NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

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NCERT Solutions Class 11 Physics Thermodynamics – Access and Download For Free

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics: If you are a Class 11 CBSE student and searching for Thermodynamics Class 11 Physics NCERT solutions, this is where it comes in handy. In Thermodynamics Class 11 chapter, you will study how heat is converted to work and vice versa. For example, electrical energy can be utilized to boil water and similarly, steam can be utilized to produce electricity. The CBSE NCERT solutions for Class 11 Physics Chapter 12 Thermodynamics carry detailed explanations to exercise questions. Try to solve all these questions yourself before looking at the solutions. NCERT solutions help to improve conceptual knowledge.

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NCERT Solutions Class 11 Physics Thermodynamics – Access and Download For Free
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics
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NCERT Solutions for Class 11 Subject wise
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The NCERT solution for thermodynamics class 11 physics is prepared by our highly experienced Subject Matter which gives all the answers to the NCERT Book. Utilizing the NCERT solutions for Class 11 , specifically thermodynamics class 11 physics questions and answers, can aid students in practising questions and grasping important syllabus concepts before their exams. These solutions are available for free and can be accessed below.

** This chapter has been renumbered as Chapter 11 in accordance with the CBSE Syllabus 2023–24.

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NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

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NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics Exercise

Q12.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0\times 10^4\ J/g$ ?

Answer:

The volumetric flow of water is

$\\\frac{dV}{dt}=3\ litres\ minute^{-1}\\$

Density of water = 1000 g/litre

The mass flow rate of water is

$\\\frac{dm}{dt}=\rho \frac{dV}{dt}\\ \frac{dm}{dt}=3000\ g\ min^{-1}$

Specific heat of water, c = 4.2 J g ^-1 ^o C ^-1

The rise in temperature is $\Delta T=77-27=50\ ^{o}C$

Rate of energy consumption will be

$\\\frac{dQ}{dt}=\frac{dm}{dt}c\Delta T\\ \frac{dQ}{dt}=3000\times 4.2\times 50\\ \frac{dQ}{dt}=6.3\times 10^{5}\ J\ min^{-1}$

The heat of combustion of fuel $=4.0\times 10^{4}J/g$

Rate of consumption of fuel is

$\\\frac{6.3\times 10^{5}}{4\times 10^{4}}\\ =15.75\ g\ min^{-1}$

Q12.2 What amount of heat must be supplied to $2.0\times 10^{-2}\ kg$ of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of $N = 28;\ R = 8.3\ J\ mol^{-1}K^{-1}$ .)

Answer:

Mass of nitrogen, $m=2.0\times 10^{-2}kg=20g$

Molar Mass of nitrogen, M _N = 28 g

Number of moles is n

$\\n=\frac{m}{M_{N}}\\ n=\frac{20}{28}\\ n=0.714$

As nitrogen is a diatomic gas it's molar specific heat at constant pressure C _P is as follows

$\\C_{P}=\frac{7R}{2}\\ C_{P}=\frac{7\times 8.3}{2}\\ C_{P}=29.05\ J\ mol^{-1}\ ^{o}C^{-1}$

Rise in temperature, $\\\Delta T=45 ^{o}C^{-1}$

Amount of heat Q that must be supplied is

$\\Q=nC_{P}\Delta T\\ Q=0.714\times 29.05\times 45\\ Q=933.38\ J$

Q12.3 (a) Explain why

(a) Two bodies at different temperatures $T_1$ and $T_2$ if brought in thermal contact do not necessarily settle to the mean temperature $(T_1 + T_2) / 2$ .

Answer:

As we know the heat will flow from the hotter body to the colder body till their temperatures become equal. That temperature will be equal to the mean of the initial temperatures of the bodies only if the two thermal capacities of the two bodies are equal.

Q12.3 (b) Explain why

The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

Answer:

The coolant should have high specific heat so that it can absorb large amounts of heat without itself getting too hot and its temperature lies in the permissible region. Higher is the specific heat more will be the heat absorbed by the same amount of the material for a given increase in temperature.

Q12.3(c) Explain why

Air pressure in a car tyre increases during driving.

Answer:

As the car is driven the air inside the tyre heats due to frictional forces. Thus the temperature of the air inside the tyre increases and this, in turn, increases the pressure inside the tyre.

Q12.3 (d) Explain why

(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Answer:

The climate of a harbour town is more temperate than that of a town in a desert at the same latitude because of the formation of sea breezes.

