NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

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NCERT Solutions Class 11 Physics Chapter 11 – Access and Download For Free

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter: If you're a Class 11 student seeking thermal properties of matter exercise solutions, you're in the right place for NCERT solutions. On this Careers360 page, you can access complete exercise solutions, covering questions from 11.1 to 11.20 in the main exercise, and questions 11.21 to 11.22 in the additional exercise. These accurate and precise class 11 physics chapter 11 exercise solutions are available for download in PDF format, providing convenient access for students. The answers have been crafted by the best subject matter experts, ensuring their reliability.

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Temperature is a relative measure of indication of coldness (or hotness) while heat is the form of energy transferred between two or more systems or a system and its surroundings due to the temperature difference. Thermal Properties of Matter Class 11 solutions of NCERT Physics Chapter start with the questions related to temperature conversions from one scale to another. All the questions discussed in the CBSE NCERT solutions for Class 11 Physics chapter 11 Thermal Properties of Matter are important to understand the topics covered in the chapter. ch 11 class 11 physics ncert solutions are an important tool to perform well in exams.

** This chapter has been renumbered as Chapter 10 in accordance with the CBSE Syllabus 2023–24.

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CERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

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Access NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter - Exercise Solution

Q. 11.1 The triple points of neon and carbon dioxide are $24.57\; K$ and $216.55\; K$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer:

The relation between Kelvin and Celcius scale is T _K = T _C + 273.15

Triple Point of Neon in Kelvin T _K = 24.57 K

Triple Point of Neon in Celcius T _C = T _K -273.15 = 24.57 -273.15 =-248.58 ^o C

Triple Point of carbon dioxide in Kelvin T _K = 216.55 K

Triple Point of carbon dioxide in Celcius T _C = T _K -273.15 = 216.55 - 273.15 = -56.60 ^o C

The relation between Celcius and Fahrenheit scale is $T_{F}=\frac{9}{5}T_{C} + 32$

Triple Point of Neon in Fahrenheit is

$T_{F}=\frac{9}{5}\times (-248.58) + 32$

T _F = -415.44 ^o C

Triple Point of carbon dioxide in Fahrenheit is

$T_{F}=\frac{9}{5}\times (-56.60) + 32$

T _F = -69.88 ^o C

Q. 11.2 Two absolute scales $A$ and $B$ have triple points of water defined to be $200A$ and $350B$ . What is the relation between $T_{A}$ and $T_{B}$ ?

Answer:

200 A = 273 K

$T_{K}=\frac{273}{200}T_{A}$

350 B = 273 K

$T_{K}=\frac{273}{350}T_{B}$

Equating T _K From the above two equations we have

$\\\frac{273}{200}T_{A}=\frac{273}{350}T_{B}\\ T_{A}=\frac{4}{7}T_{B}$

Q. 11.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :

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

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 $(600.5\; K).$ What is the temperature when the resistance is $123.4\; \Omega$ ?

Answer:

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

R ₀ = $101.6\Omega$ T ₀ = $273.16 K$

R = $165.5\; \Omega$ R = $600.5 K$

Putting the above values in the given equation we have

$\\\alpha =\frac{165.5-101.6}{600.5-27316}\\ \alpha =1.92\times 10^{-2}\ K^{-1}$

For R = $123.4\; \Omega$

$\\T=T_{0}+\frac{1}{\alpha }\left ( \frac{R}{R_{0}}-1 \right )\\ T=273.16+\frac{1}{1.92\times 10^{-3}}\left ( \frac{123.4}{101.6}-1 \right )\\ T=384.75K$

Q. 11.4 (a) Answer the following :
(a) 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)?

Answer:

Unlike the melting point of ice and boiling point of water, the triple point of water has a fixed value of 273.16 K. The melting point of ice and boiling point of water vary with pressure.

Q. 11.4 (b) Answer the following :

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^{\circ}C$ 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?

Answer:

The other fixed point on the Kelvin scale is 0 K. 0K is the absolute zero

Q. 11.4 (c) Answer the following :

$t_{c}=T-273.15$

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

Answer:

This is because 0 ^o C on the Celcius scale corresponding to the melting point at standard pressure is equal to 273.15 K whereas 273.16 K is the triple point of water. The triple point of water is 0.01 ⁰ C , not 0 ⁰ C

Q. 11.4 (d) Answer the following :

(d) 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?

