NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of issue are component of Class 11 Physics NCERT Solutions. Here we have actually given NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of issue.

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

Topics and also Subtopics in NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter:

 Section Name Topic Name 11 Thermal Properties of matter 11.1 Introduction 11.2 Temperature and also heat 11.3 Measurement of temperature 11.4 Ideal-gas equation and absolute temperature 11.5 Thermal expansion 11.6 Specific warmth capacity 11.7 Calorimetry 11.8 Change of state 11.9 Heat transfer 11.10 Newton’s legislation of cooling

QUESTIONS FROM TEXTBOOK

Inquiry 11. 1. The triple points of neon and also carbon dioxide are 24.57 K and also 216.55 K respectively. Express these temperatures on the Celsius and also Fahrenheit scales.Answer: The relation between kelvin range and Celsius range is TK – 273.15 =TC => TC=TK– 273.15

Inquiry 11. 2. Two absolute scales A and also B have triple points of water identified to be 200 A and also 350 B. What is the relation between TA and TB ?Answer:  As we understand, triple allude of water on absolute scale = 273.16 K, Size of one level of kelvin scale on absolute scale A

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Inquiry 11. 3. The-electrical resistance in ohms of a particular thermometer varies via temperature according to the approximate law: R = R0 <1 + α (T – T0)>. The resistances is 101.6 Ωat the triple-suggest of water 273.16 K, and 165.5 Ωat the normal melting point of lead (600.5 K). What is the temperature as soon as the resistance is 123.4 Ω ?Answer: Here, R0 = 101.6 Ω; T0 = 273.16 K Case (i) R1= 165.5 Ω; T1 = 600.5 K, Case (ii) R2 = 123.4 , T2 = ?Using the relation R = R0<1 + α (T – T0)>Case (i) 165.5 = 101.6 <1 + α (600.5 – 273.16)>

Concern 11. 4. Answer the following: (a) The triple-allude of water is a standard solved point in modem thermometry. Why ? What is wrong in taking the melting suggest of ice and the boiling allude of water as typical fixed points (as was originally done in the Celsius scale) ? (b) Tright here were 2 resolved points in the original Celsius range as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, among the solved points is the triple-point of water, which on the Kelvin absolute range is assigned the number 273.16 K. What is the various other resolved suggest on this (Kelvin) Scale ? (c) The absolute temperature (Kelvin scale) T is regarded the temperature tc on the Celsius scale tc = 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 range whose unit interval size is equal to that of the Fahrenheit scale ?Answer: (a) Triple point of water has a distinct value i.e., 273.16 K. The melting point and boiling points of ice and water respectively do not have actually unique values and readjust with the readjust in press.(b) On Kelvin’s absolute range, tbelow is just one resolved allude, namely, the triple-allude of water and tright here is no various other fixed point.(c) On Celsius range 0 °C corresponds to the melting suggest of ice at normal pressure and the worth of absolute temperature is 273.15 K. The temperature 273.16 K coincides to the triple point of water.(d)The Fahrenheit scale and Absolute range are associated as

Concern 11. 5. Two best gas thermometers A and also B use oxygen and hydrogen respectively. The complying with monitorings are made:

(a) What is the absolute temperature of normal melting allude of sulphur as check out by thermometers A and also B ? (b) What perform you think is the factor behind the slight difference in answers of thermometers A and also B ? (The thermometers are not faulty). What even more procedure is needed in the experiment to reduce the discrepancy in between the 2 readings ?Answer:
(b) The value of the melting suggest of sulphur discovered from the 2 thermometers differ slightly because of the factor that in practice, the gases do not behave actually strictly as perfect gases i.e., gases are not perfectly best.To reduce the discrepency, readings need to be taken for reduced and lower pressures and also the plot between temperature measured versus absolute pressure of the gas at triple point must be extrapolated to attain the temperature in the limit push often tends to zero (if P —> 0), as soon as the gases strategy ideal gas behaviour.

Question 11. 6. A steel tape 1 m lengthy is appropriately 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 warm day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the size of the same steel rod on a day when the temperature is 27.0 °C ? Coeffective of linear development of steel = 1.20 x 10-1K-1.Answer:  On a day once the temperature is 27 °C, the length of 1 cm department on the steel tape is precisely 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 boost in the size of 1 cm division is Δl = αlΔT = (1.2 x 10-5C-1) x 1 cm x 18 °C = 0.000216 cm As such, the size of 1 cm division on the tape becomes 1.000216 cm at 45 °C. As the size of the steel rod is read to be 63.0 cm on the steel tape at 45 °C, the actual size 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 note on the steel tape is exactly 1 cm at 27 °C.

