NCERT Solutions for Class 11 Physics Chapter 10 - Thermal Properties Of Matter

• Heat is a form of energy that flows between a body and its surrounding medium by virtue of temperature difference between them. The degree of hotness of the body is quantitatively represented by temperature.

• A temperature-measuring device (thermometer) makes use of some measurable property (called thermometric property) that changes with temperature. Different thermometers lead to different scales. To construct a temperature scale, two fixed points are chosen and assigned some arbitrary values of temperature. The two numbers fix the origin of the scale and the size of its unit.

• The Celsius temperature (tC) and the Fahrenheit temperature (tV) are related by $t_{v}=\left ( 9/5 \right )t_{c}+32$.

• The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is: $PV=\mu RT$ where $\mu$ is the number of moles and R is the universal gas constant.

• In the absolute temperature scale, the zero of the scale is the absolute zero of temperature the temperature where every substance in nature has the least possible molecular activity. The Kelvin absolute temperature scale (T) has the same unit size as the Celsius scale (TC), but differs in the origin:$T_{c}=T-273.15$.

• 1 . The coefficient of linear expansion $\alpha_{t}$ and volume expansion $\alpha_{v}$ are defined by the relations: $\frac{\Delta l}{l}=\alpha _{t}\Delta T$

$\frac{\Delta V}{V}=\alpha _{v}\Delta T$

where $\Delta l$ and $\Delta V$ denote the change in length l and volume V for a change of temperature $\Delta T$.The relation between them is : $\alpha _{v}=3\alpha_{t}$.

• The specific heat capacity of a substance is defined by s=\frac{1}{m}\frac{\Delta Q}{\Delta T}.

where m is the mass of the substance and $\Delta Q$ is the heat required to change its temperature by $\Delta T$ . The molar specific heat capacity of a substance is defined by C=\frac{1}{\mu }\frac{\Delta Q}{\Delta T}

where $\mu$ is the number of moles of the substance.

• The latent heat of fusion (Lf) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure. The latent heat of vaporization (Lv) is the heat per unit mass required to change a substance from liquid to the vapour state without change in the temperature and pressure.

The three modes of heat transfer are conduction, convection and radiation.

• In conduction, heat is transferred between neighbouring parts of a body through molecular collisions, without any flow of matter. For a bar of length L and uniform cross section A with its ends maintained at temperatures TC and TD, the rate of flow of heat H is: $H=KA\frac{T_{c}-T_{d}}{L}$.

where K is the thermal conductivity of the material of the bar.

• Newton’s Law of Cooling says that the rate of cooling of body is proportional to the excess temperature of the body over the surroundings:

where T1 is the temperature of the surrounding medium and T2 is the temperature of the body.

• Emissivity of a body at a given temperature is equal to the ratio of the total emissive power of the body $e_{\lambda }$ to the total emissive power of perfectly black body $E_{\lambda }$ at that temperature. Emissivity, $E= \frac{e_{\lambda }}{E_{\lambda }}$

• Perfectly Black Body: A body which absorbs completely the radiations of all wavelengths incident on it, is called a perfectly black body. For a perfectly black body, emissive power $E_{\lambda }$=1.

Lamp black is 96% black and platinum black is about 98% black.

• Kirchhoff’s Law of Radiation: This law states that, the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature. Mathematically, $ \frac{e_{1}}{a_{1}}=\frac{e_{2}}{a_{2}}=...=E$.

• Stefan’s Law: According to this law, the emissive power of perfectly black body (energy emitted by a black body per unit surface area per unit time) is directly proportional to the fourth power of its absolute temperature.

Mathematically, $E=\sigma T^{4}$

where $\sigma$ is a constant known as the Stefan’s constant. Value of $\sigma$ is  $5.67\times10^{-8}Wm^{-2}K^{-4}$.

where, b is known as Wien’s constant and its value is $2.89\times10^{-3}m-K$.

Access NCERT Solutions for Physics  Chapter 10 – Thermal Properties of Matter

1. The triple points of neon and carbon dioxide are $24.57$ K and $216.55$ K respectively. Express these temperatures on Celsius and Fahrenheit scales.

Ans: The relation between Kelvin and Celsius scales:

${{T}_{C}}={{T}_{K}}-273.15$

The relation between Celsius and Fahrenheit scales:

${{T}_{F}}=\frac{9}{5}{{T}_{c}}+32$

For neon:

${{T}_{K}}=24.57K$ 

$\Rightarrow {{T}_{C}}=24.57-273.15=-248.58{}^\circ C$ 

As ${{T}_{F}}=\frac{9}{5}{{T}_{C}}+32$ 

$\Rightarrow {{T}_{F}}=\frac{9}{5}\left( -248.58 \right)+32$ 

$\Rightarrow {{T}_{F}}=-415.44{}^\circ F$ 

For carbon dioxide:

${{T}_{K}}=216.55K$ 

$\Rightarrow {{T}_{C}}=216.55-273.15=-56.60{}^\circ C$ 

As ${{T}_{F}}=\frac{9}{5}{{T}_{C}}+32$

$\Rightarrow {{T}_{F}}=\frac{9}{5}\left( -56.60 \right)+32$ 

$\Rightarrow {{T}_{F}}=-69.88{}^\circ F$ 

Therefore, the temperatures on the Celsius scale and Fahrenheit scale are $-248.58{}^\circ C$  and $-415.44{}^\circ F$ respectively for neon and $-56.60{}^\circ C$ and $-69.88{}^\circ F$ respectively for carbon dioxide.

2. Two absolute scales A and B have triple points of water defined to be $200$ A and $350$ B. What is the relation between ${{T}_{A}}$and ${{T}_{B}}$ ?

Ans: Given that,

Triple point of water on absolute scale A, ${{T}_{1}}=200$A

Triple point of water on absolute scale B, ${{T}_{2}}=350$B

Triple point of water on Kelvin scale, ${{T}_{K}}=273.15$K

The temperature $273.15$ K on Kelvin scale is equivalent to $200$ A on an absolute scale A.