Answer:

Let at a certain temperature the reading on Fahrenheit and Kelvin Scale be T _F and T _K respectively

$T_{F}-32=\frac{9}{5}(T_{K}-273)$ $(i)$

Let at another temperature the reading on Fahrenheit and Kelvin Scale be T' _F and T' _K respectively

$T'_{F}-32=\frac{9}{5}(T'_{K}-273)$ $(ii)$

Subtracting equation (ii) from (i)

$T_{F}-T'_{F}=\frac{9}{5}(T_{K}-T'_{K})$

For T _K - T' _K = 1 K, T _F - T' _F = 9/5

Therefore corresponding to 273.16 K the absolute scale whose unit interval size is equal to that of the Fahrenheit scale

$\\T_{F}=\frac{9}{5}\times 273.16\\ T_{F}=491.688$

Q. 11.5 (a) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

Temperature	Pressure Thermometer A	Pressure Thermometer B
Triple-point of water	$1.250\times 10^{5}Pa$	$0.200\times 10^{5}Pa$
Normal melting point of sulphur	$1.797\times 10^{5}Pa$	$0.287\times 10^{5}Pa$

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

Answer:

$\\PV=nRT\\$

$\frac{P}{T}=\frac{nR}{V}$

As the moles of oxygen and hydrogen inside the thermometers and the volume occupied by the gases remain constant P/T would remain constant.

The triple point of water(T ₁ ) = 273.16 K

Pressure in thermometer A at a temperature equal to the triple point of water (P ₁ ) = $1.250\times 10^{5}Pa$

Pressure in thermometer A at a temperature equal to Normal melting point of sulphur (P ₂ ) = $1.797\times 10^{5}Pa$

The normal melting point of sulphur as read by thermometer A T ₂ would be given as

$T_{2}=\frac{P_{2}T_{1}}{P_{1}}$

$\\T_{2}=\frac{1.797\times 10^{5}\times 273.16}{1.250\times 10^{5}}\\ T_{2}=392.69\ K$

Pressure in thermometer B at a temperature equal to the triple point of water (P' ₁ ) = $0.200\times 10^{5}Pa$

Pressure in thermometer B at a temperature equal to Normal melting point of sulphur (P' ₂ ) = $0.287\times 10^{5}Pa$

The normal melting point of sulphur as read by thermometer B T' ₂ would be given as

$\\T'_{2}=\frac{P'_{2}T_{1}}{P'_{1}}\\ T'_{2}=\frac{0.287\times 10^{5}\times 273.16}{0.200\times 10^{5}}\\ T'_{2}=391.98\ K$

Q. 11.5 (b) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

Temperature	Pressure Thermometer A	Pressure Thermometer B
Triple-point of water	$1.250\times 10^{5}Pa$	$0.200\times 10^{5}Pa$
Normal melting point of sulphur	$1.797\times 10^{5}Pa$	$0.287\times 10^{5}Pa$

(b) 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?

Answer:

The slight difference in answers of thermometers A and B occur because the gases used in the thermometers are not ideal gases. To reduce this discrepancy the experiments should be carried out at low pressures where the behaviour of real gases tend close to that of ideal gases.

Q. 11.6 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 $=1.20\times 10^{-5}K^{-1}.$

Answer:

At 27 ^o C the 63 cm (l ₁ ) mark on the steel tape would be measuring exactly 63 cm as the tape is calibrated at 27 ^o C

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

Actual length when the scale is giving a reading of 63 cm on at 45 ^o C is l ₂

$\\l_{2}=l_{1}(1+\alpha \Delta T)\\ =63\times (1+1.20\times 10^{-5}\times (45-27))\\ =63.013608cm$

The actual length of the steel rod on a day when the temperature is 45 ^o C is 63.013608 cm.

Length of the same steel rod on a day when the temperature is 63 cm.

Q. 11.7 A large steel wheel is to be fitted on to 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 : $\alpha _{steel}=1.20\times 10^{-5}K^{-1}$ .

Answer:

Diameter of the steel shaft at 27 ^o C (T ₁ ) d ₁ = 8.70 cm

The diameter of the central hole in the wheel d ₂ = 8.69 cm

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

The wheel will slip on the shaft when the diameter of the steel shaft becomes equal to the diameter of the central hole in the wheel.

