• AglaSem
  • Schools
  • Admission
  • Career
  • News
  • Hindi
  • Mock Test
  • Docs
  • ATSE
aglasem
  • CBSE
    • Date Sheet
    • Syllabus
    • Sample Papers
    • Question Papers
  • ICSE / ISC
  • State Boards
    • Date Sheet
    • Admit Card
    • Result
    • Sample Paper
    • Question Paper
  • NCERT
    • NCERT Books
    • NCERT Solutions
    • NCERT Exempler
    • Teaching Material
  • Mock Tests
  • Study Material
    • Notes
    • Solved Sample Papers
    • Maps
    • Writing Skill Format
  • Olympiads
    • ATSE
    • KVPY
    • NTSE
    • NMMS
  • School Admission
  • Entrance Exams
    • JEE Main
    • NEET
    • CLAT
  • Students Guide
    • Careers Opportunities
    • Courses & Career
    • Courses after 12th
  • Others
    • RD Sharma Solutions
    • HC Verma Solutions
    • Teaching Material
    • Classes Wise Resources
    • Videos
No Result
View All Result
aglasem
  • CBSE
    • Date Sheet
    • Syllabus
    • Sample Papers
    • Question Papers
  • ICSE / ISC
  • State Boards
    • Date Sheet
    • Admit Card
    • Result
    • Sample Paper
    • Question Paper
  • NCERT
    • NCERT Books
    • NCERT Solutions
    • NCERT Exempler
    • Teaching Material
  • Mock Tests
  • Study Material
    • Notes
    • Solved Sample Papers
    • Maps
    • Writing Skill Format
  • Olympiads
    • ATSE
    • KVPY
    • NTSE
    • NMMS
  • School Admission
  • Entrance Exams
    • JEE Main
    • NEET
    • CLAT
  • Students Guide
    • Careers Opportunities
    • Courses & Career
    • Courses after 12th
  • Others
    • RD Sharma Solutions
    • HC Verma Solutions
    • Teaching Material
    • Classes Wise Resources
    • Videos
No Result
View All Result
aglasem
No Result
View All Result

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

by aglasem
August 26, 2021
in Extras
Reading Time: 1 min read
0
NCERT Solutions

NCERT Solutions Class 11 Physics Chapter 11 Thermal Properties of matter – Here are all the NCERT solutions for Class 11 Physics Chapter 11. This solution contains questions, answers, images, explanations of the complete chapter 11 titled Of Thermal Properties of matter taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 11 Thermal Properties of matter After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of matter in one place.

Go to Exercise

NCERT Solutions Class 11 Physics Chapter 11 Thermal Properties Of matter

Subscribe For Latest Updates

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Physics for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 11 Thermal Properties Of matter , Physics, Class 11.

Class 11
Subject Physics
Book Physics Part I
Chapter Number 11
Chapter Name

Thermal Properties Of matter

NCERT Solutions Class 11 Physics chapter 11 Thermal Properties Of matter

ATSE 2022, Olympiad Registration Open.

Class 11, Physics chapter 11, Thermal Properties Of matter solutions are given below in PDF format. You can view them online or download PDF file for future use.

Thermal Properties Of matter Download

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

Did you find NCERT Solutions Class 11 Physics chapter 11 Thermal Properties Of matter helpful? If yes, please comment below. Also please like, and share it with your friends!

NCERT Solutions Class 11 Physics chapter 11 Thermal Properties Of matter- Video

You can also watch the video solutions of NCERT Class11 Physics chapter 11 Thermal Properties Of matter here.

Video – will be available soon.

If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.

Download NCERT Solutions Class 11 Physics chapter 11 Thermal Properties Of matter In PDF Format

You can also download here the NCERT Solutions Class 11 Physics chapter 11 Thermal Properties Of matter in PDF format.

Click Here to download NCERT Solutions for Class 11 Physics chapter 11 Thermal Properties Of matter

Question & Answer

Q.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.

Ans : \(\begin{array}{l}{\text { Kelvin and Celsius scales are related as: }} \\ {T_{\mathrm{c}}=T_{\mathrm{K}}-273.15 \ldots(i)} \\ {\text { Celsius and Fahrenheit scales are related as: }} \\ {T_{\mathrm{F}}=\frac{9}{5} T_{\mathrm{C}}+32}\end{array}\) \(\begin{array}{l}{\text { Celsius and Fahrenheit scales are related as: }} \\ {T_{\mathrm{F}}=\frac{9}{5} T_{\mathrm{C}}+32}\end{array}\) \(\begin{array}{l}{\frac{\text { For neon: }}{T_{k}=24.57 \mathrm{K}}} \\ {\therefore T_{\mathrm{c}}=24.57-273.15=-248.58^{\circ} \mathrm{C}} \\ {T_{\mathrm{F}}=\frac{9}{5} T_{\mathrm{C}}+32}\end{array}\) \(\begin{array}{l}{=\frac{9}{5}(-248.58)+32} \\ {=415.44^{\circ} \mathrm{F}}\end{array}\) \(\begin{array}{l}{\text { For carbon dioxide: }} \\ {T_{k}=216.55 \mathrm{K}} \\ {\therefore T_{\mathrm{c}}=216.55-273.15=-56.60^{\circ} \mathrm{C}} \\ {T_{\mathrm{F}}=\frac{9}{5}\left(T_{\mathrm{C}}\right)+32}\end{array}\) \(\begin{array}{l}{=\frac{9}{5}(-56.60)+32} \\ {=-69.88^{\circ} \mathrm{C}}\end{array}\)