Q12.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer:

As the walls of the cylinder and the piston is insulated the process will be adiabatic. i.e. $PV^{\gamma }$ would be constant.

Hydrogen is a diatomic gas and therefore $\gamma =1.4$

Let the initial and final pressure be P ₁ and P ₂ respectively.

Let the initial and final volume be T ₁ and T ₂ respectively.

$\\P_{1}V_{1}^{\gamma }=P_{2}V_{2}^{\gamma }\\ \frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{\gamma }\\ \frac{P_{2}}{P_{1}}=2^{1.4}\\ \frac{P_{2}}{P_{1}}=2.639$

The pressure thus increases by a factor 2.639

Q12.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take $1 cal = 4.19 J$ )

Answer:

In the first case, the process is adiabatic i.e. $\Delta Q=0$

22.3 J work is done on the system i.e. $\Delta w=-22.3\ J$

$\\\Delta Q=\Delta u+\Delta w\\ 0=\Delta u-22.3\\ \Delta u=22.3\ J$

Since in the latter process as well the initial and final states are the same as those in the former process $\Delta u$ will remain the same for the latter case.

In the latter case net heat absorbed by the system is 9.35 cal

$\\\Delta Q=9.35\times 4.2\\ \Delta Q=39.3\ J$

$\\\Delta w=\Delta Q-\Delta u\\ \Delta w=39.3-22.3\\ \Delta w=17.0\ J$

The network done by the system in the latter case is 17.0 J

Q12.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :

(a) What is the final pressure of the gas in A and B ?

(b) What is the change in internal energy of the gas?

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Answer:

As the entire system is thermally insulated and as free expansion will be taking place the temperature of the gas remains the same. Therefore PV is constant.

Initial Pressure P ₁ = 1 atm

Initial Volume, V ₁ = V

Final Volume, V ₂ = 2V

Final Pressure P ₂ will be

$\\P_{2}=\frac{P_{1}V_{1}}{V_{2}}\\ P_{2}=\frac{P_{1}}{2}\\ P_{2}=0.5\ atm$

The final pressure of the gas in A and B is 0.5 atm.

b) Since the temperature of the gas does not change its internal energy would also remain the same.

c) As the entire system is thermally insulated and as free expansion will be taking place the temperature of the gas remains the same.

d) The intermediate states of the system do not lie on its P-V-T surface as the process is a free expansion, it is rapid and the intermediate states are non-equilibrium states.

Q12.7 A steam engine delivers $5.4\times 10^8\ J$ of work per minute and services $3.6\times 10^9\ J$ of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Answer:

Output = $5.4\times 10^8\ J$

Input = $3.6\times 10^9\ J$

Efficiency is $\eta$

$\\\eta =\frac{output}{input}\\ \eta =\frac{5.4\times 10^{8}}{3.6\times 10^{9}}\\ \eta =0.15$

The efficiency of the engine is 0.15.

The percentage efficiency of the engine is 15%.

Heat wasted per minute = Heat produced per minute - useful work done per minute

$\\=3.6\times 10^{9}-5.4\times 10^{8}\\ =3.06\times 10^{9}\ J\ min^{-1}$

Q12.8 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

Answer:

The rate at which heat is supplied $\Delta Q=100\ W$

The rate at which work is done $\Delta W=75\ J s^{-1}$

Rate of change of internal energy is $\\\Delta u$

$\\\Delta u=\Delta Q-\Delta w\\ \Delta u=100-75\\ \Delta u=25\ J\ s^{-1}$

The internal energy of the system is increasing at a rate of 25 J s ^-1

Q12.9 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)

screenshot123

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F

Answer:

The work done by the gas as it goes from state D to E to F is equal to the area of triangle DEF

DF is change in pressure = 300 N m ^-2

FE is change in Volume = 3 m ³

$\\area(DEF)=\frac{1}{2}\times DF\times FE\\ area(DEF)=\frac{1}{2}\times300\times 3\\ area(DEF)=450\ J$

Work done is therefore 450 J.

Q12.10 A refrigerator is to maintain eatables kept inside at 9 ^o C. If room temperature is 36 ^o C, calculate the coefficient of performance.

