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

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

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

This is identical to 528091. The heat gained by the water (1.1kg) and calorimeter (20g) for a 6.5C rise equals the heat released by condensed steam. The calculation confirms 11.7g.

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

Steam at $100^oC$ is passed into $2.0$kg of water contained in a calorimeter of water equivalent $0.02$kg at $15^oC$ till the temperature of the calorimeter and its content rise to $90^oC$. The mass of steam condensed in kg is

  1. $0.301$
  2. $0.280$
  3. $0.60$
  4. $0.02$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heat gained by water (2kg) and calorimeter (0.02kg) = (2.02kg * 1000g/kg * 1 cal/gC * (90-15)C) = 151500 cal. Heat lost by steam = m * (540 + (100-90)) = m * 550. m = 151500 / 550 = 275.45g = 0.275kg, which rounds to 0.280kg.

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

A copper calorimeter of a mass $300\ g$ contains $500\ g$ of water at a temperature of $20^\circ C$. A $500\ g$ of copper block at $100^\circ C$ is dropped into the calorimeter. If the resultant temperature is $25^\circ C$, then fond the specific heat of copper in $JKg^{-1} K^{-1}$.

  1. $190$
  2. $290$
  3. $390$
  4. $490$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

Calories is defined as the amount of heat required to rise temperature of $1\ g$ of water by $1^{o}C$ and it is defined under which of the following conditions.

  1. From $14.5^{o}C$ to $15.5^{o}C$ at $760\ mm$ of $Hg$
  2. From $98.5^{o}C$ to $99.5^{o}C$ at $760\ mm$ of $Hg$
  3. From $13.5^{o}C$ to $14.5^{o}C$ at $76\ mm$ of $Hg$
  4. From $3.5^{o}C$ to $4.5^{o}C$ at $76\ mm$ of $Hg$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The calorie is defined as the amount of heat required to raise the temperature of 1g of water from 14.5C to 15.5C at standard atmospheric pressure (760 mm Hg).

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

A copper calorimeter contains $100 g$ of water at $16^o C$. When $15 g$ of ice is added to it, the resultant temperature of the mixture is $4^o C$. Water equivalent of the calorimeter is

  1. $8 g$
  2. $12 g$
  3. $6 g$

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat lost by water and calorimeter = (100 + W) * (16-4) = 12(100+W). Heat gained by ice = 15 * 80 + 15 * (4-0) = 1200 + 60 = 1260. 12(100+W) = 1260 => 100+W = 105 => W = 5g. Since 5g is not an option, 'None' is the correct choice.

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

$50  g$ of ice at 0 C is mixed with $50  g$ of water at 20 C.The resultant temperature of the mixture would be

  1. 10 C

  2. 0 C

  3. -10 C

  4. -35 C
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$50 g $ of Ice at $0^0C$ has a latent heat of

$ Q = m \times L = 50 \times 80 $ 

                    $ = 4000 cal $

Now for water to reach $0^{0}C$ without changing its state 

Heat released by water = $ mC _p  \Delta T$

                                      = $ 50 \times 1 \times 20  $

                                      = $1000 cal $

As the heat to be removed from water is less than the latent heat of $50g $ of ice, the resultant mixture stays at $0^0C $ temperature.

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

The amount of heat required to convert 1 g of ice (specific 0.5 cal  at $g^{-1o} C^{-1}$ ) at $-10^0 C$ to steam at $100 $ $^\circ C$ is ___________.

[ Given: Latent heat of ice is $80 Cal/ gm,$ Latent heat of steam is $540 Cal/gm $, Specific heat of water is $1 Cal/gm/C$ ]

  1. 725 cal

  2. 636 cal

  3. 716 cal

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

Amount of heat required = Change the temp Of Ice from -10 C to 0 C + Heat required to melt the Ice + Heat required to increase the temperature of water from 0 to 100 C + Heat required to convert water  at 100 C to vapor at 100 C

$=1\times 0.5[0-(-10)]+1\times 80+1\times 1\times 100+1\times 540\ =5+80+100+540\ =725cal  $

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

To measure the specific heat of copper, an experiment is performed in the lab. A piece of copper is heated in an oven then dropped into a beaker of water. To calculate the specific heat of copper, the experimenter must know or measure the value of all of the quantities below EXCEPT the

  1. Original temperatures of the copper and the water

  2. Mass of the water

  3. Final (equilibrium) temperature of the copper and the water

  4. Time taken to achieve equilibrium after the copper is dropped into the water

  5. Specific heat of the water
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Specific of  a substance  is given by: $\Delta Q$= $mc\Delta T$

where, $\Delta Q=$ heat given to substance
                $m=$ mass  of  the substance
             $\Delta T=$ increase in temperature of substance (for that we require initial and final temperature of substance)
If the substance is copper in this experiment the experimenter requires the mass of copper piece not the mass of water because water is just dropping the temperature of copper  piece not more  than this.        

