NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matters are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Physics |

Chapter |
Chapter 10 |

Chapter Name |
Thermal Properties of Matter |

Number of Questions Solved |
22 |

Category |
NCERT Solutions |

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

**NCERT Exercises**

**Question 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:**

Here, Triple point of neon, T, = 24.57 K

Triple point of CO_{2,} T_{2} = 216.55 K

Relation between kelvin scale and Celsius scales

T_{c}=T_{k}– 273.15

where

T_{c} = Temperature on Celsius scale

T_{k} = Temperature on kelvin scale

For neon T_{c} = 24.57 – 273.15 = -248.58°C

For CO_{2}, T_{c} = 216.55 – 273.15 = -56.60°C

Relation between kelvin and Fahrenheit scales

F=C+32

For neon T_{F} = x – 248.58 + 32 = -415.44°F

For CO_{2}T_{F} = x – 56.60° + 32 = – 69.88°F 5

**Question 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_{e}?

**Answer:**

Here, triple point of water on scale

A = 200 A

triple point of water on scale B = 350 B triple point of water on kelvin scale = 273.16 K

According to question

200 A = 350 B = 273.16 K

**Question 3.**

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R_{0} (1 + α (T-T_{0})] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

**Answer:**

Here, R_{0} = 101.6 Ω; T_{0} = 273.16 K

Case (i) R_{1} = 165.5 Ω; T, = 600.5 K

Case (ii) R_{2} = 123.4 Ω; T_{2} = ?

Using the relation R = R_{0} [1 + α(T – T_{0})]

Case (i) 165.5 = 101.6 [1 + a (600.5 – 273.16)]

**Question 4.**

**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)?

**(b)**There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°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?

**(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?

**(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:**

**(a)**This is because the triple point of water has a unique value i.e. 273.16 K at a unique point, where exists unique values of pressure and volume. On other hand, the melting point of ice and boiling point of water do not have unique set of values as they change with the change in pressure and volume.

**(b)**The kelvin absolute scales also have the fixed points as the Celsius scales have. The other fixed point is absolute zero. It corresponds to the temperature, when the volume and pressure of a gas will become zero.

**(c)**Triple point of water on Celsius scale is 01°C and on kelvin scale is 273.16 and the size of degree on the two scale is same, so t

_{c}– 0.01 = T- 273.16

∴ t

_{c}= T- 273.15

**(d)**The unit interval size of Fahrenheit scale is 212 – 32 = 180 divisions Also we know that the unit interval size of absolute scale is 100.

∴ Triple point of water on an absolute scale having 180 divisions is given by

**Question 5.**

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

**(a)** What is the absolute temperature of normal melting point of sulphur as read by thermometers

A and B?

**(b)** What do you think is the reason behind the slightly difference in answers of thermometers A and 6? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

**Answer:**

Let T be the melting point of sulphur.

The triple point of water, T_{tr} = 273.16 K

For thermometer A : P_{tr} = 1.25 x 10^{5} Pa

P = 1.797 x 10^{5} Pa

**Question 6.**

A steel tape 1 m long is correctly calibrated for a temperature of 27.0°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°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°C? Coefficient of linear expansion of steel = 1.20 x 10-^{5 }K-^{1
}**Answer:**

Here, length of the steel tape at 27°C is 100 cm.

∴ L = 100 cm, T = 27°C

The length of steel tape at 45°C is V = L + ΔL

L’ = L + αLΔT (∵AL = aLΔT)

L’ = L + αL(T_{2} – T_{1})

L’ = 100 + (1.20 x 10^{-5}) x 100 x (45° – 27°)

= 100.0216 cm

Length of 1 cm mark at 27°C on this scale, at

45°c=

Length of 63 cm measured by this tape at 45°C will be = x 63 = 63.0136 cm.

When temperature is 27°C, the size of 1 cm mark on the steel tape will be exactly 1 cm as the steel tape has been calibrated at 27°C. Therefore, length of the steel of the rod at 27°C = 63 x 1 = 63 cm

**Question 7.**

A large steel wheel is to be fitted on to a shaft of the same material. At 27°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: α_{steel} = 1.20 x 10 ^{5} K^{-1}.

