroundings at a constant temperature takes $2$ minutes to cool from ${30}^{o}C$ to ${25}^{o}C$. A mass of $100g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from ${30}^{o}C$ to ${25}^{o}C$. The specific heat of the liquid is : (The water equivalent of the vessel is $30g$)

  1. $2.0kcal/kg$
  2. $7kcal/g$
  3. $3kcal/kg$
  4. $0.5kcal/kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from $30^{\circ} C$ to $25^{\circ} C$) is same 2 minutes.

$\therefore \left(\dfrac{dQ}{dt} \right) _{water} = \left(\dfrac{dQ}{dt} \right) _{liquid}$

Or $\dfrac{(m _w c _w + W) \Delta T _1}{t _1}  = \dfrac{(m _l c _l + W) \Delta T _2}{t _2}$

$\therefore \Delta T _1=\Delta T _2, \, t _1=t _2$

(w = water equivalent of the vessel)

or $m _w c _w = m _l c _l$

$\therefore$ specific heat of liquid,

$C _l = \dfrac{m _w c _w}{m _l} = \dfrac{50 \times 1}{100} = 0.5 kcal/kg$
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

'n' number of liquids of masses m,2m,3m,4m, .......... having specific heats S, 2S, 3S, 4S, ...... at temperatures t, 2t, 3t, 4t, ........ are mixed. The resultant temperature of the mixture is

  1. $\frac{3n}{2n+1} t$
  2. $\frac{2n(n+1)}{3(2n+1)} t$
  3. $\frac{3n(n+1)}{2(2n+1)} t$
  4. $\frac{3n(n+1)}{(2n+1)} t$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a weighted average problem for temperature. The total heat gained/lost must sum to zero. The calculation involves summing the products of mass, specific heat, and temperature for each component.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

$1$ $\mathrm { g }$ of a steam at $100 ^ { \circ } \mathrm { C }$ melts how much ice at $\mathrm { CC }$ (Latent heat of ice $= 80$ cal/gm and latent heat of steam $ = 540 \mathrm { cal/gm }$



  1. $1 gm$
  2. $2gm$
  3. $4 gm$
  4. $8 gm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat released by 1g of steam at 100C condensing to water at 100C is 540 cal. Heat released by 1g of water cooling from 100C to 0C is 100 cal. Total heat = 640 cal. Heat required to melt m grams of ice at 0C is m * 80 cal/g. Thus, m = 640 / 80 = 8g.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The quantity of heat (in J) required to raise the temperature of $1.0\, kg$ of ethanol from $293.45\, K$ to the boiling point and then change the liquid to vapor at that temperature is closest to 
[Given : Boiling point of ethanol $351.45\, K$
              Specific heat capacity of liquid ethanol $2.44\, J\, g^{-1}\, K^{-1}$
               Latent heat of vaporization of ethanol $855\, J \, g^{-1}$]

  1. $1.42\, \times 10^2$
  2. $9.97\, \times 10^2$
  3. $1.42\, \times 10^5$
  4. $9.97\, \times 10^5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat to raise temperature: Q1 = m * c * delta T = 1000g * 2.44 J/gK * (351.45 - 293.45)K = 1000 * 2.44 * 58 = 141,520 J. Heat to vaporize: Q2 = m * L = 1000g * 855 J/g = 855,000 J. Total Q = 141,520 + 855,000 = 996,520 J, which is approximately 9.97 * 10^5 J.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

When water is heated from $0^{\circ}C$ to $4^{\circ}C$ and $C _{p}$ and $C _{v}$ are its specific heated at constant pressure and constant volume respectively, then:

  1. $C _{p} >C _{v}$
  2. $C _{p}< C _{v}$
  3. $C _{p}=C _{v}$
  4. $C _{p}-C _{v}=R$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Water has highest density at $4^{\circ}C$. This changes its properties from other simple fluids.

When water is heated from $0^{\circ}C$ to $4^{\circ}C$, the volume of liquid decreases.
Thus for this transition, $P\Delta V$ is negative.
$\int C _PdT=\int C _VdT+P\Delta V$
$\implies C _P<C _V$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A mass of $50$ g of a certain metal at $150^0C$ is immersed in $100$ g of water at $11^0C.$ The final temperature is $20^0C$. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{-1}K^{-1}$.

  1. $0.682 J g^{-1}K^{-1}$
  2. $582 J g^{-1}K^{-1}$
  3. $0.582 J g^{-1}K^{-1}$
  4. $0.0582 J g^{-1}K^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass if the solid , $m _s=50 g$

Initial temperature of the solid, $t _s=150^0 C=150+273=423 k$
Mass of water, $m _w=100 g$
Temperature of the water, $t _w=11^0 C=11+273=293 k$
According to  principle of calorie meter,
Heat gained by water =Heat lost by the solid 
$\begin{array}{l} \therefore { m _{ w } }{ C _{ w } }\left( { t-{ t _{ w } } } \right) ={ m _{ s } }{ C _{ s } }\left( { { t _{ s } }-t } \right)  \ \Rightarrow 100\times 4.2\left( { 293-284 } \right) =50\times { C _{ s } }\times \left( { 423-293 } \right)  \ \Rightarrow 3780=6500\, { C _{ s } } \ \therefore { C _{ s } }=0.582\, \, J/g\, \, k \end{array}$
Hence, Option $C$ is correct.

Multiple choice chemistry metals physical properties of metals and non-metals some physical properties of metals general characteristics and uses of metals

The density of ice is $0.921\;g\ cm^{-3}$. Calculate the mass of a cubic block of ice which is $76\;mm$ on each side.

  1. $49 \:g$
  2. $4\times 10^2\;g$
  3. $0.04 \:g$
  4. $4\times 10^3\;g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Now, 1 $mm $ =   0.1 $cm$

$ \therefore $ 76$mm $ =   7.6 $cm$
Volume of cube $ =  $ $ (7.6)^3$ $ =  $ 438. 98 $ cm^3$
Now, mass $ =  $ density $ \times $ volume $ =  $ 0.921 $ \times $ 438.98 $ =  $ 404 $g$ $ =  $ $ 4.04 \times10^2 :g$