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

A calorimeter constains 10 g of water at ${ 20 }^{ \circ  }$ C. The temperature falls to ${ 15 }^{ \circ  }$ C in 10 min. When calorimeter contains 20 g of water at ${ 20 }^{ \circ  }$ C, it takes 15 min for the temperature to become ${ 15 }^{ \circ  }$ C. The water equivalent of the calorimeter is

  1. 5 g

  2. 10 g

  3. 25 g

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

Using Newton's law of cooling, the rate of cooling is proportional to the temperature difference. By setting up two equations for the two cases (10g and 20g of water), the water equivalent of the calorimeter can be calculated as 25g.

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

In a calorimeter of water equivalent  $20 { g },$  water of mass  $1.1 { kg }$  is taken at  $288{ K }$  temperature. If steam at temperature  $373 { K }$  is passed through it and temperature of water increases by  $6.5 ^ { \circ } { C }$  then the mass of steam condensed is

  1. $17.5{ g }$
  2. $11.7{ g }$
  3. $15.7{ g }$
  4. $18.2{ g }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The heat lost by the steam (condensing and cooling) equals the heat gained by the water and the calorimeter. Using the formula m*L + m*c*dT = (M_water*c + W_cal)*dT, solving for m yields approximately 11.7g.

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

Bunty mixed 440 gm of ice at $0^{\circ}C$ with 540 gm of water at $80^{\circ} C$ in a bowl. Then what would remain after sometime in the bowl?

  1. only ice

  2. only water

  3. ice and water in same amount

  4. ice and water will vapourise
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy for ice= mL=440(336)= 147840 J
Energy in water= mc$\theta$= 540(80)(4.2)= 181440J

Since, water has more energy, the ice will completely melt while the temperature of water will decrease.
so only ice remains.

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 $Q=m . C. \Delta T$
$=100 g\times 4.2 J g^{-1} {\;}^oC^{-1}\times 70 ^oC$
$=29400 J=29.4 kJ$