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

 

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter: 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. The solutions of NCERT class 11 physics chapter 11 thermal properties of matter 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 concept studied in the chapter. NCERT solutions are an important tool to perform well in exams. Consider a day in the winter season. Let the temperature at 6 am is 5 degree and 9 am be 10 degrees. We will say that 6 am is much colder than 9 am or we can say 9 am is hotter than 6 am which means cold and hot are relative terms.  We know that a glass of ice left on a table on a hot day eventually warms up whereas a cup of hot coffee on the same table cools down. It means that when the temperature of the body and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium until the body and the surrounding medium are at the same temperature. So hot coffee becomes cold and cold water becomes warmer. Questions based on the following topics are discussed in the NCERT solutions for class 11 physics chapter 11 thermal properties of matter.

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

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

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. 

 

NCERT solutions for class 11 physics chapter 11 thermal properties of matter exercise

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 TK = TC + 273.15

Triple Point of Neon in Kelvin TK = 24.57 K

Triple Point of Neon in Celcius TC = TK -273.15 = 24.57 -273.15 =-248.58 oC

Triple Point of carbon dioxide in Kelvin TK = 216.55 K

Triple Point of carbon dioxide in Celcius T= TK -273.15 = 216.55 - 273.15 = -56.60 oC

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

TF = -415.44 oC

Triple Point of carbon dioxide in Fahrenheit is

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

TF = -69.88 oC

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 TK 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. 11.4 (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 (c)  Answer the following :

(c) The absolute temperature (Kelvin scale) T is related to the temperature t_{c} on the Celsius scale by

 t_{c}=T-273.15

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

Answer:

This is because  0oC 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 0.010C , not 00C

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 and TK respectively

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

Let at another temperature the reading on Fahrenheit and Kelvin Scale be T' 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- T'K = 1 K, TF - 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(T1) = 273.16 K

Pressure in thermometer A at a temperature equal to the triple point of water (P1) = 1.250\times 10^{5}Pa 

Pressure in thermometer A at a temperature equal to Normal melting point of sulphur (P2) = 1.797\times 10^{5}Pa

The normal melting point of sulphur as read by thermometer A T2 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. 11.6 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 oC the 63 cm (l1) mark on the steel tape would be measuring exactly 63 cm as the tape is calibrated at 27 oC

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 oC is l2

\\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 oC 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 oC (T1) d1 = 8.70 cm

The diameter of the central hole in the wheel d2 = 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 oC (d1) = 4.24 cm 

Area of the hole at 227 oC 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 oC be d2

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

Change in diameter is d-d1 = 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 T1 = 27 oC

Final Temperature T2 = -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 T1 = 40.0 oC

Final Temperature T2 = 250 oC

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 oC 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, T1 = 500 oC

Final Temperature of Copper block, T2 = 0 oC

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 equivalent0.025\; 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 oC

Final Temperature of metal block = 40 oC

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 oC

Final Temperature of the calorimeter and water = 40 oC

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.

Thermal Properties of Matter Excercise:

Question:

 

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 oF

Final Temperature of the boy = 98 oF

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 oC-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 = 6s2

A=6(30)2

A=5400 cm2 = 0.54 m2

Temperature difference = 45 oC

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 m2

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 oC

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 oC

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 oC contains more energy available for heat dissipation than the same amount of water at 100 oC 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 T1

Let its final Temperature be T2

Let the surrounding temperature be T0

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 oC to 50 oC in 5 minutes when the surrounding temperature is 20 oC.

T2 = 50 oC

T1 = 80 oC

T0 = 20 oC

t = 5 min = 300 s.

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

For T1 = 60 oC and T2 = 30 oC 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 oC to 30 oC at the surrounding temperature of 20 oC. 

NCERT solutions for class 11 physics chapter 11 thermal properties of matter additional exercise

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 oC 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 CO2 decrease with decrease in pressure. This we can see from the solid lines in the P-T phase diagram of CO2.

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

(c) What are the critical temperature and pressure for  CO_{2} ? What is their significance?

Answer:

The critical temperature and pressure for CO2 are 31.1 oC and 73.0 atm respectively. If the temperature exceeds this critical value of temperature CO2 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:

CO2 is vapour at -70 oC 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:

CO2 is solid at -60 oC 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 

 (c) 15^{\circ}C under 56\; atm\; ?

Answer:

CO2 is liquid at 15 oC 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 oC lies to the left of the triple point of water i.e. in the region of solid and vapour phases. Once we start compressing CO2 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?

Answer:

At room temperature (27 oC) and 4 atm pressure CO2 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 CO2 from room temperature at this constant pressure CO2 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 oC under 10 atm pressure CO2 is in the solid phase. At room temperature (27 oC) under 10 atm pressure CO2 is in the vapour phase. At 10 atm pressure, CO2 can exist in all three phases depending upon the temperature. Therefore as CO2 is heated from -65 oC 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 ofCO_{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 oC is above the critical temperature of CO2. Once CO2 is isothermally compressed at this temperature it would not liquefy irrespective of how high the pressure is but at very high pressures CO2 will not behave as an ideal gas.

NCERT solutions for class 11 physics chapter wise 

Chapter 1

NCERT solutions for class 11 physics chapter 1 Physical world

Chapter 2

Solutions of NCERT for class 11 physics chapter 2 Units and Measurement

Chapter 3

CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line

Chapter 4

NCERT solutions for class 11 physics chapter 4 Motion in a Plane

Chapter 5

Solutions of NCERT for class 11 physics chapter 5 Laws of Motion

Chapter 6

CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power

Chapter 7

NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion

Chapter 8

Solutions of NCERT for class 11 physics chapter 8 Gravitation

Chapter 9

CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids

Chapter 10

NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids

Chapter 11

NCERT solutions for class 11 physics chapter 11 Thermal Properties of Matter

Chapter 12

CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics

Chapter 13

NCERT solutions for class 11 physics chapter 13 Kinetic Theory

Chapter 14

Solutions of NCERT for class 11 physics chapter 14 Oscillations

Chapter 15

CBSE NCERT solutions for class 11 physics chapter 15 Waves

Subject wise solutions for NCERT class 11

NCERT solutions for class 11 biology

CBSE NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

Importance of NCERT solutions for class 11 physics chapter 11 thermal properties of matter:

Thermal properties of matter is an important chapter for competitive exams. Questions based on calorimetry heat transfer and thermal expansions are frequently asked in competitive exams like NEET and JEE Mains exam. The NCERT solutions for class 11 physics chapter 11 thermal properties of matter are helpful to understand these concepts and their application in numerical problems.

 

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