Let this happen at temperature T

$\\d_{2}=d_{1}(1+\alpha (T-T_{1})\)\\ 8.69=8.7(1+1.2\times 10^{5}(T-27))\\ T=-68.79\ ^{o}C$

Q. 11.8 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 $=1.70\times 10^{-5}K^{-1}.$

Answer:

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

Coefficient of superficial expansion of copper is $\beta$

$\\\beta =2\alpha \\ \beta =2\times 1.7\times 10^{-5}\\ \beta =3.4\times 10^{-5}K^{-1}$

Diameter of the hole at 27 ^o C (d ₁ ) = 4.24 cm

Area of the hole at 227 ^o C is

$\\A_{2}=A_{1}(1+\beta \Delta T)\\ A_{2}=\pi \left ( \frac{4.24}{2} \right )^{2}(1+3.4\times 10^{-5}(227-27))\\ A_{2}=14.215\ cm^{2}$

Let the diameter at 227 ^o C be d ₂

$\\\pi \left ( \frac{d_{2}}{4} \right )^{2}=14.215\\ d_{2}=4.254cm$

Change in diameter is d ₂ -d ₁ = 4.24 -4.254 = 0.014 cm.

Q. 11.9 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 $2.0\; mm\; ?$ ? Co-efficient of linear $expansion\; of \; brass=2.0\times 10^{-5}K^{-1};$ $Young's\; modulus \; of\; brass=0.91\times 10^{11}Pa.$

Answer:

Youngs Modulus of Brass, $Y=0.91\times 10^{11}$

Co-efficient of linear expansion of Brass, $\alpha =2.0\times 10^{-5}K^{-1}$

The diameter of the given brass wire, d = 2.0 mm

Length of the given brass wire, l = 1.8 m

Initial Temperature T ₁ = 27 ^o C

Final Temperature T ₂ = -39 ^o C

$\\Y=\frac{Stress}{Strain}\\ Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\\ F=\frac{\Delta lAY}{l}\\ F=\frac{(l\alpha \Delta T)AY}{l}\\ F=\alpha \Delta TAY\\ F=\alpha \Delta TY\pi \frac{d^{2}}{4}\\ F=\frac{2.0\times 10^{-5}\times (-39-27)\times 0.91\times 10^{11}\times \pi \times (2\times 10^{-3})^{2}}{4}$

$F=-378N$

The tension developed in the wire is 378 N. The negative sign signifies this tension is inward.

Q. 11.10 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 $40.0^{\circ}C\; ?$ Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear $expansion\; of\; brass=2.0\times 10^{-5}K^{-1},\; steel=1.2\times 10^{-5}K^{-1}$ ).

Answer:

Length of the rods l = 50 cm

Co-efficient of linear expansion of brass, $\\\alpha _{b}=2\times 10^{-5}K^{-1}$

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

Initial Temperature T ₁ = 40.0 ^o C

Final Temperature T ₂ = 250 ^o C

Change in length of brass rod is

$\\\Delta l_{t}=l_{t}\alpha_{b} \Delta T\\ \Delta l_{t}=50\times 2.0\times 10^{-5}\times (250-40)\\ \Delta l_{t}=0.21cm$

Change in length of the steel rod is

$\\\Delta l_{s}=l_{s}\alpha_{s} \Delta T\\ \Delta l_{s}=50\times 1.2\times 10^{-5}\times (250-40)\\ \Delta l_{s}=0.126cm$

Change in length of the combined rod is

$\\\Delta l=\Delta l_{s}+\Delta l_{b}\\ \Delta l=0.126+0.21\\ \Delta l=0.336cm$

Q. 11.11 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?

Answer:

Coefficient of volume expansion of glycerine is $\gamma =49\times 10^{-5}K^{-1}$

Let initial volume and mass of a certain amount of glycerine be V and m respectively.

Initial density is

$\rho =\frac{m}{V}$

Change in volume for a 30 ^o C rise in temperature will be

$\\\Delta V=V(\gamma \Delta T)\\ \Delta V=V(49\times 10^{-5}\times 30)\\ \Delta V=0.0147V$

Final Density is

$\\\rho' =\frac{m}{V+\Delta V}\\ \rho'=\frac{m}{1.0147V}\\ \rho'=\frac{0.986m}{V}$

Fractional Change in density is

$\\\frac{\rho'-\rho}{\rho}\\ =\frac{\frac{0.986m}{v}-\frac{m}{v}}{\frac{m}{v}}\\ =-0.014$

The negative sign signifies with an increase in temperature density will decrease.

