# NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics

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NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics deals with the concept of temperature and heat conversion to another form of energy. Students can solve all the numerical and also learn about the theories. Our experts have prepared the NCERT solution with utmost concern to make it easy for students to prepare for exams. Important concepts discussed in the chapter are

• Thermodynamics – Thermodynamics is the science that deals with the relationship between heat, work, temperature, and energy. The law of thermodynamics deals with how the energy in a system changes and how the system can perform useful work in its surroundings.
• Thermal equilibrium – when two body which are in physical contact with each other and there is no heat transfer between them then the two bodies are said to be in thermal equilibrium.
• Zeroth law of thermodynamics - Zeroth law of thermodynamics states that if two isolated systems are in thermal equilibrium with the third system then they are in thermal equilibrium with each other.
• Heat - The form of energy that can be transferred between two materials of different temperatures is called heat. The symbol of heat is "Q".
• Internal energy – internal energy of a system is an isolated system is defined as total potential energy associated with the vibrational motion and electrical energy of atoms within molecules and kinetic energy due to the motion of molecule.
• Work – the process of transfer of energy transferred to or from an object due to the application of force along a displacement is called work.
• First law of thermodynamics - The first law of thermodynamics is the law of conservation of energy.
• Specific heat capacity - The amount of heat needed to change the heat content of a 1-mole substance due to 1 Kelvin or 1 °C.
• Thermodynamic processes – change from one equilibrium macrostate to another macrostate is called thermodynamics. Type of thermodynamics are
• Isobaric process
• Isochoric process
• Isothermal process
• Quasi-static process
• Heat engines – an engine in a system that converts heat to mechanical energy.
• Refrigerators and heat pumps
• Second law of thermodynamics – when energy is transferred or transformed from one form to another more and more of it is wasted.
• Reversible and irreversible processes – the processes in which the change is not permanent and can be reversed are called reversible change while irreversible changes are those where changes are permanent.
• Carnot heat engine - A Carnot heat engine is a heat engine that operates on the Carnot cycle.

In our NCERT Solutions Class Physics, we have articulated all the solutions of the particular chapter in the pointwise method. All the tough concepts are explained in simple language. It is advised by our experts that students must refer to our NCERT solution for their studies; it is the best way to secure good marks in their exams.

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NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics

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 × 104 J/g?

Solution:

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 = $\frac {6.3 \times10^5}{4\times10^4}$ = 15.75 g/min

2. What amount of heat must be supplied to 2.0 × 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.)

Solution:

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

$\frac {2.0 \times10^{-2} \times10^3}{28} = 0.714$

Molar specific heat at constant pressure for nitrogen CP = $\frac 72R = \frac 72 \times 8.3$ = 29.05 J mol-1K-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.

3. Explain why

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.

Solution:

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.

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?

Solution:

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.

Initial pressure inside the cylinder = P1

Final pressure inside the cylinder = P2

Initial volume inside the cylinder = V1

Final volume inside the cylinder = V2

Ratio of specific heats, γ = 1.4

For an adiabatic process, we have:

$P_1V^T_1 = P_2V^T_2$

The final volume is compressed to half of its initial volume. Hence, the pressure increases by a factor of 2.639.

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)

Solution:

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 × 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.

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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.

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

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

c) What is the change in the temperature 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?

Solution:

a) 0.5 atm

The volume available to the gas is doubled as soon as the stopcock between cylinders A and B is opened. Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5 atm.

b) Zero

The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

c) Zero

Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

d) No

The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in nonequilibrium states, they do not lie on the P-V-T surface of the system.

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

Solution:

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

Heat supplied from the boiler, H = 3.6 × 109

Efficiency of the engine = $\frac {Output \, energy}{Input\, energy}$

∴ η = $\frac WH = \frac {5.4\times10^8}{3.6\times10^9} = 0.15$

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

Amount of heat wasted = 3.6 × 109 – 5.4 × 10

= 30.6 × 108 = 3.06 × 109

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

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?

Solution:

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.

9. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Figure. 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

Solution:

Total work done by the gas from D to E to F = Area of ΔDEF

Area of ΔDEF = $\frac 12DE \times EF$

Where,

DF = Change in pressure

= 600 N/m2 – 300 N/m

= 300 N/m

FE = Change in volume

= 5.0 m3 – 2.0 m3

= 3.0 m

Area of ΔDEF =$\frac 12$ x 300 + 3 = 450 J

Therefore, the total work done by the gas from D to E to F is 450 J.

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

Solution:

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

Room temperature, T2 = 36°C = 309 K

Coefficient of performance = $\frac {T_1}{T_2-T_1} = \frac {282}{309-282} = 10.44$

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

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