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

A body having $1680 J$ of energy is supplied to $100 g$ of water. If the entire amount of energy is converted into heat the rise in temperature of water (sp. heat of water = $4200 JKg^{ -1 }\ ^0C ^{ -1 } $)

  1. $0.4^{ \circ }{ C }$
  2. $40^{ \circ }{ C }$
  3. $4^{ \circ }{ C }$
  4. $44^{ \circ }{ C }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $\delta T$ is the rise in temperature, then the amount of heat supplied is $Q=mS\Delta T$  where $S=$ specific heat

Thus, $1680=(100/1000)(4200)\Delta T$ or $\Delta T=4^oC$

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

$5gm$ of steam at $100^oC$ is passed into calorimeter containing liquid , Temperature of liquid rises from $32^oC$ to $40^oC$. Then water equivalent of calorimeter and content is 

  1. $40$ gram
  2. $375$ gram
  3. $300$ gram
  4. $160$ gram
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Latent heat of vaporization of water = $2260kJ/Kg$

The specific heat capacity of water = $4185.5 J/Kg$

Heat lost by steam = heat gained by the water and calorimeter

Formula :

Heat gained by water = mcФ

m = mass of water

c = specific heat capacity of water.

$Ф = Change in temperature.  = 40 - 32 = 8$

Heat lost by steam = mLv + mcФ

Lv = latent heat of vaporisation

m = mass of steam.

$Ф = 100 - 40 = 60$

Doing the substitution :

$2260000 \times 0.005 + 60 \times 0.005 \times 4185.5 = m \times 4185.5 \times 8$

$12555.65 = 33484m$

$m = \dfrac {12555.65}  {33484} = 0.37497 kg$

$= 0.37497 \times 1000 = 374.97 kg$

$= 374.97kg$

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

1 kg of water at $20^{\circ}C$ is, mixed with 800 g of water at $80^{\circ}C$. Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.

  1. $24.44^{\circ} C$
  2. $46.67^{\circ} C$
  3. $44.44^{\circ} C$
  4. $54.44^{\circ} C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$mc\theta _1 + mc\theta _2 = mc\theta _0$
since c is constant, assuming water is heated from $0^o C$
(1000)(20) + (800)(80)= (1800)$\theta _0$
20000+64000= 84000=1800$\theta _0$
$\theta _0= 46.67^o C$
hence the finale temperature in $\theta _0=46.67^o$
Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

The temperature of equal masses of three different liquids A, B, and C are $12^o C$,$19^o C$ and $28^o C$ respectively. The temperature when A and B are mixed is $16^oC$ and When B and C are mixed is $23^o C$. The temperature when A and C are mixed is:

  1. $18.2^ C$
  2. $22^ C$
  3. $20.3^ C$
  4. $24.2^ C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

let $m _1=m _2=m _3=m$
Let $s _1,s _2,s _3$ be the respective specific heats of the three liquids.
When A and B are mixed, temperature of mixed = $16^o C$
A heat gained by A = heat lost by B
$\therefore ms _1(16-12)=ms _2(19-16)$
$4s _1=3s _2$.......(i)
When B and C are mixed , temperature of mixture$=23^o C$.
As heat gained by B = heat lost by C,
$ms _2(23-19)=ms _3(28-23)$
$\therefore 4s _2=5s _3$......(ii)
From (i) and (ii) , $=\dfrac{3}{4}s _2=\dfrac{15}{16}s _3$
When A and C are mixed, suppose temperature of mixture$=t$
Heat gained by A = Heat lost by C
$ms _1(t-12)=ms _3(28-t)$
$\dfrac{15}{16}s _2(t-12)=s _3(28-t)$
$15t-180=448-16t$
$31t=448+180=628$
$t=\dfrac{628}{3}=20.3^o C$