NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

NCERT Solutions for Class 11 Physics Chapter 12 - Thermodynamics

Welcome to the Chapter 12 - Thermodynamics, Class 11 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 12 - Thermodynamics. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Thermodynamics and excel in their exams. By going through these Thermodynamics question answers, you can strengthen your foundation and improve your performance in Class 11 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

Q 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 x 10⁴ J/g?

Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water, raising the temperature from 27°C to 77°C.

Initial temperature, T₁ = 27°C

Final temperature, T₂ = 77°C

∴Rise in temperature, ΔT = T₂ - T₁ = 77 - 27= 50°C

Heat of combustion = 4 × 10⁴ J/g

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

Mass of flowing water, m = 3.0 litre/min = 3000 g/min

Total heat used, ΔQ = mc ΔT

= 3000 × 4.2 × 50

= 6.3 × 10⁵ J/min

∴Rate of consumption = = 15.75 g/min

Q 10:

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

Temperature inside the refrigerator, T₁ = 9°C = 282 K

Room temperature, T₂ = 36°C = 309 K

Coefficient of performance = T₁ / T₂ - T₁

= 282 / 309 - 282

= 282 / 27

= 10.44

Therefore, the coefficient of performance of the given refrigerator is 10.44.

Q 2:

What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N² = 28; R = 8.3 J mol^-1 K^-1.)

Mass of nitrogen, m = 2.0 × 10^-2 kg = 20 g

Rise in temperature, ΔT = 45°C

Molecular mass of N₂, M = 28

Universal gas constant, R = 8.3 J mol^-1 K^-1

Number of moles, n = m / M

= 2.0 x 10^-2 x 10³ / 28 = 0.714

Molar specific heat at constant pressure for nitrogen, C_P = 7/2R

= 7/2 x 8.3

= 29.05 J mol^-1 K^-1

The total amount of heat to be supplied is given by the relation:

ΔQ = nCP ΔT

= 0.714 × 29.05 × 45

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

Q 3:

(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 +T2)/2. (b) 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. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂)/2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Q 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)

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.

This is an adiabatic process. Hence, change in heat is zero.

∴ ΔQ = 0

ΔW = -22.3 J (Since the work is done on the system)

From the first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where,

ΔU = Change in the internal energy of the gas

∴ ΔU = ΔQ - ΔW = - (- 22.3 J)

ΔU = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

ΔQ = 9.35 cal = 9.35 x 4.19 = 39.1765 J

Heat absorbed, ΔQ = ΔU + ΔQ

∴ΔW = ΔQ - ΔU = 39.1765 - 22.3 = 16.8765 J

Therefore, 16.88 J of work is done by the system.

Q 7:

A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Work done by the steam engine per minute, W = 5.4 × 10⁸ J

Heat supplied from the boiler, H = 3.6 × 10⁹ J

Efficiency of the engine = Output energy / Input energy

∴ n = W / H

= 5.4 × 10⁸ / 3.6 × 10⁹

Hence, the percentage efficiency of the engine is 15 %.

Amount of heat wasted = 3.6 × 10⁹ - 5.4 × 10⁸

= 30.6 × 10⁸ = 3.06 × 10⁹ J

Therefore, the amount of heat wasted per minute is 3.06 × 10⁹ J.

Q 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?

Heat is supplied to the system at a rate of 100 W.

∴Heat supplied, Q = 100 J/s

The system performs at a rate of 75 J/s.

∴Work done, W = 75 J/s

From the first law of thermodynamics,we have:

Q = U + W

Where, U = Internal energy

∴U = Q - W

= 100 - 75

= 25 J/s

= 25 W

Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Frequently Asked Questions about Thermodynamics - Class 11 Physics

- 1. How many questions are covered in Thermodynamics solutions?
- All questions from Thermodynamics are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
- 2. Are the solutions for Thermodynamics helpful for exam preparation?
- Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
- 3. Can I find solutions to all exercises in Thermodynamics?
- Yes, we provide solutions to all exercises, examples, and additional questions from Thermodynamics with detailed explanations.
- 4. How do these solutions help in understanding Thermodynamics concepts?
- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
- 5. Are there any tips for studying Thermodynamics effectively?
- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Thermodynamics

The Thermodynamics is an important chapter of 11 Physics. This chapter’s important topics like Thermodynamics are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

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