Q.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 TA and TB ?

Ans : \(\begin{array}{l}{\text { Triple point of water on absolute scaleA, } T_{1}=200 \mathrm{A}} \\ {\text { Triple point of water on absolute scale } \mathrm{B}, T_{2}=350 \mathrm{B}} \\ {\text { Triple point of water on Kelvin scale, } T_{\kappa}=273.15 \mathrm{K}} \\ {\text { The temperature } 273.15 \mathrm{K} \text { on Kelvin scale is equivalent to } 200 \mathrm{A} \text { on absolute scale } \mathrm{A} \text { . }}\end{array}\) \(\begin{array}{l}{T_{1}=T_{\mathrm{k}}} \\ {200 \mathrm{A}=273.15 \mathrm{K}} \\ {\therefore \mathrm{A}=\frac{273.15}{200}} \\ {\text { The temperature } 273.15 \mathrm{K} \text { on Kelvin scale is equivalent to } 350 \mathrm{B} \text { on absolute scale B. }}\end{array}\) \(\begin{array}{l}{T_{2}=T_{\kappa}} \\ {350 \mathrm{B}=273.15} \\ {\therefore \mathrm{B}=\frac{273.15}{350}} \\ {T_{\mathrm{A}} \text { is triple point of water on scale } \mathrm{A} \text { . }} \\ {T_{\mathrm{B}} \text { is triple point of water on scale B. }}\end{array}\) \(\begin{array}{l}{\therefore \frac{273.15}{200} \times T_{\mathrm{A}}=\frac{273.15}{350} \times T_{\mathrm{B}}} \\ {T_{\mathrm{A}}=\frac{200}{350} T_{\mathrm{B}}} \\ {\text { Therefore, the ratio } T_{\mathrm{A}} : T_{\mathrm{B}} \text { is given as } 4 : 7 .}\end{array}\)

Q.3: The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = Ro [1 + α (T – To )] 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 ?

Ans : \(\begin{array}{l}{\text { It is given that: }} \\ {R=R_{0}\left[1+\mathrm{a}\left(T-T_{0}\right)\right] \ldots(i)} \\ {\text { Where, }} \\ {R_{0} \text { and } T_{0} \text { are the initial resistance and temperature respectively }} \\ {R \text { and } T \text { are the final resistance and temperature respectively }} \\ {\text { a is a constant }}\end{array}\) \(\begin{array}{l}{\text { At the triple point of water, } T_{0}=273.15 \mathrm{K}} \\ {\text { Resistance of lead, } R_{0}=101.6 \Omega} \\ {\text { At normal melting point of lead, } T=600.5 \mathrm{k}} \\ {\text { Resistance of lead, } R=165.5 \Omega} \\ {\text { Substituting these values in equation (i), we get: }}\end{array}\) \(\begin{array}{l}{R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]} \\ {165.5=101.6[1+\alpha(600.5-273.15)]} \\ {1.629=1+\alpha(327.35)} \\ {\therefore \alpha=\frac{0.629}{327.35}=1.92 \times 10^{-3} \mathrm{K}^{-1}} \\ {\text { For resistance, } R_{1}=123.4 \Omega} \\ {\mathrm{R}_{1}=\mathrm{R}_{0}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\right]}\end{array}\) \(\begin{array}{l}{\text { Where, T is the temperature when the resis tan ce of lead is } 123.4 \Omega} \\ {123.4=101.6\left[1+1.92 \times 10^{-3}(\mathrm{T}-273.15)\right]} \\ {1.214=1+1.92 \times 10^{-3}(\mathrm{T}-273.15)} \\ {\frac{0.214}{1.92 \times 10^{-3}}=\mathrm{T}-273.15} \\ {\therefore \mathrm{T}=384,61 \mathrm{K}}\end{array}\)

Q.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 tc on the Celsius scale by 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 scale whose unit interval size is equal to that of the Fahrenheit scale ?

