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

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

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter is designed and prepared by the best teachers across India. All the important topics are covered in the exercises and each answer comes with a detailed explanation to help students understand concepts better. These NCERT solutions play a crucial role in your preparation for all exams conducted by the CBSE, including the JEE.

NCERT TEXTBOOK QUESTIONS SOLVED

1. 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 10-1K-1.

Ans. On a day when the temperature is 27 °C, the length of 1 cm division on the steel tape is exactly 1 cm, because the tape has been calibrated for 27 °C.When the temperature rises to 45 °C (that is, ΔT = 45 – 27 = 18 °C), the increase in the length of 1 cm division is Δl = αlΔT = (1.2 x 10-5C-1) x 1 cm x 18 °C = 0.000216 cm Therefore, the length of 1 cm division on the tape becomes 1.000216 cm at 45 °C. As the length of the steel rod is read to be 63.0 cm on the steel tape at 45 °C, the actual length of the rod at 45 °C is 63.0 x 1.000216 cm = 63.0136 cm The length of the same rod at 27 °C is 63.0 cm, because 1 cm mark on the steel tape is exactly 1 cm at 27 °C.

2. Answer the following questions based on the P – T phase diagram of CO2 (Fig. of question 17 given above)
(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?
(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure ?
(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature – 65 °C as it is heated up at room temperature at constant pressure.
(d) CO2 is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe ?

Ans. (a) No, the CO2 does not go through the liquid phase. The point (1.00 atm, – 60°C) is to the lift of the triple-point O and below the sublimation curve OA. Therefore, when CO2 is compressed at this point at constant temperature, the point moves perpendicular to the temperature-axis and enters the solid phase region. Hence, the CO2 vapour condenses to solid directly without going through the liquid phase.
(b) CO2 at 4.0 atm pressure and room temperature (say, 27 °C) is in vapour phase. This point (4.0 atm, 27°C) lies below the vaporation curve OC and to the right of the triple point O. Therefore, when CO2 is cooled at this point at constant pressure, the point moves perpendicular to the pressure-axis and enters the solid phase region. Hence, the CO2 vapour condenses directly to solid phase without going through the liquid phase.
(c) When the solid CO2 at – 65 °C is heated at 10 atm pressure, it is first converted into liquid. A further increase in its temperature brings it into the vapour phase. If a horizontal line at P = 10 atm is drawn parallel to the T-axis, then the points of intersection of line with the fusion and vaporization curve give the fusion and boiling points at 10 atm.
(d) Above 31.1°C, the gas cannot be liquefied. Therefore, on being compressed isothermally at 70°C, there will be no transition to the liquid region. However, the gas will depart, more and more from its perfect gas behaviour with the increase in pressure.

3. 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) heat systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Ans. (a) According to Kirchh off’s law of black body radiations, good emitters are good absorbers and bad emitters are bad absorbers. A body with large reflectivity is a poor absorber of heat and consequently, it is also a poor emitter.
(b) Brass is a good conductor of heat, while wood is a bad conductor. When we touch the brass tumbler on a chilly day, heat starts flowing from our body to the tumbler and we feel it cold. However, when the wooden tray is touched, heat does not flow from our hands to the tray and we do not feel cold.
(c) An optical pyrometer is based on the principle that the brightness of a glowing surface of a body depends upon its temperature. Therefore, if the temperature of the body is less than 600°C, the image formed by the optical pyrometer is not brilliant and we do not get the reliable result. It is for this reason that the pyrometer gives a very low value for the temperature of red hot iron in the open.
(d) The lower layers of earth’s atmosphere reflect infrared radiations from earth back to the surface of earth. Thus the heat radiation received by the earth from the sun during the day are kept trapped by the atmosphere. If atmosphere of earth were not there, its surface would become too cold to live.
(e) Steam at 100°C possesses more heat than the same mass of water at 100°C. One gram of steam at 100°C possesses 540 calories of heat more than that possessed by 1 gm of water at 100°C. That is why heating systems based on circulation of steam are more efficient than those based on circulation of hot water.

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