Answer:

Room Temperature, T ₁ = 36 ^o C = 309 K

The temperature which has to be maintained inside the fridge, T ₂ = 9 ^o C = 282 K

Coefficient of performance is E

$\\E=\frac{T_{2}}{T_{1}-T_{2}}\\ E=\frac{282}{309-282}\\ E=\frac{282}{27}\\ E=10.4$

The thermodynamics class 11 physics questions and answers for Thermodynamics in Class 11 Physics are highly beneficial for students as they provide a solid foundation in understanding the principles of heat, work, energy, and the laws of thermodynamics. This knowledge not only aids in performing well in exams but also simplifies the comprehension of subsequent chapters in physics, particularly those related to heat and energy transfer. Additionally, problem-solving skills developed through these thermodynamics class 11 physics ncert solutions are transferable and valuable in a wide range of academic and practical scenarios.

NCERT Solutions for Class 11 Physics Chapter Wise

Chapter 1	Physical world
Chapter 2	Units and Measurement
Chapter 3	Motion in a straight line
Chapter 4	Motion in a Plane
Chapter 5	Laws of Motion
Chapter 6	Work, Energy and Power
Chapter 7	System of Particles and Rotational motion
Chapter 8	Gravitation
Chapter 9	Mechanical Properties of Solids
Chapter 10	Mechanical Properties of Fluids
Chapter 11	Thermal Properties of Matter
Chapter 12	Thermodynamics
Chapter 13	Kinetic Theory
Chapter 14	Oscillations
Chapter 15	Waves

NCERT solutions of thermodynamics physics 11: Important Formulas and Diagrams

Thermodynamics First Law

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Where:

ΔQ is heat provided to the system by the environment, ΔW denotes the amount of work done by the system in relation to the environment, and ΔU is the change in the system's internal energy

Thermodynamics Second Law

ds >= 0

Where: S is entropy

Thermodynamics Third Law

S →0 as T →0

Where: s is entropy and T is the temperature

Molar Specific Heat Capacity

Here: μ= the mass of material in moles, C is the substance's molar specific heat capacity, ΔQ is the quantity of heat that a substance absorbs or rejects and ΔT stands for temperature change.

These thermodynamics physics class 11 formulas are valuable tools that students can utilize to calculate and analyze various thermodynamic parameters, making it easier to solve complex problems and gain a deeper understanding of heat, energy, and the laws of thermodynamics. By using these ncert solutions of thermodynamics physics 11 formulas effectively, students can excel in exams and develop strong problem-solving skills, which are transferable to a wide range of academic and real-world scenarios

Importance of NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

As the Class 11 exams and competitive exams like NEET and JEE Main are considered the NCERT solutions for Class 11 are important.
At least one question is expected for JEE Mains and two questions are expected for NEET from the chapter Thermodynamics.

Key features of class 11 physics ch 12 ncert solutions

Comprehensive Coverage: These numericals of thermodynamics class 11 physics offer a thorough explanation of the chapter's concepts, ensuring students have a comprehensive understanding of thermodynamics.

Exam Preparation: Designed to help students excel in exams, these solutions provide step-by-step guidance for problem-solving and concept revision.

Enhanced Learning: The detailed class 11 physics thermodynamics questions and answers pdf facilitate effective revision, making it easier for students to grasp and retain key concepts.

Homework Assistance: These class 11 physics chapter 12 exercise solutions simplify homework problems by providing clear guidance, ensuring students can complete assignments confidently.

Access to Support: Students can seek clarification for any doubts or uncertainties not addressed in the solutions, which can contribute to better exam performance.

NCERT Solutions for Class 11 Subject wise

NCERT solutions for Class 11 Biology

NCERT solutions for Class 11 Maths

NCERT solutions for Class 11 Chemistry

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

If you have any queries or uncertainties that are not addressed in the class 11 physics chapter 12 exercise solutions or any other chapter, you can contact us. You will get answers to your questions, which can aid you in achieving better scores in your exams.

Frequently Asked Questions (FAQs)

1. What is the weightage of Thermodynamics for JEE Main

One question from Class 11 chapter Thermodynamics is expected for JEE Main. Students should practice previous year papers of JEE Main to get a good score.

2. What is the weightage of thermodynamics for NEET

Two questions from class 11 chapter Thermodynamics is expected for NEET. The competition level of NEET exam is very high. So practicing more questions is important. To practice more questions refer to NCERT exemplar and NEET previous year papers

3. Is Thermodynamics helpful in higher studies

Yes, the concepts of thermodynamics are used in various fields of Engineering and Science. Not only Thermodynamics chapter, all the chapters in NCERT book are basics of some field of higher studies.

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