Ans : \(\begin{array}{l}{\text { (a) The triple point of water has a unique value of } 273.16 \mathrm{K} \text { . At particular values of }} \\ {\text { volume and pressure, the triple point of water is always } 273.16 \mathrm{K} \text { . The melting point of }} \\ {\text { ice and bolling point of water do not have particular values because these points depend }} \\ {\text { on pressure and temperature. }}\end{array}\) (b) The absolute zero or O K is the other fixed point on the Kelvin absolute scale. \(\begin{array}{l}{\text { (c) The temperature } 273.16 \mathrm{K} \text { is the triple point of water. It is not the melting point of }} \\ {\text { ice. The temperature } 0^{\circ} \mathrm{C} \text { on Celsius scale is the melting point of ice. Its corresponding }} \\ {\text { value on Kelvin scale is } 273.15 \mathrm{K} \text { . }} \\ {\text { Hence, absolute temperature (Kelvin scale) } T, \text { is related to temperature } t_{c \prime} \text { on Celsius }} \\ {t_{c}=T-273.15}\end{array}\) \(\begin{array}{l}{\text { (d) Let } T_{F} \text { be the temperature on Fahrenheit scale and } T_{\kappa} \text { be the temperature on }} \\ {\text { absolute scale. Both the temperatures can be related as: }} \\ {\frac{T_{F}-32}{180}=\frac{T_{K}-273.15}{100}}\end{array}\) \(\begin{array}{l}{\text { Let } T_{F_{1}} \text { be the temperature on Fahrenheit scale and } T_{\kappa 1} \text { be the temperature on absolute }} \\ {\text { scale. Both the temperatures can be related as: }} \\ {\frac{T_{r 1}-32}{180}=\frac{T_{\mathrm{k} 1}-273.15}{100}}\end{array}\) \(\begin{array}{l}{\text { It is given that: }} \\ {T_{\mathrm{k} 1}-T_{\mathrm{k}}=1 \mathrm{K}}\end{array}\) \(\begin{array}{l}{\text { Subtracting equation (i) from equation (ii), we get: }} \\ {\frac{T_{\mathrm{FI}}-T_{\mathrm{F}}}{180}=\frac{T_{\mathrm{KI}}-T_{\mathrm{K}}}{100}=\frac{1}{100}}\end{array}\) \(\begin{array}{l}{T_{\mathrm{F} 1}-T_{\mathrm{F}}=\frac{1 \times 180}{100}=\frac{9}{5}} \\ {\text { Triple point of water }=273.16 \mathrm{K}} \\ {\text { ¿Triple point of water on absolute scale }=273.16 \times \frac{9}{5}=491.69}\end{array}\)

Q.5: Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

Ans : \(\begin{array}{l}{\text { (a) Triple point of water, } T=273.16 \mathrm{K} \text { . }} \\ {\text { At this temperature, pressure in thermometer } A, P_{A}=1.250 \times 10^{5} \text { Pa }} \\ {\text { Let } T_{1} \text { be the normal melting point of sulphur. }} \\ {\text { At this temperature, pressure in thermometer } A_{r} P_{1}=1.797 \times 10^{5} \text { Pa }} \\ {\text { According to Charles' law, we have the relation: }}\end{array}\) \(\begin{array}{l}{\frac{P_{\mathrm{A}}}{T}=\frac{P_{1}}{T_{1}}} \\ {\therefore T_{1}=\frac{P_{1} T}{P_{\mathrm{A}}}=\frac{1.797 \times 10^{5} \times 273.16}{1.250 \times 10^{5}}} \\ {=392.69 \mathrm{K}} \\ {\text { Therefore, the absolute temperature of the normal melting point of sulphur as read by }} \\ {\text { thermometer } \mathrm{A} \text { is } 392.69 \mathrm{K} \text { . }}\end{array}\) \(\begin{array}{l}{\text { Therefore, the absolute temperature of the normal melting point of sulphur as read by }} \\ {\text { thermometer } A \text { is } 392.69 \mathrm{K} \text { . }} \\ {\text { At triple point } 273.16 \mathrm{K} \text { , the pressure in thermometer } \mathrm{B}, P_{\mathrm{B}}=0.200 \times 10^{5} \mathrm{Pa}} \\ {\text { At temperature } T_{1}, \text { the pressure in thermometer } \mathrm{B}, P_{2}=0.287 \times 10^{5} \mathrm{Pa}} \\ {\text { According to Charles' law, we can write the relation: }}\end{array}\) \(\begin{array}{l}{\frac{P_{\mathrm{B}}}{T}=\frac{P_{1}}{T_{1}}} \\ {\frac{0.200 \times 10^{5}}{273.16}=\frac{0.287 \times 10^{3}}{T_{1}}} \\ {\therefore T_{1}=\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \times 273.16=391.98 \mathrm{K}}\end{array}\) \(\begin{array}{l}{\text { Therefore, the absolute temperature of the normal melting point of sulphur as read by }} \\ {\text { thermometer } B \text { is } 391.98 \mathrm{K} \text { . }}\end{array}\) \(\begin{array}{l}{\text { (b) The oxygen and hydrogen gas present in thermometers } A \text { and } B \text { respectively are not }} \\ {\text { perfect ideal gases. Hence, there is a slight difference between the readings of }} \\ {\text { thermometers } A \text { and } B \text { . }}\end{array}\) \(\begin{array}{l}{\text { To reduce the discrepancy between the two readings, the experiment should be carried }} \\ {\text { under low pressure conditions. At low pressure, these gases behave as perfect ideal }} \\ {\text { gases. }}\end{array}\)

