NCERT Solutions for Class 11 Physics Chapter 12

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 × 10⁴ Jg^-1?

Ans.

Here,initial temperature of  water, T 1 = 27°C Final temperature of  water, T 2 = 77°C ∴ Rise in temperature, ΔT = T 2 –T 1 ΔT= 77°C – 27°C = 50°C Heat of combustion = 4×10 4 Jg -1 Specific heat of water, c = 4 .2 J g –1 °C –1 Mass of water heated, m = 3 .0 litremin -1 = 3000 gmin -1 Total amount of heat used is given as: ΔQ = mc ΔT ΔQ = 3000 gmin -1 × 4 .2 J g –1 °C –1 × 50°C = 6 .3×10 5   Jmin -1    ∴Rate of consumption of fuel = 6 .3 × 10 5   Jmin -1 4 × 10 4 Jg -1 ∴Rate of consumption of fuel = 15 .75 gmin -1

Q.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.)

Ans.

Here,mass of nitrogen, m = 2 .0 ×10 –2 kg = 20 g Increase in the temperature, ΔT = 45°C Molecular mass of nitrogen, M = 28 g Universal gas constant, R = 8 .3 J mol –1 K –1 Number of moles is given as: n =  m M =  20 g 28 g = 0.714 As nitrogen is a diatomic gas, At constant pressure, molar specific heat for nitrogen, C P = 7 2 R C P = 7 2 × 8 .3 J mol –1 K –1 = 29.05  J mol -1 K -1 Total amount of heat to be supplied is given as: ΔQ =  nC P ΔT ΔQ = 0.714 × 29.05  J mol -1 K -1 × 45 K ΔQ = 933.38 J ∴Total amount of heat to be supplied = 933.38 J

Q.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.

Ans.

(a) When we bring two bodies at different temperatures and in thermal contact, heat starts flowing from the body at the higher temperature to the body at the lower temperature. This process goes on till the temperatures of both the bodies become equal. This is due to the fact the fact that heat flows from higher temperature to lower temperature. The equilibrium temperature of any two bodies can be equal to the mean temperature only if they have the same thermal capacities.

(b) Since the heat-absorbing capacity of the substance is directly proportional to its specific heat, therefore, the specific heat of the coolant in a chemical or nuclear plant must be high.

(c) While driving, the motion of air molecules inside the car increases. Due to the increased motion of air molecules inside the car, the temperature inside the car increases. According to Gay-Lussac’s law, at constant volume, temperature is directly proportional to pressure. Therefore, due to the increased temperature inside the car, pressure inside it also increases.

(d) Since the relative humidity in a harbour town is more than the relative humidity in a desert town, therefore, the climate of a harbour town is more temperate than that of a town in the desert at the same latitude.

Q.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?

Ans.

Since the cylinder is insulated from its surroundings, the exchange of heat does not take place between the system ( cylinder ) and its surroundings. Therefore, it is an adiabatic process . Consider P i and P f be the initial and final pressure of cylinder respectively. Let V i and V f respectively are the initial and final volume of the cylinder. Ratio of specific heats, γ = 1.4 In an adiabatic process, P i V i γ = P f V f γ (i) Since the final volume is compressed to half of its initial volume, ∴ V f = V i 2 (ii) Substituting the value of  V f  from equation (ii) in equation (i), we obtain: P i ( V i ) γ = P f ( V i 2 ) γ P f P i = ( V i ) γ ( V i 2 ) γ = ( 2 ) γ = ( 2 ) 1.4 = 2.639 ∴Pressure increases by a factor of 2.639.

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)

Ans.

Here, work done ( W ) on the system = 22.3 J As it is an adiabatic process,  ∴ΔQ = 0 Since the work is done on the system,  ∴ΔW = –22.3 J According to first law of thermodynamics, ΔQ = ΔU + ΔW Here, ΔU = Change in the internal energy of the gas ∴ΔU = ΔQ – ΔW ΔU = 0 – ( – 22.3 J ) ΔU = +22.3 J For the movement of gas from state A to state B via a process, ΔQ = 9.35 cal = 9.35 × 4.19 J = 39.1765 J Heat absorbed is given as: ΔQ = ΔU + ΔW ∴ΔW = ΔQ – ΔU ΔW = 39.1765 J – 22.3 J = 16.8765 J ∴Total work done by the system = 16.88 J

Q.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?
(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?

Ans.

(a) As we remove the stopcock between cylinders A and B is suddenly, the volume available to the gas is doubled. According to Boyle’s law, at constant temperature, volume is inversely proportional to pressure. Therefore, as the volume available to the gas is doubled, the pressure will lowered to half of its original value. Since the initial pressure of the gas is 1 atm, therefore, the final pressure in each cylinder will be 0.5 atm.

(b) Since in this case, no work is done by or on the gas and no heat is supplied by an external source, therefore, the internal energy of the gas will not change at all.

(c) Since the internal energy of the gas does not change at all, therefore, the temperature of the gas will not change.

(d) In this process, the free and rapid expansion of gas takes place and it cannot be controlled. In this case, the intermediate states of the gas are non-equilibrium states and they do not satisfy the ideal gas equation. It shows that the given process is not quasi-static. Therefore, the intermediate states of the system do not lie on its P-V-T surface.

Q.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?

Ans.

Here, work done by steam engine per minute, W = 5.4 ×  10 8  J Heat supplied from hot  reservoir, H 1  = 3 .6 ×10 9  J Efficiency of the steam engine =  Output energy Input energy ∴η = W H η = 5.4 ×  10 8 J 3.6 ×  10 9 J = 0.15 ∴The percentage efficiency of the engine is 15 %. Heat wasted per  minute, H 2  =  H 1  - W H 2  = 3 .6 × 10 9 J – 5 .4 ×10 8 J H 2  = 30 .6 × 10 8 = 3 .06 ×10 9 J ∴The amount of heat wasted per minute is 3.06 ×  10 9 J.

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

Ans.

Here, rate of heat supplied, ΔQ = 100 W                  Work done by the  system, ΔW = 75 Js -1 According to first law of thermodynamics, we have: ΔQ = ΔU + ΔW Here, ΔU = Change in internal energy ∴ΔU = ΔQ – ΔW ∴ΔU =100 Js-1- 75 Js-1= 25 WTherefore, change in the internal energy of the given electric heater = 25 W

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

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

Ans.

From the given figure, Change in pressure from E to F, dP = EF = 5.0 – 2 .0 Nm -2 = 3 .0 Nm -2 Change in volume from F to D, dV = FD = 600 – 300 = 300 m 3 From the given figure, it can be observed that Total work done by the gas from D to E to F, W = Area of ΔDEF W = Area of ΔDEF =  1 2  × FD × EF W = 1 2 × 300 m 3 × 3  Nm -2 = 450 J ∴Total work done by the gas from D to E to F = 450 J

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

Ans.

\begin{array}{l}\text{Given,}{\text{temperature inside the refrigerator, T}}_{\text{1}}\text{= 9°C = 282 K}\\ {\text{Room temperature, T}}_{\text{2}}\text{= 36 °C = 309 K}\\ \text{Coefficient of performance}\text{of}\text{refrigerator,}\alpha \text{=}\frac{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}{\text{– T}}_{\text{1}}}\\ \alpha \text{=}\frac{\text{282 K}}{\text{(309 – 282) K}}\text{= 10}\text{.44}\\ \therefore \text{Coefficient of performance of the given refrigerator = 10}\text{.44}\end{array}