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 measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

When 60 calories of heat are supplied to 15 g of water, the rise in temperature is

  1. <font><font>$75^{\circ}C$</font></font>
  2. <font><font>$90^{\circ}C$</font></font>
  3. <font><font>$4^{\circ}C $</font></font>
  4. <font><font>$0.25^{\circ}C$</font></font>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Heat = m S \Delta T$
$\Rightarrow 60=15\times 1\times\triangle T \ \Rightarrow \triangle T=4^oC$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

When 60 calories of heat are supplied to 15 g of water, the rise in temperature is

  1. $75^\circ C$
  2. $900^\circ C$
  3. $4^\circ C$
  4. $0.25^\circ C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that, $heat=mass\times specific \;heat\times change\;in\;temperature \ \Rightarrow 60=15\times 1\times \triangle T \ \Rightarrow \triangle T=4^oC$

Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

Which of the following properties must be known in order to calculate the amount of heat needed to melt 1.0kg of ice at $0^oC$? 
I. The specific heat of water 
II. The latent heat of fusion for water 
III. The density of water.

  1. I only

  2. I and II only

  3. I, II, and III

  4. II only

  5. I and III only
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The latent heat is the heat required to change the state of unit mass of substance ,  therefore heat required to change the mass $m$ of substance  is given by ,

            $Q=mL$ ,  where $m=$ mass of substance , $L=$ latent heat
 here we have $m=1.0kg$ but we don't have value of $L$ (latent heat of fusion for water) so it is required .
    Density and specific heat of water are not required here , as it is clear from formula mentioned above .

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

How much heat is required to raise the temperature of $100 g$ of water from ${ 5 }^{ \circ  }C$ to ${ 95 }^{ \circ  }C$?

  1. $900 kcal$
  2. $90 kcal$
  3. $10 kcal$
  4. $9 kcal$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given ,  $m=100g ,  \theta _{1}=5^{0}C , \theta _{2}=95^{0}C$

We have , specific heat of water $c=1cal/g-^{o}C$
Now , heat required to raise the temperature of water is given by ,
                    $Q=mc\Delta \theta=mc(\theta _{2}-\theta _{1})$
or                 $Q=100\times1\times(95-5)=9000cal=9kcal$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

2000 cal of heat is supplied to 200 g of water. Find the rise in temperature. (Specific heat of water = 1 cal ${ { g }^{ -1 } }^{ \circ  }{ C }^{ -1 }$)

  1. ${ 10 }^{ \circ }C$
  2. ${ 20 }^{ \circ }C$
  3. ${ 30 }^{ \circ }C$
  4. ${ 40 }^{ \circ }C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ,  $m=200g , Q=2000cal ,  \Delta\theta=? , $ , specific heat of water $c=1cal/g-^{o}C$

Now , heat required to raise the temperature of water is given by the definition of specific heat ,
                    $Q=mc\Delta \theta$
or                 $\Delta \theta=Q/(mc)=2000/(200\times1)=10^{o}C$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

What will be the amount of heat required to convert $50 g$ of ice at ${ 0 }^{ \circ  }C$ to water at ${ 0 }^{ \circ  }C$?

  1. $400 cal$
  2. $4000 cal$
  3. $3000 cal$
  4. $300 cal$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amount of heat required to convert unit mass of ice into water is called latent heat ($L$) of fusion of ice  i.e.

                       $Q=mL$ ,
   given ,          $m=50g$ ,

   we have ,     $L=80cal/g$

Hence ,           $Q=50\times80=4000cal$ 

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

Calculate the quantity of heat required to convert 1.5 kg of ice at ${ 100 }^{ \circ  }C$ to water at ${ 15 }^{ \circ  }C$. (${ L } _{ ice }\quad =\quad 3.34\quad \times \quad { 10 }^{ 5 }\quad J{ \quad kg }^{ -1 }$, ${ C } _{ water }\quad =\quad 4180\quad J{ \quad kg }^{ -1 }\quad ^{ \circ  }{ { C }^{ -1 } }$)

  1. $5.85\quad \times \quad { 10 }^{ 5 }\quad J$
  2. $5.95\quad \times \quad { 10 }^{ 5 }\quad J$
  3. $3.95\quad \times \quad { 10 }^{ 5 }\quad J$
  4. $4.95\quad \times \quad { 10 }^{ 5 }\quad J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have ,  $L _{ice}=3.34\times10^{5}J/kg , m=1.5kg$

From the definition of latent heat , heat required to convert ice into water at constant temperature $0^{o}C$ ,
             $Q _{1}=mL _{ice}$
or          $Q _{1}=1.5\times3.34\times10^{5}=5.01\times10^{5}J$
Now , heat required to heat up the water at $0^{o}C$ to$15^{o}C$ ,
              $Q _{2}=mc(15-0)=1.5\times4180\times15=0.94\times10^{5}J$  , where $c=4180J/kg-^{o}C$
 Total heat required ,
             $Q=Q _{1}+Q _{2}$ 

or          $Q=5.01\times10^{5}+0.94\times10^{5}=5.95\times10^{5}J$ 

Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

One calorie is defined as the heat required to raise the temperature of $1$ gm of water by $1^o$C in a certain interval of temperature and at certain pressure. The temperature interval and pressure is?

