Class 12 Physics Second law of thermodynamics Notes

Class 12 Physics Second law of thermodynamics Notes
UNIT 2
CLASS 12 PHYSICS • HEAT AND THERMODYNAMICS

Second Law of Thermodynamics

Chapter 5
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Second Law of Thermodynamics

1) Kelvin’s Statement

It is impossible to get a continuous supply of work from a body by cooling it to a temperature lower than that of its surroundings.

2) Clausius Statement

It is impossible to make flow of heat from the body at lower temperature to the body at higher temperature without any external workdone.

Heat Engine

A heat engine is a device which converts heat energy into mechanical work. It consists of three parts:

1) Source

The source is a hot body at constant high temperature from which the heat engine can draw heat.

2) Working Substance

Working substance is an ideal gas which on being supplied with heat performs mechanical work.

3) Sink

Sink is a cold body at low temperature to which any amount of heat is rejected.

Efficiency of Heat Engine

Block diagram of a heat engine between a hot source and cold sink Source (T₁) Heat Engine Sink (T₂) Q₁ Q₂ Work (W)
Fig: Heat engine

The efficiency of heat engine is defined as the ratio of amount of workdone to the total heat taken from source.

Efficiency (η) = amount of workdone / heat taken from source

If Q1 is the amount of heat taken from source and Q2 is the heat rejected to the sink:

Workdone (W) = Q1 − Q2
η = W/Q1
η = (Q1 − Q2)/Q1
η = 1 − Q2/Q1
η% = (1 − Q2/Q1) × 100%

No engine will convert all the heat absorbed from the source into work. So, the efficiency of an engine is always less than 1 or 100%.

Refrigerator

A refrigerator is a device which transfers heat from a cold body to a hot body by means of an external agency.
Block diagram of a refrigerator transferring heat from cold sink to hot source using external work Hot Body Refrigerator Cold Body Q₁=Q₂+W Q₂ Work (W)
Fig: Block diagram of refrigerator

Let Q2 be the amount of heat taken from the body at lower temperature and W be the workdone on it. So, the amount of heat absorbed by the hot body is:

Q1 = Q2 + W
W = Q1 − Q2

Coefficient of Performance of a Refrigerator

The coefficient of performance of refrigerator is defined as the ratio of amount of heat transfer from cold body to hot body to the amount of workdone on it.
β = Q2/W
[∵ W = Q1 − Q2]
β = Q2/(Q1 − Q2)

Relation between Coefficient of Performance and Efficiency

For a heat engine working between Q1 and Q2:

η = 1 − Q2/Q1   — (i)

For refrigerator working between Q1 and Q2:

β = Q2/(Q1 − Q2)
β = 1/(Q1/Q2 − 1)   — (ii)

From equation (i):

Q2/Q1 = 1 − η
Q1/Q2 = 1/(1 − η)   — (iii)

Using equation (iii) in equation (ii):

β = 1/[1/(1 − η) − 1]
β = (1 − η)/η

This is the required relation between coefficient of performance and efficiency.

Carnot Engine and Carnot Cycle

Carnot Engine

Carnot performs an ideal heat engine having maximum efficiency which can work in a cycle of operation is called Carnot engine.

Carnot Cycle

The Carnot engine performs work in four-cycle of operations having two isothermal and two adiabatic process; such a cycle of operation is called Carnot cycle.

Consider n mole of an ideal gas enclosed in a cylinder having initial pressure, volume and temperature P1, V1 and T1 respectively. The point corresponding to this pressure, volume and temperature is A in the P-V diagram as shown in figure. The different processes in a Carnot cycle are discussed below.

Pressure-volume diagram of Carnot cycle with two isothermal and two adiabatic processes A (P₁,V₁) B (P₂,V₂) C (P₃,V₃) D (P₄,V₄) P V Isothermal T₁ Isothermal T₂
Fig: P-V diagram of Carnot cycle

I) Isothermal Expansion

Initially the cylinder is placed on hot source and gas absorbed Q1 amount of heat at constant temperature and gas is expanded isothermally from initial state A(P1,V1) to final state B(P2,V2) which is represented by curve AB in P-V diagram.