${{T}_{1}}={{T}_{K}}$ 

$\Rightarrow 200A=273.15K$ 

$\Rightarrow A=\frac{273.15}{200}$ 

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.

${{T}_{2}}={{T}_{K}}$ 

$\Rightarrow 350B=273.15$ 

\[\Rightarrow B=\frac{273.15}{350}\] 

${{T}_{A}}$ is the triple point of water on scale A.

${{T}_{B}}$ is the triple point of water on scale B.

Therefore,

$\frac{273.15}{200}\times {{T}_{A}}=\frac{273.15}{350}\times {{T}_{B}}$ 

$\Rightarrow {{T}_{A}}=\frac{200}{350}{{T}_{B}}$ 

Therefore, the ratio ${{T}_{A}}$ : ${{T}_{B}}$ is given as $4$: $7$.

3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law $R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$

The resistance is $101.6\Omega $ at the triple point of water $273.16$K, and  $165.5\Omega $at the normal melting point of lead $\left( 600.5K \right)$. What is the temperature when the resistance is  $123.4\Omega $?

Ans: It is given that:

$R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$ \[\] …

Where, ${{R}_{0}}$ and ${{T}_{0}}$ are the initial resistance and temperature respectively;

R and T are the final resistance and temperature respectively;

$\alpha $ is constant.

At the triple point of water, ${{T}_{0}}=273.15$K

Resistance of lead, ${{R}_{0}}=101.6$$\Omega $ 

At normal melting point of lead, T $=600.5K$ 

Resistance of lead, R$=165.5$$\Omega $ 

Substituting these values in, we get:

$R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$

$\Rightarrow 165.5=101.6\left[ 1+\alpha \left( 600.5-273.15 \right) \right]$ 

$\Rightarrow 1.629=1+\alpha (327.35)$ 

$\Rightarrow \alpha =\frac{0.629}{327.35}=1.92\times {{10}^{-3}}$ 

For resistance, ${{R}_{1}}=123.4$ $\Omega $ 

$R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$

where, T is the temperature when the resistance of lead is 123.4$\Omega $

$\Rightarrow 123.4=101.6\left[ 1+1.92\times {{10}^{-3}}\left( T-273.15 \right) \right]$

$\Rightarrow 1.214=1.192\times {{10}^{-3}}\left( T-273.15 \right)$ 

$\Rightarrow \frac{0.214}{1.92\times {{10}^{-3}}}=\left( T-273.15 \right)$ 

$\Rightarrow T=384.61K$ 

Clearly, the temperature when the resistance is 123.4$\Omega $ is $384.61K$.

4. Answer the following:

1. The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

Ans: The triple point of water has a unique value of $273.16$ K.

The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

2. There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers $0{}^\circ C$ and $100{}^\circ C$ respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

Ans: The absolute zero or 0 K is itself the other fixed point on the Kelvin absolute scale.

3. The absolute temperature (Kelvin scale) T is related to the temperature ${{t}_{C}}$  the Celsius scale by ${{t}_{C}}=T-273.15$.

Why do we have 273.15 in this relation, and not 273.16?

Ans: The temperature 273.16 K is the triple point of water; not the melting point of ice.

The temperature $0{}^\circ C$  on Celsius scale is the melting point of ice. 

Its corresponding value on the Kelvin scale is 273.15 K.

Thus, absolute temperature (Kelvin scale) T, is related to temperature ${{t}_{C}}$  on Celsius scale as: ${{t}_{C}}=T-273.15$.

4. What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Ans: Let ${{T}_{F}}$  be the temperature on Fahrenheit scale and ${{T}_{K}}$ be the temperature on absolute scale. Both the temperatures have the relation:

$\frac{{{T}_{F}}-32}{180}=\frac{{{T}_{K}}-273.15}{100}$ \[\] …

Let  ${{T}_{F1}}$ be the temperature on Fahrenheit scale and  ${{T}_{K1}}$ be the temperature on absolute scale. Both the temperatures can be related as:

$\frac{{{T}_{F1}}-32}{180}=\frac{{{T}_{K1}}-273.15}{100}$     …

It is given that:

${{T}_{K1}}-{{T}_{K}}=1K$ 

Subtracting equation (a) from equation (b), we get:

$\frac{{{T}_{F1}}-{{T}_{F}}}{180}=\frac{{{T}_{K1}}-{{T}_{K}}}{100}=\frac{1}{100}$

$\Rightarrow {{T}_{F1}}-{{T}_{F}}=\frac{1\times 180}{100}=\frac{9}{5}$

Triple point of water $=273.16K$ 

Therefore, triple point of water on absolute scale $=273.16\times \frac{9}{5}=491.69K$. 

5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: 

1. What is the absolute temperature of the normal melting point of Sulfur as read by thermometers A and B?

Ans: Given that,

Triple point of water, T = 273.16 K 

At this temperature, pressure in thermometer A, \[{{P}_{A}}=1.250\times {{10}^{5}}\] Pa

 Let ${{T}_{1}}$  be the normal melting point of Sulfur. 

At this temperature, pressure in thermometer A, ${{P}_{1}}=1.797\times {{10}^{5}}$Pa

According to Charles’ law, we have the relation:

$\frac{{{P}_{A}}}{T}=\frac{{{P}_{1}}}{{{T}_{1}}}$ 

$\Rightarrow {{T}_{1}}=\frac{{{P}_{1}}T}{{{P}_{A}}}=\frac{1.797\times {{10}^{5}}\times 273.16}{1.25\times {{10}^{5}}}$ 

$\Rightarrow {{T}_{1}}=392.69K$ 

Therefore, the absolute temperature of the normal melting point of Sulfur as read by thermometer A is 392.69 K.

Now, at triple point 273.16 K, the pressure in thermometer B,  \[{{P}_{B}}=0.200\times {{10}^{5}}\]Pa

At temperature ${{T}_{1}}$ , the pressure in thermometer B, \[{{P}_{2}}=0.287\times {{10}^{5}}\]Pa

According to Charles’ law, we can write the relation: 

$\frac{{{P}_{B}}}{T}=\frac{{{P}_{1}}}{{{T}_{1}}}$

$\Rightarrow {{T}_{1}}=\frac{0.287\times {{10}^{5}}\times 273.16}{0.200\times {{10}^{5}}}=391.98K$

Therefore, the absolute temperature of the normal melting point of Sulfur as read by thermometer B is 391.98 K.

2. What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Ans: Oxygen, as well as hydrogen gas contained in thermometers A and B respectively, are not perfect ideal gasses.

There is a minor difference between the readings of thermometers A and B.

To lessen the discrepancy between the two readings, the experiment must be carried under low pressure conditions so that gasses can behave as ideal gas.

6. A steel tape 1m long is correctly calibrated for a temperature of $27.0{}^\circ C$ . The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is $45.0{}^\circ C$.  What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is $27.0{}^\circ C$?  Coefficient of linear expansion of steel  $=1.20\times {{10}^{-5}}{{K}^{-1}}$ .

Ans: Given that,

Length of the steel tape at temperature $27.0{}^\circ C$, $l=1m=100cm$ 

At temperature  ${{T}_{1}}=45{}^\circ C$,  the length of the steel rod, ${{l}_{1}}=63cm$ 

Coefficient of linear expansion of steel, $\alpha =1.20\times {{10}^{-5}}{{K}^{-1}}$ 

 Let ${{l}_{2}}$  be the actual length of the steel rod and ${l}'$ be the length of the steel tape at  $45{}^\circ C$.

\[{l}'=l+al\left( {{T}_{1}}-T \right)\] 

\[\Rightarrow {l}'=100+1.20\times {{10}^{-5}}\times 100\left( 45-27 \right)\]

\[\Rightarrow {l}'=100.0216cm\] 

Hence, the actual length of the steel rod measured by the steel tape at $45{}^\circ C$ can be calculated as: 

\[{{l}_{2}}=\frac{100.0216}{100}\times 63=63.0136cm.\]

Therefore, the actual length of the rod at $45{}^\circ C$  is 63.0136 cm. Its length at $27.0{}^\circ C$ is 63.0 cm.

7. A large steel wheel is to be fitted onto a shaft of the same material. At $27{}^\circ C$  , the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range $:{{\alpha }_{steel}}=1.20\times {{10}^{-5}}{{K}^{-1}}$.

Ans: Here,

The given temperature, $T=~27{}^\circ C$  can be written in Kelvin as \[27+273=300\text{ }K\].

Outer diameter of the steel shaft at T,  ${{d}_{1}}=8.70cm$ 

Diameter of the central hole in the wheel at T,  ${{d}_{2}}=8.69cm$  

Coefficient of linear expansion of steel, $:{{\alpha }_{steel}}=1.20\times {{10}^{-5}}{{K}^{-1}}$

After the shaft is cooled using ‘dry ice’, its temperature becomes ${{T}_{1}}$.

The wheel will slip on the shaft, if the change in diameter, \[\Delta d=8.698.70=0.01\]cm

Temperature ${{T}_{1}}$ , can be calculated from the relation: 

\[\Delta d={{d}_{1}}{{\alpha }_{steel}}\left( {{T}_{1}}-T \right)\] 

\[\Rightarrow \Delta d=8.70\times 1.20\times {{10}^{-5}}\left( {{T}_{1}}-300 \right)\] 

$\Rightarrow \left( {{T}_{1}}-300 \right)=95.78$ 

$\Rightarrow {{T}_{1}}=204.21K$ 

Or ${{T}_{1}}=204.21-273.16=-68.95{}^\circ C$ 

Clearly, the wheel will slip on the shaft when the temperature of the shaft is approximately $-69{}^\circ C.$ 

8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at $27.0{}^\circ C$ .  What is the change in the diameter of the hole when the sheet is heated to $227{}^\circ C$?  Coefficient of linear expansion of copper $=1.70\times {{10}^{-5}}{{K}^{-1}}$ .

Ans: Given that,

Initial temperature, ${{T}_{1}}=27.0{}^\circ C$ 

Diameter of the hole at ${{T}_{1}},{{d}_{1}}=4.24cm$ 

Final temperature,  ${{T}_{2}}=227{}^\circ C$ 

Diameter of the hole at ${{T}_{2}}={{d}_{2}}$ 

Coefficient of linear expansion of copper, ${{\alpha }_{Cu}}=1.70\times {{10}^{-5}}{{K}^{-1}}$ 

For coefficient of superficial expansion $\beta $, and change in temperature  ∆T, we have the relation: 

$\frac{change\text{ }in\text{ }area\left( \Delta A \right)}{original\text{ }area\left( A \right)}=\beta \Delta T$ 

$\Rightarrow \frac{\left( \pi \frac{d_{2}^{2}}{4}-\pi \frac{d_{1}^{2}}{4} \right)}{\left( \pi \frac{d_{1}^{2}}{4} \right)}=\frac{\Delta A}{A}$ 

$\Rightarrow \frac{\Delta A}{A}=\frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}$

However, $\beta =$ $2\alpha $ 

$\Rightarrow \frac{d_{2}^{2}}{d_{1}^{2}}-1=2\alpha \left( {{T}_{2}}-{{T}_{1}} \right)$ 

$\Rightarrow \frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}=2\alpha \Delta T$ 

$\Rightarrow \frac{d_{2}^{2}}{{{\left( 4.24 \right)}^{2}}}=2\times 1.7\times {{10}^{-5}}\left( 227-27 \right)+1$ 

$\Rightarrow d_{2}^{2}=17.98\times 1.0068=18.1$ 

$\Rightarrow {{d}_{2}}=4.2544cm$ 

Change in diameter  =${{d}_{2}}-{{d}_{1}}=4.2544-4.24=0.0144cm$.

Clearly, the diameter increases by $1.44\times {{10}^{-2}}cm.$ 

9. A brass wire 1.8 m long at $27{}^\circ C$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of  $-39{}^\circ C$ ,  what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass  $=2.0\times {{10}^{-5}}{{K}^{-1}}$.

Young’s modulus of brass  $=0.91\times {{10}^{11}}Pa$.

Ans: Given that,

Initial temperature,  ${{T}_{1}}=27{}^\circ C$ 

Length of the brass wire at \[{{T}_{1}},l=1.8m\] 

Final temperature,  ${{T}_{2}}=-39{}^\circ C$

Diameter of the wire,  $d=2.0mm=2\times {{10}^{-3}}m$ 

Tension developed in the wire = F 

Coefficient of linear expansion of brass,$\alpha =2.0\times {{10}^{-5}}{{K}^{-1}}$

Young’s modulus of brass,  $Y=0.91\times {{10}^{11}}Pa$ 

Young’s modulus is given by the relation:

$Y=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$ 

$\Rightarrow \Delta L=\frac{F\times L}{A\times L}$              … 

Where, F = Tension developed in the wire

A = Area of cross-section of the wire

L = Change in the length, given by the relation:

$L=\alpha \left( {{T}_{2}}-{{T}_{1}} \right)$          …

Equating equations (a) and (b), we get:

$\alpha L\left( {{T}_{2}}-{{T}_{1}} \right)=\frac{FL}{\pi {{\left( \frac{d}{2} \right)}^{2}}\times Y}$ 

$\Rightarrow F=\alpha \left( {{T}_{2}}-{{T}_{1}} \right)\pi Y{{\left( \frac{d}{2} \right)}^{2}}$

$\Rightarrow F=2\times {{10}^{5}}\times \left( -39-27 \right)\times 3.14\times 0.91\times {{10}^{11}}\times {{\left( \frac{2\times {{10}^{-3}}}{2} \right)}^{2}}$

$\Rightarrow F=-3.8\times {{10}^{2}}N$ 

(The negative sign indicates that the tension is directed inward.) 

Clearly, the tension developed in the wire is $3.8\times {{10}^{2}}N$.

10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250{}^\circ C$ , if the original lengths are at  $40{}^\circ C$?  Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass  $=2.0\times {{10}^{-5}}{{K}^{-1}}$, steel  $=1.2\times {{10}^{-5}}{{K}^{-1}}$).

Ans: Given that,

Initial Temperature, ${{T}_{1}}=40{}^\circ C$

Final temperature,  ${{T}_{2}}=40{}^\circ C$

Change in temperature $={{T}_{2}}-{{T}_{1}}=210{}^\circ C$ 

Length of the brass rod at ${{T}_{1}}$, ${{l}_{1}}=50cm$ 

Diameter of the brass rod at ${{T}_{1}}$, ${{d}_{1}}=3.0mm$

Length of the steel rod at ${{T}_{2}}$, ${{l}_{2}}=50cm$

Diameter of the steel rod ${{T}_{2}}$, ${{d}_{2}}=3.0mm$

Coefficient of linear expansion of brass, ${{\alpha }_{1}}=2.0\times {{10}^{-5}}{{K}^{-1}}$ 

Coefficient of linear expansion of steel, ${{\alpha }_{2}}=1.2\times {{10}^{-5}}{{K}^{-1}}$ 

For the expansion in the brass rod, we have:

$\frac{change\text{ }in\text{ length}\left( \Delta {{l}_{1}} \right)}{original\text{ length}\left( {{l}_{1}} \right)}={{\alpha }_{1}}\Delta T$

$\Rightarrow \Delta {{l}_{1}}=50\times \left( 2.1\times {{10}^{-5}} \right)\times 210=0.2205cm$ 

For the expansion in the steel rod, we have:

$\frac{change\text{ }in\text{ length}\left( \Delta {{l}_{2}} \right)}{original\text{ length}\left( {{l}_{2}} \right)}={{\alpha }_{2}}\Delta T$

$\Rightarrow \Delta {{l}_{2}}=50\times \left( 1.2\times {{10}^{-5}} \right)\times 210=0.126cm$

Total change in the lengths of brass and steel, $\Delta l=\Delta {{l}_{1}}+\Delta {{l}_{2}}$ 

$\Rightarrow \Delta l=0.2205+0.126=0.346cm$

Clearly, the total change in the length of the combined rod = 0.346 cm.

As the rod expands freely from both ends, no thermal stress is developed at the junction.

11. The coefficient of volume expansion of glycerine is  $49\times {{10}^{-5}}{{K}^{-1}}$ .  What is the fractional change in its density for a $30{}^\circ C$  rise in temperature?

Ans: Given that,

Coefficient of volume expansion of glycerine, ${{\alpha }_{v}}=49\times {{10}^{-5}}{{K}^{-1}}$

Rise in temperature, T$=30{}^\circ C$ 

Fractional change in its volume = $\frac{\Delta V}{V}$ 

This change is related with the change in temperature as:

$\frac{\Delta V}{V}={{\alpha }_{V}}\Delta T$

$\Rightarrow \left( {{V}_{{{T}_{2}}}}-{{V}_{{{T}_{1}}}} \right)={{V}_{{{T}_{1}}}}{{\alpha }_{V}}\Delta T$

$\Rightarrow \frac{m}{{{\rho }_{{{T}_{2}}}}}-\frac{m}{{{\rho }_{{{T}_{1}}}}}=\frac{m}{{{\rho }_{{{T}_{2}}}}}{{\alpha }_{V}}\Delta T$ 

Where,

m = Mass of glycerine

${{\rho }_{{{T}_{1}}}}$= Initial density at ${{T}_{1}}$ 

${{\rho }_{{{T}_{2}}}}$= Final density at ${{T}_{2}}$

$\Rightarrow \frac{{{\rho }_{{{T}_{1}}}}-{{\rho }_{{{T}_{2}}}}}{{{\rho }_{{{T}_{2}}}}}={{\alpha }_{V}}\Delta T$ 

Where,

$\frac{{{\rho }_{{{T}_{1}}}}-{{\rho }_{{{T}_{2}}}}}{{{\rho }_{{{T}_{2}}}}}=$Fractional change in density

Therefore, fractional change in the density of glycerin $=49\times {{10}^{-5}}\times 30=1.47\times {{10}^{-2}}$. 