**Answer:**

Here, T_{1 } = 27°C = 27 + 273 = 300 K

Let L_{1 }and L_{2} be the linear dimensions of steel at temperatures T_{1} and T_{2} respectively.

Now L, = 8.70 cm, I_{2} = 8.69 cm,

**Question 8.**

A hole is drilled in a copper sheet.The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of

copper = 1.70 x 10^{-5} K^{-1}.

**Answer:**

Here, Coefficient of linear expansion of copper, α = 1.70 x 10^{-5} °C^{-1
}ΔT = 227 -27 = 200°C

Therefore, coefficient of superficial expansion of copper

**Question 9.**

A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 x 10^{-5} K^{-1}; Young’s modulus of brass = 0.91 x 10^{11}pa.

**Answer:**

Here l_{1} = 1.8 m, t_{1}= 27°C, t_{2} = -39°C

∴t = t_{2}– t1= 39 – 27 = -66°C

_{l2} = length at t_{2}°C

For brass, α = 2 x 10^{-5} K

∴ Y = 0.91 x 10^{11} Pa

diameter of wire, d = 2.0 mm = 2.0 x 10^{-3} m

If A be the area of cross-section of the wire,

Negative sign indicates that the force is inwards due to the contraction of the wire.

**Question 10.**

A brass rod of length 50 cm and diameter 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°C, if the original lengths are at 40.0°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 x 10^{-5} °C^{-1}, steel = 1.2 x 10^{-5} °C^{-1}).

**Answer:**

**For brass rod :** α′ = 2.0 x 10^{-5} °C^{-1} ;

l′_{1} = 50 cm; ΔT = 250 – 40 = 210°C

The length of the brass rod at 250°C is given by

l′_{2} =l′_{1} (1+ α′ ΔT)

= 50(1 + 2.0 x 10^{-5} x 210) = 50.126 cm

**For steel rod**: α’ = 1.2 x 10 ^{5} °C ^{-1
}l′_{1}= 50 cm;

ΔT’ = 250 – 40 = 210°C

The length of the steel rod at 250°C is given by

l′_{2} +l′_{1} { (1 + α’ ΔT’)

= 50(1 + 1.2 x 10^{-5} x 210)

= 50.126 cm

Therefore, the length of the combined rod at 250°C

= l_{2} + l′_{2} = 50.21 + 50.126 = 100.336 cm

As the length of the combined rod at 40°C

= 50 + 50 = 100 cm

The change in length of the combined rod at 250°C = 100.336 -100.0 = 0.336 cm

No thermal stress is developed at the junction since the rod freely expand.

**Question 11.**

The coefficient of volume expansion of glycerine is 49 x 10^{-5} C^{-1}. What is the fractional change in its density for a 30°C rise in temperature?

**Answer:**

Here, γ= 49 x 10^{-5} °C^{-1}; ΔT = 30°C

Let there be m grams of glycerine and its initial volume be V. Suppose that the volume of the glycerine becomes V after a rise of temperature of 30°C then,

**Question 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% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg^{-1} °C^{-1}.

**Answer:**

Here, P = 10 kW = 10^{4} W,

mass, to = 8.0 kg = 8 x 10^{3} g

rise in temperature, ΔT = ?,

time, t = 2.5 min = 2.5 x 60 = 150 s

Sp. heat, c = 0.91 J g^{-1} °C^{-1
}Total energy, = P x t = 10^{4} x 150 = 15 x 10^{5} J

As, 50% of energy is lost,

**Question 13.**

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°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} °C^{_1}; heat of fusion of water = 335 J g^{1}).