W1 = Q1 = nRT1 ln(V2/V1)   — (i)

II) Adiabatic Expansion

Now, the cylinder is kept on an insulating stand and gas is expanded adiabatically from initial state B(P2,V2) to C(P3,V3) and temperature changes from T1 to T2, which is shown by curve BC in P-V diagram.

W2 = nR/(γ−1)(T1 − T2)   — (ii)

III) Isothermal Compression

Now, the engine is placed on a sink and is compressed isothermally from initial state C(P3,V3) to D(P4,V4) at constant temperature T2, which is shown by curve CD in P-V diagram. At this stage an amount of heat is rejected to sink.

W3 = −Q2 = −nRT2 ln(V3/V4)
W3 = nRT2 ln(V4/V3)   — (iii)

IV) Adiabatic Compression

Now, the cylinder is kept on an insulating stand and compressed the gas adiabatically from D(P4,V4) to A(P1,V1) which is shown by curve DA in P-V diagram and temperature changes from T2 to T1.

W4 = nR/(γ−1)(T2 − T1)
W4 = −nR/(γ−1)(T1 − T2)   — (iv)

Therefore total amount of workdone during one complete cycle is:

W = W1 + W2 + W3 + W4
= Q1 − Q2
Net workdone = area of loop ABCDA

Hence in Carnot engine, net workdone by the gas per cycle is numerically equal to the total area of the loop ABCDA representing the cycle.

Efficiency of Carnot Cycle

It is defined as the ratio of external workdone (W) by the engine to the amount of heat energy (Q1) absorbed from the heat source.

η = W/Q1
η = (Q1 − Q2)/Q1
η = 1 − Q2/Q1

For isothermal processes:

Q1 = nRT1 ln(V2/V1)
Q2 = nRT2 ln(V3/V4)
η = 1 − [RT2 ln(V3/V4)]/[RT1 ln(V2/V1)]   — (v)

From adiabatic curve BC:

T1V2γ−1 = T2V3γ−1
T2/T1 = (V2/V3)γ−1   — (vi)

Again from adiabatic curve DA:

T2V4γ−1 = T1V1γ−1
T2/T1 = (V1/V4)γ−1   — (vii)

From equation (vi) and (vii):

(V2/V3)γ−1 = (V1/V4)γ−1
V2/V3 = V1/V4
V2V4 = V1V3
V3/V4 = V2/V1   — (viii)

Using equation (viii) in equation (v):

η = 1 − T2/T1

This is the required expression for efficiency of Carnot engine.

Petrol Engine

Petrol engine is developed by Otto which consists of cylinder fitted with movable piston. There are two holes in cylinder fitted with inlet valve (I) and outlet valve (O). The opening and closing of these valves are controlled by motion of piston. A spark plug ‘S’ produces spark which ignites the mixture of air and petrol. The piston is attached to rod R which is connected to wheels of vehicles. The petrol engine works in following four steps:

Four strokes of a petrol engine: suction, compression, working and exhaust (a) Suction I open (b) Compression compressed (c) Working Spark (d) Exhaust O open
Fig: Four strokes of petrol engine
P-V diagram of a petrol engine cycle E A B C D P V V₁ V₂ Q₁ supplied Q₂ rejected
Fig: P-V diagram of petrol engine

I) Suction Stroke

It is shown in figure (a). In this stroke the inlet valve ‘I’ opens and mixture of air and petrol is sucked into the cylinder by outward motion of piston which is shown by curve EA.

II) Compression Stroke

It is shown in figure (b). In this stroke the inlet and outlet valve are closed. The mixture of air and petrol is compressed adiabatically to 1/5 of its original volume, hence the temperature rises up to 600°C which is shown by curve AB in P-V diagram.

III) Working Stroke

It is shown in figure (c). In this stroke both inlet and outlet valve are closed and spark is produced from spark plug ‘S’, hence temperature rises about 2000°C and pressure also rises up to 15 atmosphere which is shown by curve BC in P-V diagram and in this stage Q1 amount of heat is supplied. After ignition the piston is pushed outward so the wheels rotate, due to which temperature and pressure fall down which is shown by curve CD.

IV) Exhaust Stroke

It is shown in figure (d). In this stroke the outlet valve ‘O’ opens, hence the burnt gas escapes out and the pressure falls which is shown by curve DA in P-V diagram.