Question 11. 7. A big steel wheel is to befitted on to a shaft of the very same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and also the diameter of the main hole in the wheel is, 8.69 cm. The shaft is cooled making use of ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of direct expansion of the steel to be continuous over the compelled temperature variety αsteel= 1-20 x 10-5K-1.Answer:

Inquiry 11. 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 readjust in the diameter of the hole when the sheet is heated to 227 °C ? Coeffective of linear growth of copper = 1.70 x 10-5K-1.Answer:

Question 11. 9. A brass wire 1.8 m lengthy at 27 °C is hosted taut with bit stress in between 2 rigid supports. If the wire is coocaused a temperature of – 39 °C, what is the anxiety emerged in the wire, if its diameter is 2.0 mm ? Co-efficient of straight development of brass = 2.0 x 10-5K-1; Young’s modulus of brass = 0.91 x 1011 PaAns.

Question 11. 10. A brass rod of size 50 cm and also diameter 3.0 mm is joined to a steel rod of the same length and also diameter. What is the adjust in size of the combined rod at 250 °C, if the original lengths are at 40.0 °C ? Is tbelow a ‘thermal stress’ emerged at the junction ? The ends of the rod are totally free to expand also (Co-efficient of linear growth of brass = 2.0 x 10-5 °C-1, steel = 1.2 x 10-5 °C-3.Ans.

Concern 11. 11. The coreliable of volume expansion of glycerine is 49 x 10-5K-1. What is the fractional readjust in its density for a 30 °C increase in temperature ?Ans.

Question 11. 12. A 10 kW drilling machine is provided to drill a bore in a tiny 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 supplied up in heating the machine itself or shed to the surroundings? Specific heat of aluminium = 0.91 J g-1 K-1 .Answer: Power = 10 kW = 104 WMass, m=8.0 kg = 8 x 103 g

Concern 11. 13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then inserted on a huge ice block. What is the maximum amount of ice that deserve to melt? Specific warm of copper is 0.39 Jg-1°C-1. Heat of fusion of water = 335 Jg-1.

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Question 11. 14. In an experiment on the particular warmth of a metal, a 0.20 kg block of the steel at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The last temperature is 40° C. Compute the specific warm of the steel. If heat losses to the surroundings are not negligible, is your answer better or smaller sized than the actual value for certain warm of the metal?Answer: Mass of metal block, m = 0.20 kg = 200 gFall in the temperature of steel block,ΔT = (150 – 40) °C = 110 °CIf C be the specific warmth of steel, then warmth lost by the metal block = 200 x C x 110 cal Volume of water = 150 cm3mass of water = 150 gIncrease in temperature of water = (40 – 27) °C = 13°CHeat obtained by water = 150 x 13 cal Water indistinguishable of calorimeter, w = 0.025 kg = 25gHeat got by calorimeter,

Question 11. 15. Given below are monitorings on molar particular heats at room temperature of some common gases.

The measured molar particular heats of these gases are markedly various from those for mono atomic gases. Typically, molar specific warmth of a mono atomic gas is 2.92 cal/mol K. Explain this distinction. What have the right to you infer from the somewhat bigger (than the rest) worth for chlorine ?Answer: The gases which are listed in the over table are diatomic gases and not mono atomic gases. For diatomic gases, molar particular warm =5/2 R = 5/2 x 1.98 = 4.95, which agrees fairly well with all observations noted in the table other than for chlorine. A mono atomic gas molecule has only the translational motion. A diatomic gas molecule, acomponent from translational activity, the vibrational as well as rotational motion is additionally feasible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, warm is to be offered to increase not just translational power yet additionally rotational and also vibrational energies. Hence, molar particular heat of a diatomic gas is better than that for mono atomic gas. The higher worth of molar specific warmth of chlorine as compared to hydrogen, nitrogen, oxygen and so on shows that for chlorine molecule, at room temperature vibrational motion additionally occurs along with translational and rotational activities, whereas various other diatomic molecules at room temperature commonly have rotational activity apart from their translational activity. This is the reason that chlorine has somewhat bigger value of molar particular heat.

Question 11. 16. (a) At what temperature and pressure have the right to the solid, liquid and also vapour phases of CO2 co-exist in equilibrium ? (b) What is the effect of decrease of press on the fusion and also boiling suggest of CO2 ? (c) What are the instrumental temperature and push for CO2 ? What is their definition 1 (d) Is CO2 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) At the triple allude, temperature = – 56.6 °C and pressure = 5.11 atm.(b) Both the boiling allude and freezing allude of CO2 decrease if push decreases.(c) The instrumental temperature and pressure of CO2 are 31.1°C and also 73.0 atm respectively. Above this temperature, CO2 will certainly not liquefy/even if compressed to high pressures.(d) (i) The allude (- 70 °C, 1.0 atm) lies in the vapour area. Hence, CO2 is vapour at this point.(ii) The suggest (- 60 °C, 10 atm) lies in the solid region. Hence, CO2 is solid at this point.(iii) The suggest (15 °C, 56 atm) lies in the liquid area. Hence, CO2 is liquid at this suggest.