**Answer:**

Here, mass of copper block, m_{1} = 2.5 kg

Specific heat of copper,

C = 0.39 Jg^{-1} K-^{1} = 0.39 x 10^{3} J kg^{-1 0}C

Temperature of furnace, ΔT = 500°C Latent heat of fusion,

L = 335 Jg^{-1} = 335 x 10^{3} J kg^{-1
}If Q be the heat absorbed by the copper block,then

**Question 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°C. The final temperature is 40°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:**

Here mass of metal, m = 0.20 kg = 200 g

Fall in temperature of metal

ΔT = 150 – 40 = 110°C

If c is specific heat of the metal, then heat lost by the metal,

ΔQ = mcΔT = 200 x c x 110

Volume of water = 150 c.c.

∴ Mass of water, m’ = 150 g

Water equivalent of calorimeter,

ω= 0.025 kg = 25 g

Rise in temperature of water and calorimeter, ΔT’ = 40-27 = 13°C

Heat gained by water and calorimeter,

ΔQ’ = (m’ + w)ΔT’

= (150 + 25) x 13 = 175 x 13

As, ΔQ = ΔQ’

From (i) and (ii), 200 x c x 110 = 175 x 13

If some heat is lost to the surroundings, value of c so obtained will be less than the actual value of c.

**Question 15.**

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

The measured molar specific heats of these gases are markedly different from these for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal mol^{-1} K^{1}. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

**Answer:**

The gas which are listed in the above table are diatomic gases and not mono-atomic gases. For diatomic gases, molar specific heat

which agrees fairly well with all observations listed in the table except for chlorine. A monoatomic gas molecule has only the translational motion. A diatomic gas molecule, apart from translation motion, the vibrational as well as rotational motion is also possible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, heat is to be supplied to increase not only translational energy but also rotational and vibrational energies. Hence, molar specific heat of a diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. shows that for chlorine molecule, at room temperature vibrational motion also occurs along with translational and rotational motions, whereas other diatomic molecules at room temperature usually have rotational motion apart from their translation motion. This is the reason that chlorine has somewhat larger value of molar specific heat.

**Question 16.**

**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?

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

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

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

**(a)** -70°C under 1 atm,

**(b)** -60°C under 10 atm,

**(c)** 15°C under 56 atm?

**Answer:**

**(a)** The solid, liquid and vapour phases of C0_{2 }can exist in equilibrium at its triple point O, corresponding to which P_{tr} = 5.11 atm and T_{tr} = – 56.6°C

**(b)** From the vaporisation curve (I) and the fusion curve (II), it follows that both the boiling and fusion points of CO_{2} decrease with the decrease of pressure.

**(c)** For CO_{2}, P_{c} =0 atm and T_{c} = 31.1°C Above its critical temperature, CO_{2 }gas can not be liquified, however large pressure may be applied.

**(d)**

**(a) -70°C under 1 atm :** This point lied in vapour region. Therefore, at -70°C under 1 atm, CO_{2} is vapour.

**(b) -60°C under 10 atm :** this point lies in solid region. Therefore, CO_{2} is solid at -60°C under 10 atm.

**(c) 15°C under 56 atm :** This point lies in liquid region. Therefore, CO_{2} is liquid at 15°C under 56 atm.

**Question 17.**

**Answer the following questions based on the P-Tphase diagram of CO _{2} as given Q. 16:
(a)** CO

_{2}at 1 atm pressure and temperature -60°C is t:c ipressed isothermally. Does it go through a liquid phase?

**(b)**What happens when CO

_{2}at 4 atm pressure is cooled from room temperature at constant pressure?

**(c)**Describe qualitatively the changes in a given mass of solid CO

_{2}at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.

**(d)**CO

_{2}is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?

**Answer:**

**(a)**No. CO

_{2}at 1 atm pressure and -60°C is vapour. If it is compressed isothermally when pressure is increased without changing the temperature, it will go to solid phase directly without going through the liquid phase.

**(b)**CO

_{2}at 4 atm pressure and at room temperature (say 25°C) is vapour. If it is cooled at constant pressure, it will again condense to solid without going through the liquid phase (the horizontal line through the initial point intersects only the sublimation curve III).