12. A 10kW drilling machine is used to drill a bore in a small aluminum block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminum $=0.91J{{g}^{-1}}{{K}^{-1}}$.

Ans: Given that,

Power of the drilling machine, $P=10kW=10\times {{10}^{3}}W$ 

Mass of the aluminum block, $m=8.0kg=8\times {{10}^{3}}g$ 

Time for which the machine is used, $t=2.5\min =2.5\times 60=150s$ 

Specific heat of aluminum, $c=0.91J{{g}^{-1}}{{K}^{-1}}$ 

Rise in the temperature of the block after drilling $=\delta T$ 

Total energy of the drilling machine = Pt

$\Rightarrow Pt=10\times {{10}^{3}}\times 150=1.5\times {{10}^{6}}J$ 

It is given that only 50% of the power is useful.

Useful energy, $\Delta Q=\frac{50}{100}\times 1.5\times {{10}^{6}}J$ 

However, $\Delta Q=mc\Delta T$

$\Rightarrow \Delta T=\frac{\Delta Q}{mc}$ 

$\Rightarrow \Delta T=\frac{7.5\times {{10}^{5}}}{8\times {{10}^{3}}\times 0.91}=103{}^\circ C$ 

Clearly, in 2.5 minutes of drilling, the rise in the temperature of the block is $103{}^\circ C$.

13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of $500{}^\circ C$  and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $=0.39J{{g}^{-1}}{{K}^{-1}}$  heat of fusion of water $=335J{{g}^{-1}}$ ).

Ans: Given that,

Mass of the copper block, m $=2.5kg=2500g$ 

Rise in the temperature of the copper block, $\Delta \theta =500{}^\circ C$ 

Specific heat of copper, $C=J{{g}^{-1}}{}^\circ {{C}^{-1}}$ 

Heat of fusion of water, $L=335J{{g}^{-1}}$ 

The maximum heat the copper block can lose, $Q=mC\Delta \theta $ 

$\Rightarrow Q=2500\times 0.39\times 500=487500J$ 

Let ${{m}_{1}}$ be the amount of ice that melts when the copper block is placed on the ice block. 

The heat gained by the melted ice, $Q={{m}_{1}}L$ 

$\Rightarrow {{m}_{1}}=\frac{Q}{L}=\frac{487500}{335}=1455.22g$ 

Thus, the maximum amount of ice that can melt is 1.45 kg.

14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at $150{}^\circ C$  is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 $c{{m}^{3}}$of water at $27{}^\circ C$. The final temperature is $40{}^\circ C$ . Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Ans: Given that,

Mass of the metal, m $=0.20kg=200g$

Initial temperature of the metal, ${{T}_{1}}=150{}^\circ C$ 

Final temperature of the metal, ${{T}_{2}}=40{}^\circ C$ 

Calorimeter has water equivalent of mass, $m=0.025kg=25g$

Volume of water, $V=150c{{m}^{3}}$ 

Mass (M) of water at temperature $T=27{}^\circ C:150\times 1=150g$ 

Fall in the temperature of the metal:

$\Delta T={{T}_{1}}-{{T}_{2}}=150-40=110{}^\circ C$ 

Specific heat of water, ${{C}_{W}}=4.186J/g/{}^\circ K$ 

Specific heat of the metal = C 

Heat lost by the metal, $\theta =mc\Delta T$ 

Rise in the temperature of the water and calorimeter system:

$\Delta {T}'=40-27=13{}^\circ C$ 

Heat gained by the water and calorimeter system:

$\Delta {\theta }'={{m}_{1}}{{C}_{W}}\Delta {T}'$

$\Rightarrow \Delta {\theta }'=\left( M+{m}' \right){{C}_{W}}\Delta {T}'$

Heat lost by the metal = Heat gained by the water and calorimeter system

$mc\Delta T=\left( M+{m}' \right){{C}_{W}}\Delta {T}'$

$\Rightarrow 200\times C\times 110=\left( 150+25 \right)\times 4.186\times 13$ 

$\Rightarrow C=\frac{175\times 4.186\times 13}{110\times 200}=0.43J{{g}^{-1}}{{K}^{-1}}$ 

Clearly, when some heat is lost to the surroundings, then the value of C would be smaller than the actual value.

15. Given below are observations on molar specific heats at room temperature of some common gasses:

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92$cal/mol$ K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Ans: The gases listed in the provided table are diatomic. 

Apart from the translational degree of freedom, they also possess other degrees of freedom (modes of motion). 

Heat should be supplied to increase the temperature of these gases which would further increase the average energy of all the modes of motion.

Thus, the molar specific heat of diatomic gases turns out to be more than that of monatomic gases. 

Moreover, when only the rotational mode of motion is considered, then the molar specific heat of a diatomic gas $=\frac{5}{2}R$ 

$\Rightarrow \frac{5}{2}R=\frac{5}{2}\times 1.98=4.95$cal $mo{{l}^{-1}}{{K}^{-1}}$ 

Excluding the data of chlorine, all the other observations in the given table agree with $\left( \frac{5}{2}R \right)$.

This is because at room temperature, chlorine also has vibrational modes of motion apart from rotational and translational modes of motion.

16. Answer the following questions based on the P-T phase diagram of carbon dioxide:

1. At what temperature and pressure can the solid, liquid and vapor phases of $C{{O}_{2}}$  co-exist in equilibrium?

Ans: From the P-T phase diagram for $C{{O}_{2}}$ as shown in the figure above,

C is the triple point of the $C{{O}_{2}}$ phase diagram. 