Inquiry 11. 17. Answer the adhering to concerns based upon the P – T phase diagram of CO2 (Fig. of question 17 offered above) (a) CO2 at 1 atm push and also temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ? (b) What happens once CO2 at 4 atm push is cooled from room temperature at constant press ? (c) Describe qualitatively the changes in a provided mass of solid CO2 at 10 atm push and also temperature – 65 °C as it is heated up at room temperature at continuous pressure. (d) CO2 is heated to a temperature 70 °C and also compressed isothermally. What transforms in its properties do you suppose to observe ?Answer:  (a) No, the CO2 does not go via the liquid phase. The point (1.00 atm, – 60°C) is to the lift of the triple-suggest O and listed below the sublimation curve OA. As such, when CO2 is compressed at this suggest at consistent temperature, the suggest moves perpendicular to the temperature-axis and also enters the solid phase region. Hence, the CO2 vapour condenses to solid straight without going with the liquid phase.(b) CO2 at 4.0 atm push and also room temperature (say, 27 °C) is in vapour phase. This suggest (4.0 atm, 27°C) lies listed below the vaporation curve OC and also to the ideal of the triple suggest O. Because of this, as soon as CO2 is cooled at this allude at constant push, the point moves perpendicular to the pressure-axis and enters the solid phase area. Hence, the CO2 vapour condenses straight to solid phase without going via the liquid phase.(c) When the solid CO2 at – 65 °C is heated at 10 atm pressure, it is first converted right into liquid. A additionally rise in its temperature brings it right into the vapour phase. If a horizontal line at P = 10 atm is attracted parallel to the T-axis, then the points of intersection of line through the fusion and also vaporization curve provide the fusion and also 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 area. However before, the gas will certainly depart, even more and also even more from its perfect gas behaviour through the boost in push.

Question 11. 18. A child running a temperature of 101°F is provided an antipyrin (i.e., a medication that lowers fever) which reasons a boost in the rate of evaporation of sweat from his body. If the fever before is lugged dvery own to 98° F in 20 minutes, what is the average rate of additional evaporation brought about by the drug ? Assume the evaporation system to be the just way by which heat is lost. The mass of the boy is 30 kg. The certain heat of humale body is roughly the very same as that of water, and also latent warm of evaporation of water at that temperature is about 580 cal g-1.Answer:

Concern 11. 19. A ‘thermacole’ icebox is a cheap and reliable technique for storing small amounts of cooked food in summer in specific. A cubical icebox of side 30 cm has actually a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice staying after 6 h. The external temperature is 45°C, and also coeffective of thermal conductivity of thermacole is 0.01 Js-1 m-1 °C-1 .Answer:  Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to external air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of expocertain t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C..•. Total warm entering the icebox in 6 h is offered by

Question 11. 20. A brass boiler has a base location 0.15 m2 and also thickness 1.0 cm. It boils water at the rate of 6.0 kg/ min as soon as put on a gas cooktop. Estimate the temperature of the component of the flame in call with the boiler. Thermal conductivity of brass = 109 Js-1 m-1 K-1.(Heat of vaporization of water = 2256 x 103 J kg-1 )Answer:

Question 11. 21. Exsimple why: (a) a body through big reflectivity is a negative emitter. (b) a brass tumbler feels much chillier than a wooden tray on a chilly day. (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation provides too low a worth for the temperature of a red hot iron item in the open up, but gives a correct value for the temperature once the very same piece is in the furnace. (d) the earth without its atmosphere would certainly be inhospitably cold. (e) heat systems based on circulation of heavy steam are even more efficient in warming a building than those based on circulation of warm water.Answer: (a) According to Kirchh off’s law of black body radiations, good emitters are good absorbers and also negative emitters are bad absorbers. A body via huge reflectivity is a bad absorber of heat and also consequently, it is also a bad emitter.(b) Brass is a good conductor of heat, while hardwood is a negative conductor. When we touch the brass tumbler on a chilly day, heat starts flowing from our body to the tumbler and also we feel it cold. However before, when the wooden tray is touched, warm does not circulation from our hands to the tray and also we perform not feel cold.(c) An optical pyrometer is based on the principle that the brightness of a glowing surconfront of a body counts upon its temperature. Thus, if the temperature of the body is less than 600°C, the photo formed by the optical pyrometer is not brilliant and also we execute not gain the trustworthy result. It is thus that the pyrometer provides a very low worth for the temperature of red warm iron in the open.(d) The lower layers of earth’s setting reflect infrared radiations from earth back to the surface of earth. Hence the warm radiation got by the earth from the sunlight during the day are preserved trapped by the atmosphere. If environment of earth were not tbelow, its surface would come to be also cold to live.(e) Steam at 100°C possesses more warmth than the exact same mass of water at 100°C. One gram of vapor at 100°C possesses 540 calories of warm more than that possessed by 1 gm of water at 100°C. That is why heating systems based on circulation of vapor are even more reliable than those based on circulation of warm water.

Inquiry 11. 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: