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 thermal physics heat and heat transfer transfer of heat fundamentals of heat transfer

A copper ball of mass $100\ gm $ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100\ gm$, filled with $170\ gm $ of water at room temperature. Subsequently, the temperature of the system is found to be $75^{o}C$. $T$ is given by : 

  1. $825^{o}C$
  2. $800^{o}C$
  3. $885^{o}C$
  4. $1250^{o}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Heat given $=$ Heat taken
$\left(100\right)\left(0.1\right))\left(T–75\right))=\left(100\right))\left(0.1\right))\left(45\right))+\left(170\right))\left(1\right))\left(45\right))$
$ 10\left(T−75\right))=450+7650=8100$
$T−75=810$
$T={885}^{\circ}$C
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 1000 g of water. If the entire amount of energy is converted into heat, the rise in temperature of water (sp. heat of water $=4200 J kg^{-1} {\;}^oC^{-1})$

  1. $0.4 ^oC$
  2. $40 ^oC$
  3. $4 ^oC$
  4. $44 ^oC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Q=1680 J$
$m=1000 g$ or 1 kg


$\Delta T=\dfrac {Q}{mc}$

$=\dfrac {1680}{1\times 4200}=0.4^oC.$

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 of $5 ^oC$ to $95 ^oC$?

  1. 900 kcal

  2. 90 kcal

  3. 10 kcal

  4. 9 kcal
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$m=100 g, C=1 cal g^{-1} {\;}^oC^{-1}$
$\Delta T=(95-5)^oC=90^oC$
$Q=mC\Delta T$
$=100 g\times 1 cal g^{-1} {\;}^oC^{-1}\times 90^oC$
$=100\times 1\times 90 cal$
$=9000 cal=9 k cal$.

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

5 kg of water at $80^oC$ is taken in a bucket of negligible heat capacity, 15 kg of water at $20^oC$ is added to it. What is the temperature of the mixture?

  1. $45^oC$
  2. $65^oC$
  3. $85^oC$
  4. $35^oC$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hot water, $m _h=5 kg, T _h=80^oC$
Cold water, $m _c=15 kg, T _c=20^oC$
$T=?$
If the temperature of mixture is T
Heat lost by hot water
$5\times C\times (80-T)$
Heat gained by cold water
$15\times C\times (T-20)$
According to the principle of calorimetry, Heat lost $=$ Heat gained
$5\times C\times (80-T)=15\times C\times (T-20)$
$\therefore 80-T=3(T-20)$

$T=35^oC$.

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

5 g of water at $30^oC$ and 5 g of ice at $-20^oC$ are mixed together in a calorimeter. What is the final temperature of the mixture. Given specific heat of ice $=0.5 cal g^{-1} (^oC)^{-1}$ and latent heat of fusion of ice $=80 cal g^{-1}$.

  1. $0^oC$
  2. $1^oC$
  3. $10^oC$
  4. $20^oC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Heat lost by 5 g of water at $30^oC$ to water at $0^oC$.
$Q _L=5\times 1\times 30=150 cal$
Heat required by 5 g of ice at $-20^oC$
$(Q _1)=5\times 0.5\times (2.0)=50 cal$
Heat required by 5 g of ice at $0^oC$  into water at $0^oC$.
$(Q _2)=5\times 80=400 cal$
$Q _L < (Q _1+Q _2)$
$\therefore$ The final temperature of the mixtuure is $0^oC$.

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

In a process 10 g of ice at $-5^oC$ is converted into the steam at $100^oC$. If specific heat of ice is $0.5 \ cal g^{-1} {\;}^oC^{-1}$, then the amount of heat required to convert 10 g of ice from $-5^oC$ to $0^oC$ is :

  1. 15 cal

  2. 25 cal

  3. 50 cal

  4. 100 cal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$C=0.5 cal g^{-1} {\;}^oC^{-1}$


Heat required to rise the temperature of ice at $-5^oC$ to $0^oC$ is
$Q=mC\Delta T$
$=(10g)(0.5 \ cal g^{-1} {\;}^oC^{-1})$$(0^oC)-(-5^oC)$
$=25 cal$

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

400 g of ice at 253 K is mixed with 0.05 kg of steam at $100^oC$. Latent heat of vaporisation of steam $=540 cal g^{-1}$. Latent heat of fusion of ice $=80 cal g^{-1}$. Specific heat of ice $=0.5 cal g^{-1} {\;}^oC^{-1}$. Find the resultant temperature of the mixture.

  1. 253 K

  2. 260 K

  3. 273 K

  4. 290 K
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat lost by 0.05 kg of steam at $100^oC$ to water at $0^oC$
$Q _L=(50\times 540)+(50\times 1\times 100)$
$=27000+5000=32000 cal$
Heat required by 400 g of ice at 253 K $(-20^oC)$ to convert into water at 273 K $10^oC) Q _1=(400\times 0.5\times 20)+(400\times 80)$
$=(4000+32000)=36000 cal$
$\therefore Q _L < Q _1$, only part of ice melts and final temperature remains $0^oC$ or 273 K.