  1. $13.5^o$ C to $14.5^o$ C & $76$ mm of Hg
  2. $6.5^o$ C to $7.5^o$ C & $76$ mm of Hg
  3. $14.5^o$ C to $15.5^o$ C & $760$ mm of Hg
  4. $98.5^o$ C to $99.5^o$ C & $760$ mm of Hg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

One calories is defined as the amount of heat required to raise the temp of $1\ gm$ of water from $14.5^oC$ to $15.5^oC$

in $760\ mm$ of $Hg$.

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

2000 cal of heat is supplied to 200 g of water. Find the rise in temperature. (Specific heat of water $=1 cal g^{-1} {\;}^oC^{-1})$

  1. $10 ^oC$
  2. $20 ^oC$
  3. $30 ^oC$
  4. $40 ^oC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity of heat supplied
$Q=2000 cal$
Mass of water, $m=200 g$
Specific heat of water
$C=1 cal g^{-1} {\;}^oC^{-1}$
Rise in temperature $=?$
From relation, $Q=m C\Delta T$
$\Rightarrow \Delta T=\frac {Q}{m.C}$
$=\frac {2000 cal}{200 g\times 1cal g^{-1} {\;}^oC^{-1}}=10^oC$
So, the temperature of water rises by $10^oC$.

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

500 g of hot water at $60^oC$ is kept in the open till its temperature falls to $40^oC$. Calculate the heat energy lost to the surroundings by the water. (Specific heat of water $=4200 J kg^{-1} {\;}^oC^{-1})$

  1. 2400 J

  2. 5000 J

  3. 40000 J

  4. 42000 J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$m=500, g=0.5 kg$
$\Delta T=60-40=20^oC$
$C=4200 J kg^{-1} {\;}^oC^{-1} Q=?$
Using the formula, $Q=m C \Delta T$
$=0.5\times 4200\times 20=42000 J$
$\therefore \text {Heat lost}=42,000 J$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

How much amount of heat is required to raise the temperature of 100 g of water from $30 ^oC$ to $100 ^oC$? The specific heat of water $=4.2 J g^{-1} {\;}^oC^{-1}$.

  1. 25.5 kJ

  2. 29.4 kJ

  3. 30 kJ

  4. 40 kJ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mass of water, $m=100 g$
Rise in temperature $(\Delta T)$
$=(100^oC-30^oC)=70 ^oC$
Specific heat of water
$C=4.2 J g^{-1} {\;}^oC^{-1}$
Then $Q=m . C. \Delta T$
$=100 g\times 4.2 J g^{-1} {\;}^oC^{-1}\times 70 ^oC$
$=29400 J=29.4 kJ$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

A liquid P if specific heat capacity $2400 J kg^{-1} K^{-1}$ and at $70^oC$ is mixed with another liquid R of specific heat capacity $1000 J kg^{-1} K^{-1}$ at $30^oC$. After mixing, the final temperature of the mixture is $40^oC$. Find the ratio of the mass of the liquids mixed?

  1. 4 : 5

  2. 8 : 5

  3. 40 : 5

  4. 48 : 5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let mass of the liquids P and R be $m _P$ and $m _R$ respectively.
According to calorimeter,
Heat lost by liquid P $=$ Heat gained by liquid R.
$\therefore m _PC _P(70-40)=m _RC _R(40-30)$
$m _P(2400)(40)=m _R(1000)(10)$
$\frac {m _P}{m _R}=\frac {2400\times 40}{1000\times 10}=\frac {48}{5}$
$\frac {m _P}{m _R}=48 : 5$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

Calculate the amount of heat required to convert 5 kg of ice to $0^oC$ to vapour at $100^oC$.

  1. $1.5\times 10^7 J$
  2. $2.5\times 10^7 J$
  3. $3.5\times 10^7 J$
  4. $4.5\times 10^7 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m=5 kg, L _f=3.36\times 10^5 J kg^{-1}$
$L _V=2.25\times 10^6 J kg^{-1}$
$C=4200 J kg^{-1} {\;}^oC^{-1}$ Then $Q=?$
$Q=$ Heat required to convert ice at $^oC$ to water $0^oC +$ Heat required to convert water at $100^oC$ to vapour at $100^oC$
$=mL _f+ms(100-0)+mL$
$=5\times 3.36\times 10^5+5\times 4200\times (100-0)+5\times 2.25\times 10^6$
$=16.8\times 10^5+21\times 10^5+112.5\times 10^5$
$=1.5\times 10^7J$