NCERT / CBSE Book for Class 11 Physics

You can download the NCERT Book for Class 11 Physics in PDF format for free. Otherwise you can also buy it easily online.

  • Click here for NCERT Book for Class 11 Physics
  • Click here to buy NCERT Book for Class 11 Physics

All NCERT Solutions Class 11

  • NCERT Solutions for Class 11 Accountancy
  • NCERT Solutions for Class 11 Biology
  • NCERT Solutions for Class 11 Chemistry
  • NCERT Solutions for Class 11 Maths
  • NCERT Solutions for Class 11 Economics
  • NCERT Solutions for Class 11 History
  • NCERT Solutions for Class 11 Geography
  • NCERT Solutions for Class 11 Political Science
  • NCERT Solutions for Class 11 Sociology
  • NCERT Solutions for Class 11 Psychology
  • NCERT Solutions for Class 11 English
  • NCERT Solutions for Class 11 Hindi
  • NCERT Solutions for Class 11 Physics
  • NCERT Solutions for Class 11 Business Studies
  • NCERT Solutions for Class 11 Statistics

All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 Class 2 Class 3
Class 4 Class 5 Class 6
Class 7 Class 8 Class 9
Class 10 Class 11 Class 12

Download the NCERT Solutions app for quick access to NCERT Solutions Class 11 Physics Chapter 11 Thermal Properties Of matter. It will help you stay updated with relevant study material to help you top your class!


Previous Next

To get fastest exam alerts and government job alerts in India, join our Telegram channel.

Tags: NCERT Physics SolutionsNCERT SolutionsNCERT Solutions Class 11NCERT Solutions PDFPhysics

aglasem

Related Posts

PSEB 8th Result 2021
Extras

PSEB 8th Result 2022 – Check Roll Number Wise at pseb.ac.in

CBSE Class 12 Answer Key
Extras

Class 12 Business Studies Answer Key / Solution for Term 2 CBSE Board Exam 2022 (Code 054) – Available, Download PDF

ISC Class 12 Answer Key
Extras

ISC Maths Answer Key 2022 Class 12 Sem 2 – Check Solved Paper

SOSE Result
Extras

SOSE Result 2022 – Date, Link @ edudel.nic.in

Next Post
NCERT Solutions

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

Discussion about this post

Registration Open!!

ATSE 2022, Olympiad Registration Open.

Term 2 Solved Sample Paper 2022

  • Class 8 Sample Paper Term 2
  • Class 9 Sample Paper Term 2
  • Class 10 Sample Paper Term 2
  • Class 11 Sample Paper Term 2
  • Class 12 Sample Paper Term 2

CBSE Board Quick Links

  • CBSE Date Sheet
  • CBSE Result
  • CBSE Syllabus
  • CBSE Sample Papers
  • CBSE Question Papers
  • CBSE Notes
  • CBSE Practice Papers
  • CBSE Mock Tests

Class Wise Study Material

  • Class 1
  • Class 2
  • Class 3
  • Class 4
  • Class 5
  • Class 6
  • Class 7
  • Class 8
  • Class 9
  • Class 10
  • Class 11
  • Class 12

For 2022

  • Solved Sample Papers
  • Maps
  • Revision Notes
  • CBSE
  • State Board

Study Material

  • Class Notes
  • NCERT Solutions
  • NCERT Books
  • HC Verma Solutions
  • Courses After Class 12th

Exam Zone

  • JEE Main 2022
  • NEET 2022
  • CLAT 2022
  • Fashion & Design
  • Latest
  • Disclaimer
  • Terms of Use
  • Privacy Policy
  • Contact

© 2019 aglasem.com

No Result
View All Result
  • CBSE
  • ICSE / ISC
  • State Board
  • NCERT
  • Mock Test
  • Study Material
  • Olympiads
  • Schools Admission
  • Entrance Exams
  • Student Guide
  • HC Verma Solutions
  • Videos

© 2019 aglasem.com

Registration Open. ATSE 2022 Scholarship Apply Now!!