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

A piece of copper weighing 500 g is heated to $100^oC$ and dropped into 200g of water at $25^oC$. Find the temperature of the mixture. The specific heat of Cu is $0.42 J g^{-1} {\;}^oC^{-1}$.

  1. $30^oC$
  2. $40^oC$
  3. $50^oC$
  4. $60^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, mass of copper, $m _1=500 g$
Mass of water, $m _2=200 g$
Initial temperature of copper, $t _1=100^oC$
Initial temperature of water, $t _2=25^oC$
Sp. heat of copper, $C _1=0.42 J g^{-1} {\;}^oC^{-1}$
Sp. heat of water, $C _2=4.2 J g^{-1} {\;}^oC^{-1}$
Final temperature of the mixture $=t^oC$
Then,
Heat lost by the copper piece
$=m _1C _1(t _1-t)$
Heat gained by water $=m _2C _2(t-t _2)$
We know, Heat lost $=$ Heat gained
$\Rightarrow m _1C _1(t _1-t)=m _2C _2(t-t _2)$
$\Rightarrow 500\times 0.42\times (100-t)$
$=200\times 4.2\times (t-25)$
$\Rightarrow (100-t)=\frac {200\times 4.2}{500\times 0.42}\times (t-25)$
$=4(t-25)$
This given, $5t=200$
$\Rightarrow t=\frac {200}{5}^oC=40^oC$
Thus, the final temperature of the mixture is $40^oC$.

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

$10\ kg$ of hot water in a bucket at $70^oC$ is cooled for taking a bath adding to it $20\ kg$ water at $20^oC$. What is the temperature of the mixture? (Neglect the thermal capacity of the bucket)

  1. $30.67^oC$
  2. $36.67^oC$
  3. $60.67^oC$
  4. $46.67^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

m (hot water) $=10 kg$,
T (hot water) $=70^oC$
m (cold water) $=20 kg$
T (cold water) $=20^oC, T (final)=?$
Using the formula $Q=mC\Delta t$
We get heat lost by hot water
$=10\times C\times (70-T _f)$
Where $T _f$ is the final temperature
Heat gained by cold water
$=20\times C\times (T _f-20)$
Using the principle of calorimetry
Heat lost $=$ Heat gained
We get $10\times C\times (70-T _f)$
$=20\times C\times (T _f-20)$
$\therefore 700-10T _f=20T _f-400$
or $30T _f=1100 \therefore T _f=36.67^oC$.

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

What is the final temperature of the mixture of 300 g of water at $25^oC$ added to 100 of ice at $0^oC$.

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

Heat lost by a hot body $=$ Heat gained by a

$\therefore 300 (25-\theta)=100\times 80+100\times 0.5\theta$
$\therefore \theta=-\frac {5}{3.5}$
Since $\theta$ is negative
Heat lost is utilised to melt only same part of ice. Hence equilibrium temperature is $0^oC$.

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

500 g of water at $100^oC$ is mixed with 300 g at $30^oC$. Find the temperature of the mixture. Specific heat of water $=4.2 J g^{-1} {\;}^oC^{-1}$.

  1. $73.8^oC$
  2. $53.8^oC$
  3. $40^oC$
  4. $60^oC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass of hot water, $m _1=500 g$
Mass of cold water, $m _2=300 g$
Temp. of hot water, $t _1=100^oC$
Temp. of cold water, $t _2=30^oC$
Sp. heat of water, $C=4.2 J g^{-1} {\;}^oC^{-1}$
Let temp. of mixture be $t^oC$. Then, Heat gained by cold water
$=m _2\times C\times (t-t _2)$
According to the principle of calorimetry, Heat lost $=$ Heat gained
$500\times 4.2\times (100-t)$
$=300\times 4.2\times (t-30)$
$\Rightarrow 5(100-t)=3(t-30)$
$\Rightarrow -3t-5t=-90-500$
$\Rightarrow -8t=-590$
$\Rightarrow t=\frac {590}{8}=73.8^oC$
So, the final temperature of the mixture is $73.8^oC$.

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

A calorimeter contains $70.2 \,g$ of water at $15.3^o C$. If $143.7 \,g$ of water at $36.5^o C$ in mixed it with the common temperature is $28.7^o C$. The water equivalent of the calorimeter is:

  1. $15.6 \,g$
  2. $9.4 \,g$
  3. $6.3 \,g$
  4. $13.4 \,g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Assume water equilient is $= ngm$
So, $(70.2) \times 1 \times (28.7 - 15.3) + n (28.7 - 15.3) = 143.7 \times 1 (36.5 - 28.7)$
$n (13.4) = 180.18$
$n = 13.4 \,g$

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.