This suggests that at the temperature and pressure corresponding to this point (i.e., at $56.6{}^\circ C$  and 5.11 atm), the solid, liquid, and vaporous phases of $C{{O}_{2}}$ co-exist in equilibrium.

2. What is the effect of decrease of pressure on the fusion and boiling point of $C{{O}_{2}}$?

Ans: The fusion and boiling points of $C{{O}_{2}}$ decrease as pressure decreases.

3. What are the critical temperature and pressure for $C{{O}_{2}}$ ? What is their significance?

Ans: The critical temperature and critical pressure of $C{{O}_{2}}$ are $31.1{}^\circ C$  and 73 atm respectively.

Even when there is compression to a pressure greater than 73 atm, $C{{O}_{2}}$ would not liquify above the critical temperature.

4. Is $C{{O}_{2}}$ solid, liquid or gas at a) $-70{}^\circ C$ under 1 atm, b) $60{}^\circ C$ under 10 at m, c) $15{}^\circ C$ under 56 atm?

Ans: With reference to the P-T phase diagram of $C{{O}_{2}}$ ;

a) $C{{O}_{2}}$is gaseous at $-70{}^\circ C$, under 1 atm pressure.

b) $C{{O}_{2}}$is solid at $60{}^\circ C$, under 10 atm pressure. 

c) $C{{O}_{2}}$is liquid at $15{}^\circ C$, under 56 atm pressure.

17. Answer the following questions based on the P–T phase diagram of $C{{O}_{2}}$: 

1. $C{{O}_{2}}$ at 1 atm pressure and temperature $-60{}^\circ C$ is compressed isothermally. Does it go through a liquid phase?

Ans: The P-T phase diagram for $C{{O}_{2}}$ is considered as shown in the question.

At 1 atm pressure and at $-60{}^\circ C$ , $C{{O}_{2}}$ positions itself to the left of $-56.6{}^\circ C$ (triple point C). Thus, it places itself in the region of vaporous and solid phases. 

Clearly, $C{{O}_{2}}$ condenses into the solid state directly without going through the liquid state.

2. What happens when $C{{O}_{2}}$ at 4 atm pressure is cooled from room temperature at constant pressure?

Ans: At 4 atm pressure, $C{{O}_{2}}$ places itself below 5.11 atm (triple point C). 

Thus, it positions itself in the region of vaporous and solid phases. 

Clearly, it condenses into the solid-state directly without passing through the liquid state.

3. Describe qualitatively the changes in a given mass of solid $C{{O}_{2}}$ at 10 atm pressure and temperature $-65{}^\circ C$as it is heated up to room temperature at constant pressure.

Ans: If the temperature of a mass of solid $C{{O}_{2}}$ (at 10 atm pressure and at $-65{}^\circ C$ ) is increased, it changes to the liquid phase and then to the vaporous phase.

It projects a line parallel to the temperature axis at 10 atm. 

Clearly, the fusion and boiling points are provided by the intersection point where this parallel line cuts the fusion and vaporization curves.

4. $C{{O}_{2}}$is heated to a temperature $70{}^\circ C$ and compressed isothermally. What changes in its properties do you expect to observe?

Ans: When $C{{O}_{2}}$ is heated to $70{}^\circ C$ and compressed isothermally, it would not show any transition to the liquid state. 

This is because $70{}^\circ C$is higher than the critical temperature of $C{{O}_{2}}$. 

Clearly, it would remain in the vapor state but would depart from its ideal behavior as pressure is increased.

18. A child running a temperature of $101{}^\circ F$  is given an antipyrine (i.e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to  $98{}^\circ F$in 20 min, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of the human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal ${{g}^{-1}}$ .

Ans: Given that,

Initial temperature of the body of the child, ${{T}_{1}}=101{}^\circ F$ 

Final temperature of the body of the child, ${{T}_{2}}=98{}^\circ F$

Change in temperature, $\Delta T=\left[ \left( 101-98 \right)\times \frac{5}{9} \right]{}^\circ C$ 

Time taken to reduce the temperature, t = 20 min 

Mass of the child, $m=30kg=30\times {{10}^{3}}g$ 

Specific heat of the human body = Specific heat of water $=c=100cal/kg/{}^\circ C$ 

Latent heat of evaporation of water,  $L=580cal$ ${{g}^{-1}}$ 

The heat lost by the child is given as:

$\Delta \theta =mc\Delta T$

$\Rightarrow \Delta \theta =30\times 1000\times \left( 101-98 \right)\times \frac{5}{9}=50000cal$

Let ${{m}_{1}}$  be the mass of the water evaporated from the child’s body in 20 min. 

Loss of heat through water is given by:

$\Delta \theta ={{m}_{1}}L$

$\Rightarrow {{m}_{1}}=\frac{\Delta \theta }{L}$

$\Rightarrow {{m}_{1}}=\frac{50000}{580}=86.2g$

Therefore, average rate of extra evaporation caused by the drug $=\frac{{{m}_{1}}}{t}$ 

$\Rightarrow \frac{{{m}_{1}}}{t}=~~\frac{86.2}{200}=4.3g/\min $

19. A ‘thermocol’ ice box is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of $5.0$ cm. If $4.0$  kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is$45{}^\circ C$ , and coefficient of thermal conductivity of thermocol is 0.01 J ${{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$ . [Heat of fusion of water $=335\times {{10}^{3}}Jk{{g}^{-1}}$]

Ans: Given that,

Side of the given cubical ice box, $s=30cm=0.3m$ 

Thickness of the ice box, $l=5.0cm=0.05m$ 

Mass of ice kept in the ice box, m = 4 kg

Time gap, $t=6h=6\times 60\times 60s$

Outside temperature, $T=45{}^\circ C$ 

Coefficient of thermal conductivity of thermacole, $K=0.01J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$   

Heat of fusion of water,  $L=335\times {{10}^{3}}Jk{{g}^{-1}}$ 

Assume ${m}'$ be the total amount of ice that melts in 6 h. 

