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NCERT Class 11 Physics Textbook Chapter 12 With Answer PDF Free Download
Chapter 12: Thermodynamics
In the previous chapter, we studied the thermal properties of matter. In this chapter, we shall study laws that govern thermal energy.
We shall study the processes where work is converted into heat and vice versa. In winter, when we rub our palms together, we feel warmer; here work done in rubbing produces the ‘heat’.
Conversely, in a steam engine, the ‘heat’ of the steam is used to do useful work in moving the pistons, which in turn rotate the wheels of the train.
In physics, we need to define the notions of heat, temperature, work, etc. more carefully. Historically, it took a long time to arrive at the proper concept of ‘heat’.
Before the modern picture, heat was regarded as a fine invisible fluid filling in the pores of a substance.
On contact between a hot body and a cold body, the fluid (called caloric) flowed from the colder to the hotter body!
This is similar to what happens when a horizontal pipe connects two tanks containing water up to different heights.
The flow continues until the levels of water in the two tanks are the same.
Likewise, in the ‘caloric’ picture of heat, heat flows until the ‘caloric levels’ (i.e., the temperatures) equalize.
In time, the picture of heat as fluid was discarded in favor of the modern concept of heat as a form of energy. An important experiment in this connection was due by Benjamin Thomson (also known as Count Rumford) in 1798.
He observed that the boring of a brass cannon generated a lot of heat, indeed enough to boil water.
More significantly, the amount of heat produced depended on the work done (by the horses employed for turning the drill) but not on the sharpness of the drill.
In the caloric picture, a sharper drill would scoop out more heat fluid from the pores; but this was not observed.
A most natural explanation of the observations was that heat was a form of energy and the experiment demonstrated the conversion of energy from one form to another–from work to heat.
Thermodynamics is the branch of physics that deals with the concepts of heat and temperature and the inter-conversion of heat and other forms of energy.
Thermodynamics is a macroscopic science. It deals with bulk systems and does not go into the molecular constitution of matter.
In fact, its concepts, and laws were formulated in the nineteenth century before the molecular picture
of the matter was firmly established.
Thermodynamic description involves relatively few macroscopic variables of the system, which are suggested by common sense and can be usually measured directly.
A microscopic description of a gas, for example, would involve specifying the coordinates and velocities of the huge number of molecules constituting the gas.
The description of the kinetic theory of gases is not so detailed but it does involve the molecular distribution of velocities. Thermodynamic description of a gas, on the other hand, avoids the molecular description altogether.
Instead, the state of a gas in thermodynamics is specified by macroscopic variables such as pressure, volume, temperature, mass, and a composition that is felt by our sense perceptions and is measurable.
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NCERT Solutions Class 11 Physics Chapter 12 Thermodynamics
1. A geyser heats water flowing at the rate of 3.0 liters 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?
Water is flowing at a rate of 3.0 litre/min
The geyser heats the water, raising the temperature from 270 C to 770 C.
Initial temperature, T1 = 270 C
Final temperature, T2 = 770 C
Rise in temperature, T = T2 – T1
= 77 – 27
= 500 C
Heat of combustion = 4 x 104 J / g
Specific heat of water, C = 4.2 J / g 0C
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 105 J / min
Rate of consumption = 6.3 x 105 / (4 x 104)
= 15.75 g/min
Therefore, rate of consumption is 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.)
Mass of nitrogen = 2 x 10-2 kg
= 20 g
Rise in temperature = ΔT
= 450 C
Heat required = Q =?
Q = nCT
C = 7R / 2 (diatomic molecule)
C = 7 x 8.3 / 2
n (no. of moles) = w / m
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
Q = 933.75 J
Thermodynamics Textbook With Solutions PDF Free Download