## NCERT Solutions for Class 11 Physics Chapter 12 - ThermodynamicsIn this article we will be covering NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics. Students will gain the self-confidence to give the exam without anxiety. The laws governing thermal energy will be covered in this chapter for the students. In this chapter, a quick explanation of how work can be transformed into heat and vice versa is given. The chapter-by-chapter NCERT Solutions has been created by Javatpoint in accordance with the revised CBSE Syllabus 2023-2024 to aid students in understanding the fundamental concepts of this chapter. The ideas of temperature, heat, and its transformation into different kinds of energy are all covered by the notion of thermodynamics in this chapter. This chapter also includes numericals, which should be regularly practised in order to do well in the exam. To assist students in reviewing the chapter a few days before the test, the solutions include formulas, definitions, and essential topics that have been emphasised in plain language. ## NCERT Solutions for Class 11 Physics Chapter 12
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 × 10
The flowing rate of water is given as 3.0 litre per minute The temperature of the water increases from 27° C to 77° C,because of the geyser Thus,Initial temperature of the water,T Final temperature of the water,T ∴ The rise in temperature will be given as,∆T=T =77° C-27° C=50° C Heat of combustion is given as=4×10 Specific heat of water,c=4.2 J g Mass of the flowing water,m=3.0 litre/min=3000 g/min Total heat used is given as, ∆Q=mc∆T =3000×4.2×50 =6.3×10
What amount of heat must be supplied to 2.0 × 10
The molecular mass of the nitrogen is given as - 20×10 The temperature rise in nitrogen is = 45° C The molecular mass of N And the universal gas constant=8.3 J mol Let the number of moles be,n Where,m=mass of the nitrogen and M=molecular mass of nitrogen Substituting values in above equation we get, For a diatomic gas,let us consider the molar specific heat at constant pressure is C where,R=Universal gas constant Substituting the values in the above equation,we get, Now,let the amount of heat supplied to the nitrogen gas be Q, Q=nC Substituting the values in the above equation,we get, Q=0.714×29.05×45 Q=934 J Thus, the amount of heat supplied to the nitrogen is 934 J.
Explain why (a) Two bodies at different temperatures T (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.) If two bodies are at T b.) The capacity to quickly absorb heat from various plant components is a necessary quality of the coolant. This requires a high coolant specific heat. The capacity of the coolant to absorb heat will increase as the heat capacity increases. c.) When the car is moving, the air molecules inside the tyre become hotter. Charles' Law states that the relationship between temperature and pressure is inverse. Pressure will therefore rise as the temperature within the tyre rises. d.) The relative humidity in Harbour Town is higher than in the desert, hence the temperature there will be higher.
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?
The cylinder's surrounds are totally isolated from it. As a result, there is no heat transfer between the system (cylinder) and its environment. The process is hence adiabatic. Let, The initial pressure inside the cylinder = P The final pressure inside the cylinder = P The initial volume inside the cylinder = V The final volume inside the cylinder = V For an adiabatic process,we have: (P Half of the original volume is compressed into the final volume. Hence, As a result, the pressure rises by 2.639 times.
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 by the system is=22.3 J The quantity of heat absorbed by the system=9.35 J We know that 1 cal is equivalent to 4.19 J Let Q denote the change in heat,then ∆Q=0 This is because the work done is in an adiabatic process. Let ∆W be the amount of work transferred from the system. ∆W=-22.3 J [Negative sign indicates that the work is done on the system] Let ∆U denote the change in internal energy of the system. Thus,from the first law of thermodynamics, ∴ ∆Q=∆U+∆(W) . . .(i) Substituting values in above equation, 0=∆U+(-23) ∆U=22.3 J When the system absorbs 9.35 cal of heat to transition from state A to state B, then, ∆Q'=9.35 cal=9.35×4.19=39.176 J Substitute the values according to equation (1)'s first law of thermodynamics. ∆Q'=∆U+∆W' ∆W'=39.1765-22.3 =16.9 J As a result, the system's work done is 16.9 J.
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?
a.) The gas will now begin to flow from cylinder P to cylinder Q, which is totally evacuated, as soon as the stop cock is opened. As a result, the volume of the gas will double because both cylinders have an equal capacity. Hence, the pressure will be halved from its initial value because the relationship between pressure and volume is inverse. Each cylinder will now have a pressure of 0.5 atm because the gas in cylinder P had a starting pressure of 1 atm. b.) In this instance, the gas's internal energy will remain constant, or ΔU = 0. This is due to the fact that only when work is performed by or on the system can the internal energy change. As there is no work being done by the gas or on the gas in this instance. As a result, the gas's internal energy will remain constant. c.) The temperature of the gas will remain unchanged. This is due to the fact that no effort is done by the gas during expansion. As a result, during this operation, the gas's temperature won't change. d.) The aforementioned scenario clearly illustrates unrestrained expansion, which is uncontrollable, quick, and unpredictable. The intermediate states do not satisfy the gas equation and do not lie on the system's pressure-volume-temperature surface because they are in non-equilibrium states.
A steam engine delivers 5.4×10
The work done by the system engine per minute,W=5.4×10 The amount of heat supplied from the boiler,H=3.6×10 Hence, the engine's efficiency is 15% as a percentage. Amount of heat wasted=3.6×10 =30.6×10 =3.6×10 Hence, 3.06 x 109 J of heat is lost per minute.
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?
According to the law of conservation of energy,. Total Energy=Work Done+Internal Energy. ∆Q=∆W+∆U. Heat is supplied at a rate of,∆Q=100 W. Work is done at a rate of,∆W=75 J s The rate of changing internal energy is ∆U. ∆U=∆Q-∆W. =100-75. =25 J s The system's internal energy is rising at a rate of 25 W.
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
Total work done by the gas from D to E to F=Area of ∆DEF DF= change in pressure =600-300 =300 N m FE=Change in volume =5.0-2.0 =3.0 m Substituting values in equation 1, =450 J 450 J is the total work performed by the gas from D to E to F.
A refrigerator is to maintain eatables kept inside at 90 C. If room temperature is 360 C, calculate the coefficient of performance.
The temperature inside the refrigerator,T_1=9° C=273+9=282 K The room temperature is,T Substituting the values in the above equation, we get, As a result, the refrigerator's performance coefficient is 10.44. |