Q. 11.12 A $10\; kW$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0\; kg.$ How much is the rise in temperature of the block in $2.5$ minutes, assuming $50^{o}/_{o}$ of power is used up in heating the machine itself or lost to the surrounding. $Spcific\; heat\; of\; aluminium=0.91\; j\; g^{-1}K^{-1}.$

Answer:

Power of the drilling machine, P = 10 kW

Time. t = 2.5 min

Total energy dissipated E is

$\\E=Pt\\ E=10\times 10^{3}\times 2.5\times 60\\ E=1.5\times 10^{6}J$

Thermal energy absorbed by aluminium block is

$\\\Delta Q=\frac{E}{2}\\ \Delta Q=\frac{1.5\times 10^{6}}{2}\\ \Delta Q=7.5\times 10^{5}J$

Mass of the aluminium block, m = 8.0 kg

Specific heat of aluminium, c = 0.91 J g ^-1 K ^-1

Let rise in temperature be $\Delta T$

$\\mc\Delta T=\Delta Q\\ \Delta T=\frac{\Delta Q}{mc}\\ \Delta T=\frac{7.5\times 10^{5}}{0.91\times 8\times 10^{3}}\\ \Delta T=103.02\ ^{o}C$

Q. 11.13 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 $=0.39\; j\; g^{-1}K^{-1};$ heat of fusion of water $=335\; j\; g^{-1}$ ).

Answer:

Mass of copper block m = 2.5 kg

Initial Temperature of the copper block, T ₁ = 500 ^o C

Final Temperature of Copper block, T ₂ = 0 ^o C

Specific heat of copper, c = 0.39 J g ^-1 K ^-1

Thermal Energy released by the copper block is $\Delta Q$

$\\\Delta Q=mc\Delta T\\ \Delta Q=2.5\times 10^{3}\times 0.39\times 500\\ \Delta Q=487500\ J$

Latent heat of fusion of water, L = 335 j g ^-1

Amount of ice that can melt is

$\\w=\frac{\Delta Q}{L}\\ w=\frac{487500}{335}\\ w=1455.22g$

1.455 kg of ice can melt using the heat released by the copper block.

Q. 11.14 In an experiment on the specific heat of a metal, a $0.20\; kg$ block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent $0.025\; kg$ ) containing $150\; cm^{3}$ 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?

Answer:

Let the specific heat of the metal be C.

Mass of metal block m = 200 g

Initial Temperature of metal block = 150 ^o C

Final Temperature of metal block = 40 ^o C

The heat released by the block is

$\\\Delta Q=mc\Delta T\\ \Delta Q=200\times c\times (150-40)\\ \Delta Q=22000c$

Initial Temperature of the calorimeter and water = 27 ^o C

Final Temperature of the calorimeter and water = 40 ^o C

Amount of water = 150 cm

Mass of water = 150 g

Water equivalent of calorimeter = 25 g

Specific heat of water = 4.186 J g ^-1 K ^-1

Heat absorbed by the Calorimeter and water is $\\\Delta Q'$

$\\\Delta Q'=(150+25)\times 4.186\times (40-27)\\ \Delta Q'=9523.15J$

The heat absorbed by the Calorimeter and water is equal to the heat released by the block

$\\\Delta Q=\Delta Q'\\ 22000c=9523.15\\ c=0.433\ J\ g^{-1}\ K^{-1}$

The above value would be lesser than the actual value since some heat must have been lost to the surroundings as well which we haven't accounted for.

Q. 11.15 Given below are observations on molar specific heats at room temperature of some common gases.

Gas Molar specific heat (Cv )

$(cal\; mol^{-1}K^{-1})$

Hydrogen 4.87

Nitrogen 4.97

Oxygen 5.02

Nitric oxide 4.99

Carbon monoxide 5.01

Chlorine 6.17

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?

Answer:

Monoatomic gases have only translational degree of freedom but diatomic gases have rotational degrees of freedom as well. The temperature increases with increase in the spontaneity of motion in all degrees. Therefore to increase the temperature of diatomic gases more energy is required than that required to increase the temperature of monoatomic gases by the same value owing to higher degrees of freedom in diatomic gases.

