- Chapter 1- Physical World
- Chapter 2- Units and Measurements
- 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 13- Kinetic Theory
- Chapter 14- Oscillations
- Chapter 15- Waves

Chapter 1- Physical World |
Chapter 2- Units and Measurements |
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 13- Kinetic Theory |
Chapter 14- Oscillations |
Chapter 15- Waves |

**Answer
1** :

Given

Water is flowing at a rate of 3.0 litre/min

The geyser heats the water, raising the temperature from 27^{0} C to 77^{0} C.

Initial temperature, T_{1} = 27^{0} C

Final temperature, T_{2} = 77^{0} C

Rise in temperature, T = T_{2} – T_{1}

= 77 – 27

= 50^{0} C

Heat of combustion = 4 x 10^{4} J/ g

Specific heat of water, C = 4.2 J / g^{ 0}C

Mass of flowing water, m = 3.0 litre / min

= 3000 g / min

Total heat used, Q = mcT

= 3000 x 4.2 x 50

On calculation, we get,

= 6.3 x 10^{5} J/ min

Rate of consumption = 6.3 x 10^{5 }/ (4 x10^{4})

We get,

= 15.75 g/min

Therefore, rate of consumption is 15.75 g/min

**Answer
2** :

Given,

Mass of nitrogen = 2 x 10^{-2} kg

= 20 g

Rise in temperature = ΔT

= 45^{0} C

Heat required = Q =?

Q = nCT

We know,

C = 7R / 2 (diatomic molecule)

C = 7 x 8.3 / 2

n (no. of moles) = w / m

where,

w = 20 g

m = 28 u

n = 20 / 28

n = 1/ 1.4 moles

Let the temperature be 45 K

Q = 10 / 14 x 7 / 2 x 8.3 x 45

We get,

Q = 933.75 J

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.

(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.

**Answer
3** :

(i). When two bodies having different temperatures say, T1 and T2 are brought in thermal contact with each other, there is a flow of heat from the body at the higher temperature to the body at the lower temperature till both the body reaches to an equilibrium position, i.e., both the bodies are having equal temperature. The equilibrium temperature is only equal to the mean temperature when the thermal capacities of both the bodies are equal.

(ii).The coolant used in a chemical or nuclear plant should always have a high specific heat. Higher is the specific heat of the coolant, higher is its capacity to absorb heat and release heat. Therefore, a liquid with a high specific heat value 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.

(iii). When the driver is driving a vehicle then due to the motion of air molecules the air temperature inside the tyre increases. And according to the Charles’ law, the temperature is directly proportional to pressure.

Therefore, when the temperature inside a tyre increases, then there is also an increase of air pressure.

(iv). The relative humidity in a harbour town is more than that of the relative humidity in a desert town. Humidity is a measure of water vapor in the atmosphere and the specific heat of water vapor is very high. Therefore, the climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

**Answer
4** :

The cylinder is completely insulated from its surroundings.

Therefore, no heat is exchanged between the system (cylinder)and its surroundings.

Thus, the process is adiabatic

Initial pressure inside the cylinder = P_{1}

Final pressure inside the cylinder = P_{2}

Initial volume inside the cylinder = V_{1}

Final volume inside the cylinder = V_{2}

Ratio of specific heat, γ = C_{p} / C_{v} =1.4

For an adiabatic process, we have:

P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}

The final volume is compressed to half of its initial volume

Hence,

V_{2} =V_{1} /2

P_{1}V_{1}^{γ} = P_{2}(V_{1 }/ 2)^{γ}

P_{2} /P_{1} =V_{1}^{γ} / (V_{1} /2)^{γ}

= 2^{1.4}

We get,

= 2.639

Therefore, the pressure increases by a factor of 2.639

**Answer
5** :

Given

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

This is an adiabatic process.

Therefore, change in heat is zero.

So,

ΔQ = 0

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

From first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where,

ΔU = Change in the internal energy of the gas

Hence,

ΔU = ΔQ – ΔW

= 0 – (-22.3 J)

We get,

ΔU = + 22.3 J

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

ΔQ = 9.35 cal

= 9.35 x 4.19

On calculation, we get,

= 39.1765 J

Heat absorbed, ΔQ = ΔU + ΔW

Thus,

ΔW = ΔQ – ΔU

= 39.1765 – 22.3

We get,

= 16.8765 J

Hence, 16.88 J of work is done by the system

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?

(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?

**Answer
6** :

(a). Now, as soon as the stop cock is opened the gas will start flowing from cylinder P to cylinder Q which is completely evacuated and thus the volume of the gas will be doubled because both the cylinders have equal capacity. And since the pressure is inversely proportional to volume, hence the pressure will get decreased to half of the original value.

Since, the initial pressure of the gas in cylinder P is 1 atm. Therefore, the pressure in each of the cylinder will now be 0.5 atm.

(b). Here, in this case, the internal energy of the gas will not change i.e. ΔU = 0. It is because the internal energy can change only when the work is done by the system or on the system. Since in this case, no work is done by the gas or on the gas.

Therefore, the internal energy of the gas will not change.

c) There will be no change in the temperature of the gas. It is because during the expansion of gas there is no work being done by the gas.

Therefore, there will be no change in the temperature of the gas in this process.

d). The above case is the clear case of free expansion and free expansion is rapid and it cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non – equilibrium states, they do not lie on the Pressure-Volume – Temperature surface of the system

**Answer
7** :

Given

Work done by the steam engine per minute, W = 5.4 x 10^{8} J

Heat supplied from the boiler, H = 3.6 x 10^{9} J

Efficiency of the engine = Output Energy / Input Energy

Hence,

η = W / H

= 5.4 x 10^{8} /(3.6 x 10^{9})

On simplification, we get,

= 0.15

Therefore, the percentage efficiency of the engine is 15%

Amount of heat wasted = 3.6 x 10^{9} –5.4 x 10^{8}

We get,

= 30.6 x 10^{8}

= 3.06 x 10^{9} J

Hence, the amount of heat wasted per minute is 3.06 x 10^{9} J

**Answer
8** :

According to law of conservation of energy

Total energy = work done + internal energy

ΔQ = ΔW + ΔU

Here,

Rate at which heat is supplied ΔQ = 100 W

Rate at which work is done ΔW = 75 Js^{-1}

Rate of change of internal energy is ΔU

ΔU = ΔQ – ΔW

ΔU = 100 – 75

We get,

ΔU = 25 J s^{-1}

Hence,

The internal energy of the system is increasing at a rate of 25W

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
9** :

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

Area of ∆DEF = (1/2) x DE x EF

Where,

DF = Change in pressure

= 600 N/m^{2} –300 N/m^{2}

We get,

= 300 N/ m^{2}

FE = Change in volume

= 5.0 m^{3} –2.0 m^{3}

We get,

= 3.0 m^{3}

Area of ∆DEF = (1/ 2) x 300 x 3

On further calculation, we get,

= 450 J

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

**Answer
10** :

Temperature inside the refrigerator, T_{1} =9^{0} C

= 273 + 9

= 282 K

Room temperature, T_{2} = 36^{0} C

= 273 + 36

= 309 K

Coefficient of performance = (T_{1}) / (T_{2} –T_{1})

On substituting, we get,

= 282 / (309 – 282)

We get,

= 10.44

Hence, the coefficient of performance of the given refrigeratoris 10.44

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