The amount of heat lost by the food:

$\theta =\frac{KA\left( T-0 \right)t}{l}$ 

where,

$A=$ Surface of the box $=6{{s}^{2}}=6\times {{\left( 0.3 \right)}^{2}}=0.54{{m}^{3}}$ 

$\Rightarrow \theta =\frac{0.01\times 0.54\times 45\times 6\times 60\times 60}{0.05}=104976J$ 

However, $\theta ={m}'L$ 

$\Rightarrow {m}'=\frac{\theta }{L}$

$\Rightarrow {m}'=\frac{104976}{335\times {{10}^{3}}}=0.313g$

Mass of ice left $4-0.313=3.687kg$ 

Clearly, the amount of ice remaining after 6h is 3.687kg.

20. A brass boiler has a base area of $0.15{{m}^{2}}$ and thickness $1.0$cm. It boils water at the rate of $6.0kg/\min $  when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. 

Thermal conductivity of brass $=109J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$  ;

Heat of vaporization of water $=2256\times {{10}^{3}}Jk{{g}^{-1}}.$ 

Ans: Given that,

Base area of the boiler, $A=0.15{{m}^{2}}$ 

Thickness of the boiler, \[l=1.0\text{ }cm=0.01\text{ }m\] 

Boiling rate of water, R = $6.0kg/\min $ 

Mass, m = 6 kg 

Time, t = 1 min = 60 s 

Thermal conductivity of brass, K= $=109J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$ 

Heat of vaporization, L$=2256\times {{10}^{3}}Jk{{g}^{-1}}.$

The amount of heat flowing into water through the brass base of the boiler is given by:

$\theta =\frac{KA\left( {{T}_{1}}-{{T}_{2}} \right)t}{l}$

Where, 

${{T}_{1}}$ = Temperature of the flame in contact with the boiler 

${{T}_{2}}$ = Boiling point of water $=100{}^\circ C$ 

Heat required for boiling the water: 

$\theta =mL$ 

Equating equations (a) and (b), we get:

$mL=\frac{KA\left( {{T}_{1}}-{{T}_{2}} \right)t}{l}$ 

$\Rightarrow \left( {{T}_{1}}-{{T}_{2}} \right)=\frac{mLl}{KAt}$ 

$\Rightarrow \left( {{T}_{1}}-{{T}_{2}} \right)=\frac{6\times 2256\times {{10}^{3}}\times 0.001}{109\times 0.15\times 60}=137.98{}^\circ C$

Clearly, the temperature of the part of the flame in contact with the boiler is $137.98{}^\circ C$.

21. Explain why:

1. a body with large reflectivity is a poor emitter

Ans: It is known that a body possessing a large reflectivity turns out to be a poor absorber of light radiations. 

A poor absorber would further be a poor emitter of radiation. 

Clearly, a body possessing large reflectivity is a poor emitter.

2. a brass tumbler feels much colder than a wooden tray on a chilly day

Ans: It is known that brass is a good conductor of heat. 

The moment at which one touches a brass tumbler, heat gets conducted from the body to the brass tumbler easily. 

Thus, the temperature of the body lessens to a lower value and one feels cooler.

On the other hand, the moment at which one touches a wooden tray, very little heat is conducted from the body to the wooden tray because wood is a poor conductor of heat. 

Thus, there is only a negligible drop in the temperature of the body and one does not feel cool.

Clearly, a brass tumbler feels colder than a wooden tray on a chilly day.

3. An optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red-hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.

Ans: An optical pyrometer calibrated for an ideal black body radiation provided too low a value for temperature of a red-hot iron piece kept in the open. 

Black body radiation equation is given by: 

$E=\sigma \left( {{T}^{4}}-T_{0}^{4} \right)$ 

Where, 

E = Energy radiation 

T = Temperature of optical pyrometer 

${{T}_{0}}$  = Temperature of open space 

$\sigma =$ Constant

Thus, an increase in the temperature of open space lessens the radiation energy. If the same piece of iron is placed in a furnace, the radiation energy, \[E=\sigma {{T}^{4}}\].

4. the earth without its atmosphere would be inhospitably cold

Ans: It is known that excluding its atmosphere, earth would be inhospitably cold. 

When atmospheric gases are absent, no extra heat would be trapped. 

Clearly, all the heat would be radiated back from the earth’s surface.

5. heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Ans: A heating system on the basis of the circulation of steam would be more efficient in warming a building than that on the basis of the circulation of hot water. 

This can be reasoned as steam contains surplus heat in the form of latent heat $\left( 540cal/g \right)$. 

22. A body cools from $80{}^\circ C$ to $50{}^\circ C$  in 5 minutes. Calculate the time it takes to cool from $60{}^\circ C$to $30{}^\circ C$. The temperature of the surroundings is$20{}^\circ C$.

Ans: With respect to Newton’s law of cooling;

$-\frac{dT}{dt}=K\left( T-{{T}_{0}} \right)$ 

$\Rightarrow \frac{dT}{K\left( T-{{T}_{0}} \right)}=-Kdt$

Where, 

Temperature of the body = T 

Temperature of the surroundings $={{T}_{S}}=20{}^\circ C$ 

K is a constant 

Temperature of the body falls from $80{}^\circ C$ to $50{}^\circ C$ in time, t = 5 min = 300 s 

Integrating equation (a), we get:

$\int\limits_{50}^{80}{\frac{dt}{K\left( T-{{T}_{0}} \right)}=-}\int\limits_{0}^{300}{Kdt}$

$\Rightarrow \left[ {{\log }_{e}}\left( T-{{T}_{0}} \right) \right]_{50}^{80}=-K\left[ t \right]_{0}^{300}$ 

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}\frac{80-20}{50-20}=-300$ 

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}2=-300$ 

$\Rightarrow -\frac{2.3026}{300}{{\log }_{10}}2=K$ 

The temperature of the body falls from $60{}^\circ C$ to $30{}^\circ C$ in time = t

Thus, we get:

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}\frac{60-20}{30-20}=-t$

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}4=K$

Equating equations (b) and (c), we get:

$\frac{-2.3026}{K}{{\log }_{10}}4=\frac{2.3026}{K}{{\log }_{10}}2$

$\Rightarrow t=300\times 2=600s=10\min $ 

Clearly, the time taken to cool the body from $60{}^\circ C$ to $30{}^\circ C$ is 10 minutes.