If we only consider rotational modes of freedom the molar specific heat of the diatomic gases would be given as

$\\c=\frac{fR}{2}\\ c=\frac{5}{2}\times 1.92\\ c=4.95\ cal\ mol^{-1}\ K^{-1}$

The number of degrees of freedom = 5 (3 translational and 2 rotational)

The values given in the table are more or less in accordance with the above calculated one. The larger deviation from the calculated value in the case of chlorine is because of the presence of vibrational motion as well.

Q. 11.16 A child running a temperature of $101^{\circ}F$ is given an antipyrin (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 minutes, 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 human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about $580\; cal\; g^{-1}$ .

Answer:

Initial Temperature of the boy = 101 ^o F

Final Temperature of the boy = 98 ^o F

Change in Temperature is

$\\\Delta T=3\ ^{o}F\\ \Delta T=3\times \frac{5}{9}\\ \Delta T=1.67\ ^{o}C$

Mass of the child is m = 30 kg

Specific heat of human body = 1000 cal kg ^-1 ^o C ^-1

Heat released is $\\\Delta Q$

$\\\Delta Q=mc\Delta T\\\Delta Q=30\times 1000\times 1.67\\ \Delta Q=50000\ cal$

Latent heat of evaporation of water = 580 cal g ^-1

The amount of heat lost by the body of the boy has been absorbed by water.

Let the mass of water which has evaporated be m'

$\\\Delta Q=m'L\\ m'=\frac{Q}{L}\\ m'=\frac{50000}{580}\\ m'=86.2\ g$
Time in which the water has evaporated, t = 20 min.

Rate of evaporation is m'/t

$\\\frac{m'}{t}=\frac{86.2}{20}\\ \frac{m'}{t}=4.31\ g\ min^{-1}$

Q. 11.17 A ‘thermacole’ icebox is a cheap and an 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 co-efficient of thermal conductivity of thermacole is $0.01\; j\; s^{-1}m^{-1}K^{-1}.$ $[Heat\; of\; fusion\; of\; water=335\times 103\; j\; kg^{-1}]$

Answer:

Side of the box s = 30 cm

Area available for conduction A

A = 6s ²

A=6(30) ²

A=5400 cm ² = 0.54 m ²

Temperature difference = 45 ^o C

Co-efficient of thermal conductivity of thermacole is k = 0.01 J s ^-1 m ^-1 K ^-1

Width of the box is d = 5 cm

Heat absorbed by the box in 6 hours is $\Delta Q$

$\\\Delta Q=\frac{kA\Delta T}{l}\\ \Delta Q=\frac{0.01\times 0.54\times 45\times 6\times 60\times 60}{0.05}\\ \Delta Q=104976\ J$

The heat of fusion of water is $L=335\times 10^{3}\ J\ kg^{-1}$

Amount of ice which has melted is m'

$\\m'=\frac{\Delta Q}{L}\\ m'=\frac{104976}{335\times 10^{3}}\\ m'=0.313\ kg$

Amount of ice left after 6 hours = 4 - 0.313 = 3.687 kg

Q. 11.18 A brass boiler has a base area of $0.15\; m^{2}$ and thickness $1.0\; cm.$ It boils water at the rate of $6.0 \; kg/min$ when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass $=109\; j\; s^{-1}m^{-1}K^{-1};$ Heat of vaporisation of water $=2256\times 103\; j\; kg^{-1}.$

Answer:

The rate at which water boils, R = 6.0 kg min ^-1

The heat of vaporisation of water, $L=2256\times 10^{3}\ J\ kg^{-1}$

The rate at which heat enters the boiler

$\\\frac{dQ}{dT}=RL\\ \frac{dQ}{dT}=\frac{6\times 2256\times 10^{3}}{60}\\ \frac{dQ}{dT}=2.256\times 10^{5}Js^{-1}$

The base area of the boiler, A = 0.15 m ²

Thickness, l = 1.0 cm

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

The temperature inside the boiler = Boiling point of water = 100 ^o C

Let the temperature of the flame in contact with the boiler be T

Amount of heat flowing into the boiler is

$\\\frac{dQ}{dt}=\frac{KA\Delta T}{l}\\ 2.256\times 10^{5}=\frac{109\times 0.15\times (T-100)}{1\times 10^{-2}}\\ T-100=137.98\\ T=237.98\ ^{o}C$

The temperature of the flame in contact with the boiler is 237.98 ^o C

Q. 11.19 (a) Explain why :

(a) a body with large reflectivity is a poor emitter

Answer:

A body with a large reflectivity is a poor absorber. As we know a body which is a poor absorber will as well be a poor emitter. Therefore a body with large reflectivity is a poor emitter.

Q .11.19 (b) Explain why :

(b) a brass tumbler feels much colder than a wooden tray on a chilly day

Answer:

Brass is a good conductor of heat. Therefore once someone touches brass heat from their body flows into it and it feels cold, in case of a wooden tray, no such conduction of heat from the body takes place as wood is a very poor conductor of heat.

Q. 11.19 (c) Explain why :

(c) 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

Answer:

An optical pyrometer relates the brightness of a glowing body with its temperature. In the open because of other sources of light the sensor in the optical pyrometer does not detect the true brightness of a red hot piece of iron and thus does not predict its temperature correctly whereas in the furnace the piece of iron is the only source of light and the sensor detects its brightness correctly thus giving the correct value of the temperature.

Q. 11.19 (d) Explain why :

(d) the earth without its atmosphere would be inhospitably cold

Answer:

The sun rays contain infrared radiations. These are reflected back by the lower part of the atmosphere after being reflected by the surface of the earth and are trapped inside the atmosphere thus maintaining the Earth's temperature at a hospitable level. Without these rays being trapped the temperature of the earth will go down severely and thus the Earth without its atmosphere would be inhospitably cold.

Q. 11.19 (e) Explain why :

(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Answer:

Heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water because the same amount of steam at 100 ^o C contains more energy available for heat dissipation than the same amount of water at 100 ^o C in the form of latent heat of vaporization.

Q. 11.20 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 $20^{\circ}C.$

Answer:

Let a body initially be at temperature T ₁

Let its final Temperature be T ₂

Let the surrounding temperature be T ₀

Let the temperature change in time t.

According to Newton's Law of cooling

$\\-\frac{dT}{dt}=K(T-T_{0})\\ \frac{dT}{T-T_{0}}=-Kdt\\ \int_{T_{1}}^{T_{2}}\frac{dT}{T-T_{0}}=-K\int_{0}^{t}dt\\ \left [ ln(T-T_{0}) \right ]_{T_{1}}^{T_{2}}=-K[t]_{0}^{t}\\ ln\left ( \frac{T_{2}-T_{0}}{T_{1}-T_{0}} \right )=-Kt$

where K is a constant.

We have been given that the body cools from 80 ^o C to 50 ^o C in 5 minutes when the surrounding temperature is 20 ^o C.

T ₂ = 50 ^o C

T ₁ = 80 ^o C

T ₀ = 20 ^o C

t = 5 min = 300 s.

$\\ln\left ( \frac{50-20}{80-20} \right )=-K\times 300\\ K=\frac{ln(2)}{300}\\$

For T ₁ = 60 ^o C and T ₂ = 30 ^o C we have

$\\ln\left ( \frac{30-20}{60-20} \right )=-\frac{ln2}{300}t\\ t=\frac{ln(4)\times 300}{ln(2)} \\t=\frac{2ln(2)\times 300}{ln(2)}\\ t=600\ s\\ t= 10\ min$

The body will take 10 minutes to cool from 60 ^o C to 30 ^o C at the surrounding temperature of 20 ^o C.

NCERT thermal properties of matter class 11 solutions - Additional Exercise Solution

Q. 11.21(a) Answer the following questions based on the P-T phase diagram of carbon dioxide:

(a) At what temperature and pressure can the solid, liquid and vapour phases of $CO_{2}$ co-exist in equilibrium?

Answer:

At the triple point temperature of -56.6 ^o C and pressure 5.11 atm the solid, liquid and vapour phases of $CO_{2}$ co-exist in equilibrium.

Q. 11.21(b) Answer the following questions based on the P-T phase diagram of carbon dioxide:

(b) What is the effect of decrease of pressure on the fusion and boiling point of $CO_{2}$ ?

Answer:

Both fusion and boiling point of CO ₂ decrease with decrease in pressure. This we can see from the solid lines in the P-T phase diagram of CO ₂ .

Q. 11.21 (c) Answer the following questions based on the P-T phase diagram of carbon dioxide:

Answer:

The critical temperature and pressure for CO ₂ are 31.1 ^o C and 73.0 atm respectively. If the temperature exceeds this critical value of temperature CO ₂ would not liquefy no matter how high the pressure is.