Efficiency of Petrol Engine

η = W/Q1
= (Q1 − Q2)/Q1
= 1 − Q2/Q1

For adiabatic expansion from C to D:

T2V1γ−1 = T3V2γ−1
T3/T2 = (V1/V2)γ−1

But efficiency:

η = 1 − T3/T2
η = 1 − (V1/V2)γ−1
η = 1 − (1/r)γ−1

where r = V2/V1 is compression ratio.

Diesel Engine

Diesel engine is developed by Rudolf Diesel which consists of cylinder fitted with movable piston. There are three holes in a cylinder fitted with inlet valve (I), outlet valve (E) and oil spray (O). The piston of diesel engine is connected to a wheel through the rod ‘R’. The working of diesel engine completes in 4 steps.

Four strokes of a diesel engine: suction, compression, working and exhaust (a) Suction I open (b) Compression air compressed (c) Working oil spray (d) Exhaust E open
Fig: Four strokes of a diesel engine
P-V diagram of diesel engine with adiabatic compression, constant-pressure heat addition and adiabatic expansion A B C D Q₁ supplied Q₂ rejected adiabatic compression adiabatic expansion P V V₁ V₂
Fig: P-V diagram for diesel engine

I) Suction Stroke

It is shown in figure (a). In this stroke inlet valve is open and air enters inside the cylinder by outward movement of piston. This is shown by curve CA in P-V diagram.

II) Compression Stroke

It is shown in figure (b). In this stroke all the valves are closed and the air is compressed adiabatically to 1/17th of its original volume and the temperature inside the cylinder rises up to 1000°C, which is shown by curve AB in P-V diagram.

III) Working Stroke

It is shown in figure (c). In this stroke oil spray valve ‘O’ opens and diesel enters inside the chamber where diesel burns and the volume rises at constant pressure which is shown by curve BC in P-V diagram. The work is done in this stroke. When the temperature inside the chamber reaches 2000°C, the valve O is closed. Now the volume increases due to high temperature and low pressure which is shown by curve CD in P-V diagram.

IV) Exhaust Stroke

It is shown in figure (d). In this stroke pressure decreases at constant volume, hence burnt gas will escape out from outlet valve which is shown by curve DA in P-V diagram and next cycle begins.

Efficiency of Diesel Engine

η = W/Q1
= (Q1 − Q2)/Q1
= 1 − Q2/Q1
η = 1 − (V1/V2)γ−1
η = 1 − (1/ε)γ−1

where ε = V2/V1 is called compression ratio.

Solved Numericals

Q.1 — Carnot cycle source and sink temperature

The efficiency of a Carnot cycle is 15%. If on reducing the temperature of sink by 65°C, the efficiency becomes double, find the temperature of source and sink.

Efficiency of engine (η) = 15%
Reduced efficiency (η′) = 30%

For initial Carnot efficiency:

η = (1 − T2/T1) × 100%
15/100 = 1 − T2/T1
T2/T1 = 1 − 15/100 = 0.85   — (i)

Again, after reducing sink temperature by 65°C:

η′ = [1 − (T2 − 65)/T1] × 100%
30/100 = 1 − (T2 − 65)/T1
0.70 = (T2 − 65)/T1   — (ii)

Using equation (i):

0.85 = T2/T1

Subtracting equation (ii) relation from equation (i):

0.85 − 0.70 = 65/T1
0.15 = 65/T1
T1 = 65/0.15 = 433.3 K

Using T1 = 433.3 K in equation (i):

T2/433.3 = 1 − 15/100
T2 = 433.3 × 85/100
T2 = 368.3 K

Q.2 — Efficiency of a petrol engine

A petrol engine consumes 25 kg of petrol per hour. The calorific value of petrol is 11.4×106 cal/kg. The power of the engine is 99.75 kW. Calculate the efficiency of the engine.

Output power = 99.75 kW
= 99.75 × 1000
= 99750 W

Given:

For 1 kg petrol: energy = 11.4×106 × 4.2 J
For 25 kg petrol: energy = 11.4×106 × 4.2 × 25 J in one hour

Input power:

= [11.4×106 × 4.2 × 25] / (60×60)
= 0.3325×106 W

Then:

η% = (Output power / Input power) × 100%
= [9.9750×104 / 0.3325×106] × 100%
η = 30%
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