NCERT Physics Class 11 Thermal Properties of Matter - Free PDF download

Class 11th Physics Chapter 10 talks about the effects of heat and temperature on various elements. In this chapter, students will get to know about the concept of heat, its measurement process, and how it transfers from the body of one matter to another.

During the study of this chapter, pupils will also come across real-life examples like why blacksmiths heat iron rings before fitting them on wooden wheels and the reason why beach wind changes direction after sunset, etc.

NCERT solutions available for Class 11 Physics is a vital study material for students appearing for their Class 11 exams. It simplifies the concepts and helps students grasp the topics quickly. As a result, they can learn the basics as well as advance concepts effortlessly.

Chapter 10 Physics Class 11 NCERT solutions are available for free online. Students can quickly download the pdf file for this chapter given below on this page and improve their final preparations.

Important Topics Covered in the NCERT Solutions Class 11 Physics Chapter 10 Thermal Properties of Matter

This chapter will help you to know what is heat, what is temperature, how to measure temperature, how the temperature changes the atmosphere, how every physical property responds to temperature and heat and many other information in detail. 

Important Formulas in  Class 11 Physics Chapter 10 Thermal Properties of Matter 

Here is the list of important formulae that you must remember to solve the numerical problems related to the chapter - Thermal Properties of Matter.

• Coefficient of linear thermal expansion = l1- l0 l0l

• Heat capacity C = Q/t

where, Q is the amount of heat observed

And it is the rise in temperature.

• Newton’s law of cooling dTdt = bA(T1 - T2)

• Wein’s displacement law = mT, where b = 0.2896 *10-2 mk

• Principle of calorimeter → Heat lost = Heat gain. 

Students who are appearing for the board exam and competitive exams like NEET, JEE and other state board exams can utilize the PDF of important formulas to brush up on your knowledge. 

NCERT Solutions Class 11 Physics Thermal Properties of Matter - Topic-Wise Discussion

Thermal properties of Matter Class 11 NCERT has eleven topics, and together they offer a detailed understanding of the Thermal Properties of Matter. Here is a topic-wise summary to provide you with an outline of this chapter.

1. Introduction

The chapter begins with a general introduction and helps students to get an overview of what they will learn throughout this particular chapter.

2. Temperature and Heat

The next section of NCERT solutions for Class 11 Physics Chapter 10  talks about the definition of heat and temperature. In this, you will learn the units used to measure hot and cold.

3. Measurement of Temperature

This section of Thermal Properties of Matter Class 11 NCERT solutions explains the process of measuring temperature and instruments used for that. It also discusses in detail about the thermometer, and its various scales.

4. Ideal Gas Equation and Absolute Temperature

This section of Chapter 10, talks about various properties of a gas, and how thermometers that use gas can perform better than the liquid-filled ones. It also explains the variables that describe the behavior of a gas. The ideal gas equation, along with theories of different scientists is also a part of the discussion in this chapter.

5. Thermal Expansion

Thermal expansion is one of the most important sections of ch 10 Physics Class 11; it describes how various elements expand under the influence of heat. A real-life example mentioned here is that hot water is poured over metal lids in case it gets tightened. The reason here is that this hot water expands the metal and makes it easy to open.

6. Specific Heat Capacity

Every item has a particular boiling point, and they react differently when they absorb thermal energy. For instance, a bowl of water under heat starts to pop bubbles, until it becomes turbulent when the water starts boiling. This concept of heat capacity is discussed here in-depth, and students can learn about its various aspects.

7. Calorimetry

Calorimetry stands for the measurement of heat, and this segment of Physics Class 11 Chapter 10 NCERT solutions discuss that concept in detail. With the help of real-life examples used here, students can quickly grasp this topic.

8. Change of State

There are three states to every matter, solid, liquid and gas. The transition of an element from one state to another is called change of state, and heat plays a pivotal role here. In this segment, students will also learn about the concept of triple point, latent heat and conduction.

9. Heat Transfer

The transfer of heat is a common occurrence, and this ninth topic of Chapter 10 explains this regular event further. The concepts of radiation, convection and others are a part of the discussion here.

10. Newton’s Law of Cooling

Every item, when left without any external influence, cools down gradually, and this chapter of thermal properties of matter Class 11 Solutions go on further to explain in detail.

NCERT Chapter 10 Physics Class 11 Marks Weightage

The final physics examination will be worth 100 marks. Out of these 100 marks, 30 marks will be allotted to the practical and the remaining 70 marks will be for the theoretical examination. Students should solve all NCERT theory questions and numericals before taking the final examination. This will help students to score the highest marks in their respective classes.

Why NCERT Solutions of Thermal Properties of Matter is a Must-Read?

NCERT solutions for class 11 physics chapter thermal properties of matter helps students to score better because of the following benefits –

• It's easy language and lucid explanation help students to understand any concept easily.

• Also, the in-depth explanation helps students to learn more about a particular topic.

• NCERT solutions for Class 11 Physics Chapter 10 uses bullet points to make topics more visible and evokes a more engaging learning experience.

• The use of real-life examples makes the topic easy to relate, and subsequently, students can grasp a topic better.

• The exercise questions are answered in the same way as students need to answer during an exam. It will help them prepare better for their test.

• Class 11 Physics NCERT solutions chapter thermal properties of matter follow the guidelines of CBSE. Hence, following it will not deviate students from their exam preparations and help them cover the syllabus better.

The use of NCERT solutions for Class 11 physics Chapter 10 can assist students in performing better in their final exam as well as their internals. Additionally, these books are free to download, and one can use Vedantu’s official mobile app to find them easily.

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