**(c)**CO

_{2}at 10 atm pressure and at -65°C is solid. As CO

_{2}is heated at constant pressure, it will go to liquid phase and then to the vapour phase. It is because, the horizontal line through the initial point intersects both the fusion and vapourisation curves. The fusion and boiling points can be known from the points, where the horizontal line at P-T diagram at 10 atm (initial point) intersects the respective curves.

**(d)**It will not exhibit any clear phase transition to the liquid phase. However, CO

_{2}gas will depart more and more from the ideal gas behaviour as its pressure increases.

**Question 18.**

A child running a temperature of 101°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°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 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:**

Here, fall in temperature = ΔT

**Question 19.**

A ‘thermocole’ icebox 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 6h.The outside temperature is 45°C, and co-efficient of thermal conductivity of thermocole is 0.01 J s^{-1} K^{-1}. [Heat of fusion of water = 335 x 10^{J} J kg^{-1}]

**Answer:**

Here, length of each side,

Z = 30 cm = 0.3 m

Thickness of each side, Ax = 5 cm = 0.05 m

total surface area through which heat enters into the box,

A = 6l^{2} = 6 x 0.3 x 0.3 = 0.54 m^{2
}Temp, diff., ΔT = 45 – 0 = 45°C,

K = 0.01 Jm^{-1}m^{-1} °C^{-1
}time, At = 6 hrs = 6 x 60 x 60 s

Latent heat of fusion, L = 335 x 10^{3} J kg^{-1
}Let m be the mass of ice melted in this time

**Question 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 kg min^{-1} 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 x 10^{3} J kg^{1}.

**Answer:**

Here, K = 109 J s^{-1} m^{1} K^{1}; A = 0.15 m^{2}; d = 1.0 cm = 10^{-2} m; T_{2} = 100°C, t = 1 min = 60 s Let T_{1} be the temperature of the part of boiler in contact with the stove. Therefore, amount of heat flowing per minute through the base of the boiler,

**Question 21.**

**Explain why:**

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

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

**(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

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

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

**Answer:**

**(a)** We know that a + r + t = 1 Where a, r and t are absorbance, reflectance and transmittance respectively of the surface of the body, t is also called emittance (e). Also according to Krichhoff’s law

e α a that is good absorber are good emitters and hence poor reflectors and vice-versa i.e. If r is large (i.e. large reflectively) a is smaller and hence e is smaller i.e. poor emitter.

**(b)** The thermal conductivity of brass is high e. brass is a good conductor of heat. So when a brass tumbler is touched, heat quickly flows from human body to the tumbler. Consequently, the tumbler appears colder. On the other hand, wood is a bad conductor of heat. So heat does not flow from the human body to the wooden tray, thus it appears relatively hotter.

**(c)** Let T be the temperature of the hot iron in the furnace. Thus according to Stefan’s law, heat radiated per second per unit area (E) is given by E = σT^{4} . When the body is placed in open at temperature T_{0}, then the heat radiated/sec/area (E’) is given by

E’ = σ(T_{4}-T^{4}_{0})

Clearly E’ < E, so the optical pyrometer gives too low a value for the temperature of a red hot iron piece in open.

**(d)** Gases are generally insulators. The Earth’s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the Earth. If this atmosphere is absent, then the Earth would naturally be colder as all its heat would have escaped out.

**(e)** This is because steam has much higher heat capacity (540 cal g^{_1}) than the heat capacity of water (80 cal g^{_1}) at the same temperature. Thus heating systems based on circulation of steam are more efficient than those based on circulation of hot water.

**Question 22.**

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

**Answer:**

According to Newton’s law of cooling, the rate of loss of heat ∝ cooling difference in temperature.

Here, average of 80°C and 50°C = 65°C Temperature of surroundings = 20°C

∴ Difference = 65 – 20 = 45°C

Under these conditions, the body cools 30°C in 5 minutes.

We hope the NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter, drop a comment below and we will get back to you at the earliest

## Leave a Reply