Q .11.21 d(a) Answer the following questions based on the P-T phase diagram of carbon dioxide:

(d) Is $CO_{2}$ solid, liquid or gas at

(a) $-70^{\circ}C$ under $1\; atm,$

Answer:

CO ₂ is vapour at -70 ^o C under 1 atm pressure as the point corresponding to this condition lies in vapour region in the given P-T phase diagram of carbon dioxide.

Q .11.21 d (b) Answer the following questions based on the P-T phase diagram of carbon dioxide:

(d) Is $C0_{2}$ solid, liquid or gas at

(b) $-60^{\circ}C$ under $10 \; atm,$

Answer:

CO ₂ is solid at -60 ^o C under 10 atm pressure as the point corresponding to this condition lies in the solid region in the given P-T phase diagram of carbon dioxide.

Q. 11.21 d(c) Answer the following questions based on the P-T phase diagram of carbon dioxide:

(d) Is $CO_{2}$ solid, liquid or gas at

Answer:

CO ₂ is liquid at 15 ^o C under 56 atm pressure as the point corresponding to this condition lies in the liquid region in the given P-T phase diagram of carbon dioxide.

Q. 11.22 (a) Answer the following questions based on the P – T phase diagram of $CO_{2}$ :

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

Answer:

The temperature -60 ^o C lies to the left of the triple point of water i.e. in the region of solid and vapour phases. Once we start compressing CO ₂ at this temperature starting from 1 atm pressure it will directly convert into solid without going through the liquid phase.

Q. 11.22 (b) Answer the following questions based on the P – T phase diagram of $CO_{2}$ :

(b) What happens when $CO_{2}$ at $4$ atm pressure is cooled from room temperature at constant pressure?

ptph

Answer:

At room temperature (27 ^o C) and 4 atm pressure CO ₂ exits in the vapour phase. The pressure 4 atm is less than the pressure at the triple point and therefore points corresponding to all temperatures and this pressure lie in the solid and vapour region. Once we start compressing CO ₂ from room temperature at this constant pressure CO ₂ turns from vapour to solid directly without going through the liquid phase.

Q. 11.22 (c) Answer the following questions based on the P – T phase diagram of $CO_{2}$ :

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

Answer:

At -65 ^o C under 10 atm pressure CO ₂ is in the solid phase. At room temperature (27 ^o C) under 10 atm pressure CO ₂ is in the vapour phase. At 10 atm pressure, CO ₂ can exist in all three phases depending upon the temperature. Therefore as CO ₂ is heated from -65 ^o C to room temperature at a constant pressure of 10 atm it goes from the solid phase to liquid phase and then ultimately it goes into the vapour phase.

Q. 11.22 (d) Answer the following questions based on the P – T phase diagram of $CO_{2}$ :

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

Answer:

70 ^o C is above the critical temperature of CO ₂ . Once CO ₂ is isothermally compressed at this temperature it would not liquefy irrespective of how high the pressure is but at very high pressures CO ₂ will not behave as an ideal gas.

The thermal properties of matter class 11 exercise solution are essential for board exams because questions from this chapter often appear. These solutions are also crucial for competitive exams like JEE and NEET, where a good understanding of thermal properties is frequently tested, impacting students' overall performance and future career opportunities. Also, mastering this chapter is important because it forms the basis for more advanced physics topics and is closely connected to other chapters, particularly those related to thermodynamics.

NCERT solutions for class 11 physics chapter wise

Chapter 1	Physical world
Chapter 2	Units and Measurement
Chapter 3	Motion in a straight line
Chapter 4	Motion in a Plane
Chapter 5	Laws of Motion
Chapter 6	Work, Energy and Power
Chapter 7	System of Particles and Rotational motion
Chapter 8	Gravitation
Chapter 9	Mechanical Properties of Solids
Chapter 10	Mechanical Properties of Fluids
Chapter 11	Thermal Properties of Matter
Chapter 12	Thermodynamics
Chapter 13	Kinetic Theory
Chapter 14	Oscillations
Chapter 15	Waves

Thermal properties of matter class 11 solutions: Important Formulas and Diagrams

Modulus of rigidity

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$Y=\frac{F / A}{\Delta l / l}, B=-V \frac{\Delta P}{\Delta V}, \eta=\frac{F}{A \theta}$

Poisson’s ratio

$\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}=\frac{\Delta D / D}{\Delta l / l}$

Elastic energy

U = 1/2*strain*stress*volume

Heat Capacity

Where:

ΔQ is the amount of heat supplied to the substance and ΔT change in its temperature

Specific heat capacity

These formulas and diagrams are valuable tools for students as they help in solving complex problems related to thermal properties of matter, aiding in a deeper comprehension of the subject and enhancing problem-solving skills

Highlight of NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

Highlights of thermal properties of matter class 11 ncert solutions are given below:

In ch 11 class 11 physics ncert solutions pdf, points are used to frame answers to help understand quickly.
Solution for thermal properties of matter class 11 are derived from the textbook by a highly qualified subject matter expert.
Questions and answers for ch 11 physics class 11 are given as per the latest CBSE Syllabus and guidelines.
thermal properties of matter class 11 ncert pdf links are readily available and easily accessible for free.
class 11 physics chapter 11 exercise solutions also boosts your knowledge and interest in physics. NCERT is the base of your learning and here it's easy to access.

More About Thermal Properties Of Matter Class 11 Physics Chapter 11

Thermal Properties of Matter Class 11 NCERT Topics

11.2 Temperature and heat

11.3 Measurement of temperature

11.4 Ideal-gas equation and absolute temperature

11.5 Thermal expansion

11.6 Specific heat capacity

11.7 Calorimetry

11.8 Change of state

11.9 Heat transfer

11.10 Newton’s law of cooling

Temperature And Heat

The difference between temperature and heat can be well understood from the following graph given in NCERT.

change%20of%20state

Even though heat is increased during the phase change, the temperature remains constant until the phase change has occurred. That is increasing heat always does not imply that there is an increase in temperature.

Dropped Topics –
The topics "Greenhouse Effect" and exercises 11.21 to 11.22 have been excluded from the ch 11 class 11 physics ncert solutions, "Thermal Properties of Matter," implying that they may not be part of the prescribed curriculum or are optional for students.

NCERT solutions for class 11 Subject wise

NCERT solutions for Class 11 Biology

NCERT solutions for Class 11 Maths

NCERT solutions for Class 11 Chemistry

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. How important is the chapter Thermal Properties of Matter for NEET

One question can be expected for NEET exam from the chapter Thermal Properties of Matter. Students can practice more questions from previous year NEET papers.

2. How important is Thermal Properties of Matter Class 11 for JEE Main

One question may be asked from Class 11 Physics chapter 11 Thermal Properties of Matter for JEE Main exam. This is one important chapter for KVPY and NSEP exms. Students can get more problems on the chapter from the NCERT Exemplar.

3. What are the important topics of the chapter Thermal Properties of Matter

Topics covered in NCERT book Class 11 chapter Thermal Properties of Matter are-

Temperature and heat
Measurement of temperature
Ideal-gas equation and absolute temperature
Thermal expansion
Specific heat capacity
Calorimetry
Change of state
Heat transfer
Newton’s law of cooling

4. Can you suggest a reliable source to find the NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter?

One reliable online source to access thermal properties of matter ncert solutions is Careers360. These solutions have been created by subject experts to aid students in their exam preparation. The solutions provide clear and easy-to-understand explanations to help students gain a better conceptual understanding. The solutions are available in PDF format, making it convenient for students to access and save time while finding the correct answers.

5. What is the number of topics covered in the thermal properties of matter class 11 ncert?

There are a total of 10 topics in this chapter.

6. How many questions are present in thermal properties of matter exercise?

In the Class 11 Physics Chapter 11, "Thermal Properties of Matter," exercise, there are typically 22 questions are present, out of 22 questions from 1-20( exercise) and 21-22( additional exercise).

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

NCERT Solutions Class 11 Physics Chapter 11 – Access and Download For Free

CERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

NCERT thermal properties of matter class 11 solutions - Additional Exercise Solution

NCERT solutions for class 11 physics chapter wise

Thermal properties of matter class 11 solutions: Important Formulas and Diagrams

Modulus of rigidity

Poisson’s ratio

Elastic energy

Heat Capacity

Specific heat capacity

Highlight of NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

More About Thermal Properties Of Matter Class 11 Physics Chapter 11

Thermal Properties of Matter Class 11 NCERT Topics

Temperature And Heat

NCERT solutions for class 11 Subject wise

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions