Physics : Heat and Thermodynamics : Book Back Important Questions, Answers, Solutions : Long Questions and Answer

__Heat and Thermodynamics | Physics__

__Long answer Questions:__

**1. Explain the meaning of heat and work with suitable examples.**

**Meaning
of heat and work:**

●
When we rub our hand against each other the temperature of the hands increases.
We have done some work on our hands by rubbing. The temperature of the hands
increases due to this work.

●
If we place our hands on the chin, the temperature of the chin increases.

●
This is because the hands are at higher temperature than the chin.

●
In the above example, the temperature of hands is increased due to work and
temperature of the chin is increased due to heat transfer from the hands to the
chin.

●
By doing work on the system, the temperature in the system will increase and
sometimes may not.

●
Like heat, work is also not a quantity and through the work, energy is
transferred to the system.

●
So we cannot use the word ‘the object contains more work' or ‘less work'.

●
Either the system can transfer energy to the surrounding by doing work on
surrounding or the surrounding may transfer energy to the system by doing work
on the system.

●
For the transfer of energy from one body to another body through the process of
work, they need not be at different temperatures.

**2. Discuss the ideal gas laws.**

**Boyle's
law, Charles' law and ideal gas law:**

●
For a given gas at low pressure (density) kept in a container of volume V,
experiments revealed the following information.

●
When the gas is kept at constant temperature, the pressure of the gas is
inversely proportional to the volume.

P
∝ 1/*v* known as Boyle's law.

●
When the gas is kept at constant pressure, the volume of the gas is directly
proportional to absolute temperature. V ∝ T. Known
as Charles's law.

●
By combining these two equation we have PV = CT. Here C is a positive constant.

C
is proportional to the number of particles in the gas container.

●
If we take two container of same type of gas with same volume V, same pressure
P and same temperature T, then the gas in each container obeys the above
equation. PV = CT.

●
If the two containers of gas is considered as a single system, then the
pressure and temperature of this combined system will be same but volume will
be twice and number of particles will also be double.

●
For this combined system, V becomes 2V, so C should also double to match with
the ideal gas equation P(2V)/*T* = 2C.

●
It implies that C must depend on the number of particles in the gas and also should
have the dimension of [PV /*T*]= JK^{−1}.

●
We can write the constant C as k times the number of particles N. Here k is the
Boltzmann constrit (1.381 × 10^{−23} JK^{−1}) and it is found
to be a universal constant. So the ideal gas law can be stated as follows PV =
NkT.

**3. Explain in detail the thermal expansion.**

●
Thermal expansion is the tendency of matter to change in shape, area and volume
due to a change in temperature.

●
All three states of matter (solid, liquid and gas) expand when heated. When a
solid is heated, its atoms vibrate with higher amplitude about their fixed
points. The relative change in the size of solids is small.

●
Railway tracks are given small gaps so that in the summer, the tracks expand
and do not buckle.

●
Railroad tracks and bridges have expansion joints to allow them to expand and
contract freely with temperature changes.

●
Liquids, have less intermolecular forces than solids and hence they expand more
than solids. This is the principle behind the mercury thermometers.

●
In the case of gas molecules, the intermolecular forces are almost negligible
and hence they expand much more than solids. For example in hot air ballons
when gas particles get heated, they expand and take up more space.

●
The increase in dimension of a body due to the increase in its temperature is
called thermal expansion

●
The expansion in length is called liner expansion. Similarly the expansion in
area is termed as area expansion and the expansion in volume is termed as
volume expansion.

**Linear
Expansion:**

In
solids, for a small change in temperature ∆T, the fractional change in length [∆L/*L*_{o}]
is directly proportional to ∆T. (∆L/ *L*_{o}) = α_{L} ∆T

Therefore,
α_{L} = ∆L / *L*_{o}∆*T* ; Where, α_{L} =
coefficient of linear expansion .

∆L
= Change in length ; L_{o} = Original length ; ∆T = Change in temperature,

**Area
Expansion:**

For
a small change in temperature ∆T the fractional change in area [∆A / A_{o}]
of a substance is directly proportional to ∆T and it can be written as ∆A/ A_{o}
= α_{A}∆T

Therefore,
α_{A} [∆A / A_{o}∆T];

Where
α_{A} = coefficient of area expansion;

∆A
= Change in area;

A_{o}
= Original area;

∆T
= Change in temperature

**Volume
Expansion:**

For
a small change in temperature ∆T the fractional change in volume

[∆A
/ V_{o}] of a substance is directly proportional to ∆T.

∆V
/ V_{o} = α_{V} ∆T, Therefore, α_{V} [∆V/V_{o}∆T].

Where,
α_{V} = coefficient of volume expansion ; ∆V = Change in volume ;

V_{o}
= Original volume ; ∆T = Change in temperature. Unit of coefficient of linear,
area and volumetric expansion of solids is °C^{−1} or K^{−1}.

**4. Describe the anomalous expansion of water. How is it helpful in our lives?**

**Anomalous
expansion of water:**

●
Liquids expand on heating and contract on cooling at moderate temperatures.

●
But water exhibits an anomalous behavior.

●
It contracts on heating between 0°C and 4°C.

●
The volume of the given amount of water decreases as it is cooled from room
temperature, until it reach 4°C.

●
Below 4°C the volume increases and so the density decreases.

●
The water has a maximum density at 4°C.

●
This behavior of water is called anomalous expansion of water.

●
In cold contries during the winter season, the surface of the lakes will be at
lower temperature than the bottom.

●
Since the solid water (ice) has lower density than its liquid form, below 4°C,
the frozen water will be on the top surface above the liquid water (ice floats)

●
This is due to the anomalous expansion of water.

●
As the water in takes and ponds freeze only at the top the species living in
the lakes will be safe at the bottom.

**5. Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.**

**Calorimetry:**

●
Calorimetry means the measurement of the amount of heat released or absorbed by
thermodynamic system during the heating process.

●
When a body at higher temperature is brought in contact with another body at
lower temperature, the heat lost by the hot body is equal to the heat gained by
the cold body.

●
No heat is allowed to escape to the surroundings.

●
It can be mathematically expressed as Q_{gain} = − Q_{lost }; Q_{gain}
+ Q_{lost} = 0

●
Heat gained or lost is measured with a calorimeter.

Usually
the calorimeter is an insulated container of water.

●
A sample is heated at high temperature (T_{1}) and immersed into water
at room temperature (T_{2}) in the calorimeter. After some time both
sample and water reach a final equilibrium temperature T_{f}. Since the
calorimeter is insulated, heat given by the hot sample is equal to heat gained
by the water.

Q_{gain} = − Q_{lost }

●
The heat lost is denoted by negative sign and heat gained is denoted as
positive.

●
From the definition of specific heat capacity

Q_{gain}
= m_{2}s_{2}(T_{f }−T_{2})

Q_{lost}
= m_{1}s_{1} (T_{f} – T_{1})

●
Here s1 and s2 specific heat capacity of hot sample and water respectively.

m_{2}s_{2}(T_{f
}−T_{2}) = m_{1}s_{1} (T_{f} – T_{1})

m_{2}s_{2}T_{f
}− m_{2}s_{2}T_{2 }= - m_{1}s_{1}T_{f}
+ m_{1}s_{1}T_{1}

m_{2}s_{2}T_{f
}+ m_{1}s_{1}T_{f }= m_{2}s_{2}T_{2}
+ m_{1}s_{1}T_{1}

The
final temperature T_{f }= [ m_{1}s_{1}T_{1} + m_{2}s_{2}T_{2}
] / [ m_{1}s_{1} + m_{2}s_{2 }]

**6. Discuss various modes of heat transfer.**

**Conduction:**
Conduction is the process of direct transfer of heat through matter due to
temperature difference. When two objects are in direct contact with one
another, heat will be transferred from the hotter object to the colder one.
Thermal conductivity depends on the nature of the material.

**Convection:**
Convection is the process in which heat transfer is by actual movement of
molecules in fluids such as liquids and gases. In convection, molecules move
freely from one place to another.

**Radiation:**
Radiation is a form of energy transfer from one body to another by
electromagnetic waves. Radiation which requires no medium to transfer energy
from one object to another.

**Example:** 1 Solar
energy from the sun, 2. Radiation from room heater.

**7. Explain in detail Newton’s law of cooling.**

**Newton's
law of cooling:**

●
Newton's law of cooling states that the rate of loss of heat of a body is
directly proportional to the difference in the temperature between that body
and its surroundings.

dQ/dt
∝ − (T−T_{S}) ------------- (1)

●
The negative sign indicates that the quantity of heat lost by liquid goes on
decreasing with time. Where, T = Temperature of the object Ts = Temperature of
the surrounding.

●
The rate of cooling is high initially and drecreases with falling temperature.

●
Let us consider an object of mass m and specific heat capacity s at temperature
T. Let Ts be the temperature of the surroundings. If the temperature falls by a
small amount dT in time dt, then the amount of heat lost is,

dQ
= msdT ----------------- (2)

Dividing both sides of equation (2) by

dQ
/ dt = msdT / dt ------------- (3)

●
From Newton's law of cooling dQ/dt ∝ − (T−T_{S})

dQ
/ dt = − a (T−T_{S}) ------------- 4

●
Where is some positive constant. From equation (3) and (4)

− a (T − T_{s}) = ms (dt/ dt)

dt
/ (T−T_{S}T) = − (a/ ms) dt ------------- 5

Integrating
equation (5) on both side

ʃ^{∞}_{0
}[dt / (T - T_{s})] = −ʃ^{t}_{0 }(a/ms) dt

In
(T − T_{s}) = − (a/ms)t + b_{1}

●
Where b_{1} is the constant of integration, taking exponential on both
sides, we get,

T
= T_{s}+ b_{2}e^{–(a
/ ms) t .}

Here
b_{2} = e^{b1} = Constant.

**8. Explain Wien’s law and why our eyes are sensitive only to visible rays?**

●
When's law states that, the wavelength of maximum intensity of emission of a
black body radiation is inversely proportional to the absolute temperature of
the black body.

λ_{m}
∝ 1/T

or

λ_{m}
= b / T

………….. 1

●
Where, b is known as Wien's constant. Its value is 2.898 × 10^{-3} mk

●
The Sun is approximately taken as a black body.

●
Since any object above 0 K will emit radiation. Sun also emits radiation.

●
Its surface temperature is about ; 5700 K. By substituting this value in the
equation (1).

λ_{m}
= b / T = 2.858×10^{-3} / 5700 = 508 nm

●
It is the wavelength at which maximium intensity is 508nm.

●
The visible part of the spectrum lies between 400nm to 700nm.

●
The humans evolved under the Sun by receiving its radiations. The human eye is
sensitive only in the visible not in infrared or X-ray ranges in the spectrum.

**9. Discuss the**

**a. thermal equilibrium**

**b. mechanical equilibrium**

**c. Chemical equilibrium**

**d. thermodynamic equilibrium.**

**a.
Thermal equlibrium:** Two systems are said to be in thermal equilibrium
with each other if they are at the same temperature, which will not change with
time.

**b.
Mechanical equilibrium:**

●
Consider a gas container with piston.

●
When some mass is placed on the piston, it move downward due to downward** **gravitational force and after certain, humps and
jumps, the piston will come to rest at a new position.

●
When the downward gravitational force given by the piston is balanced by the
upward force exerted by the gas, the system is said to be in mechanical
equilibrium.

●
A system is said to be in mechanical equilibrium if no unbalanced force acts on
the thermo dynamic system or on the surrounding by thermodynamic system.

**c.
Chemical equilibrium:** If there is no net chemical reaction between two
thermodynamic systems in contact with each other then it is said to be in
chemical equilibrium.

**d.
Thermodynamic equilibrium:** If two systems are set to be in
thermodynamic equilibrium, then the system are at thermal, mechanical and
chemical equilibrium with each other. In a state of thermodynamic equilibrium
the macroscopic variables such as pressure, volume and temperature will have
fixed values and do not change with time.

**10. Explain Joule’s Experiment of the mechanical equivalent of heat.**

●
Joule showed that mechanical energy can be converted into internal energy and
vice versa.

●
Two masses were attached with a rope and a paddle wheel.

●
When these masses fall through a distance h hue to gravity, both the masses
lose potential energy equal to 2mgh.

●
When the masses fall, the paddle wheel turns.

Due
to the turning of wheel inside water, frictional force comes in between the
water and the paddle wheel.

●
This causes a rise in temperature of the water. This implies that gravitational
potential energy is converted to internal energy of water.

●
The temperature of water increases due to the work done by the masses.

●
The mechanical work has the same effect as giving heat.

●
Joule found that to raise 1 g of an object by 1°C, 4.186 J of energy is
required. In earlier days the heat was measured in calorie. 1 cal = 4.186 J.
This is called Joule's mechanical equivalant of heat.

**11. Derive the expression for the work done in a volume change in a thermodynamic system.**

**Work
done in volume changes:**

●
Consider a gas contained in the cylinder fitted with a movable piston. Suppose
the gas is expanded quasi- statically by pushing the piston by a small distance
dx.

●
Since the expansion occurs quasi-statically the pressure, temperature and
internal energy will have unique values at every instant.

●
The small work done by the gas on the piston.

dW
= Fdx ……….. (1)

●
The force exerted by the gas on the piston F = PA. Here A is area of the piston
and P is pressure exerted by the gas on the piston.

●
Equation (1) can be rewritten as dW = PA dx ……. (2)

●
But Adx = dV = change in volume during this expansion process. So the samll
work done by the gas during the expansion is given by dW = PdV ………. (3)

●
dV is positive since the volume is increased. So dW is positive.

The
work done by the gas by increasing the volume from V_{i} to V_{f}
is given by

W
= ^{Vf}ʃ_{Vi }PdV ………… (4)

●
If the work is done on the system, then V_{i} > V_{f}. Then,
W is negative. It implies that while the system is doing work, the pressure
need not be constant.

To
evaluate the integration, the pressure is first expressed function of volume
and temperature using the equation of state.

**12. Derive Mayer’s relation for an ideal gas.**

**Meyer's
relation:**

●
Consider μ mole of an ideal gas in a container with volume V, pressure P and
temperature T.

●
When the gas is heated at constant volume the temperature increases by dT. As
no work is done by the gas, the heat that flows into the system will increase
only the internal energy.

●
Let the change in internal energy be dU.

●
If C_{v} is the molar specific heat capacity at constant volume.

dU
= μC_{v} dT ……… (1)

●
Suppose the gas is heated at constant pressure so that the temperature
increases by dT. If 'Q’ is the head supplied in this process and 'dv' the
change in volume of the gas.

If'
'Q' = μC_{P }dT ………. (2)

●
If W is the work done by the gas in this process, then W = PdV …………. (3)

●
But from the first law of thermodynamics.

Q
= dU + W ….. (4)

●
Substituting equations (1), (2) and (3) in (4), we get.

μC_{P }dT = μC_{v} dT + PdV ……. (5)

●
For mole of ideal gas, the equation of state is given by PV = μRT

=>
PdV + VdP = μRdT ……. (6)

Since
the pressure is constant, dP = 0

C_{P}dT
= C_{v}dT + RdT

C_{P}
= C_{v} + R (or) C_{P} − C_{v} = R

This
relation is called Meyer's relation.

**13. Explain in detail the isothermal process.**

**Isothermal
process:**

●
It is a process in which the temperature remains constant but the pressure and
volume of a thermodynamic system will change. The ideal gas equation is

PV
= μRT, Here, T is constant for this process.

●
So the equation of state for isothermal process is given by PV = constant ………..(1)

●
This implies that if the gas goes from one equilibrium state (P_{1}, V_{1})
to another equilibrium state (P_{2}, V_{2}) the following
relation holds for this process

P_{1}V_{1}=
P_{2}V_{2 }……..(2)

●
Since PV = constant, P is inversely proportional to P ∝ 1/V.

●
This implies that PV graph is a hyperbola. The pressure-volume graph for
constant temperature is also called isotherm.

●
For an ideal gas the internal energy is a function of temperature only. For an
isothermal process since temperature is constant, the internal energy is also
constant. This implies that dU or ∆U = 0.

●
For an isothermal process, the first law or thermodynamics can be written as
follows,

Q
= W ………. 3

●
The heat supplied to a gas used to do only external work.

●
The isothermal compression takes place when the piston of the cylinder is
pushed. This, will increase the internal energy which will flow out of the
system through thermal contact.

**14. Derive the work done in an isothermal process**

●
Consider an ideal gas which is allowed to expand quasi-statically at constant
temperature from initial state (P_{1}, V_{1}) to the final
state (P_{f}, V_{f}). The work done by the gas.

W
= _{Vf}ʃ^{Vf }pdV ------------(1)

●
As the process occurs quasi-statically, a every stage the gas is at equilibrium
with the surroundings. Since it is in equilibrium at every stage the ideal gas
law is valid.

●
Writing pressure in terms of volume and temperature.

P
= μRT / V ----------- (2)

●
Substituting equation (2) in (1) we get,

●
μRT is constant throughout the isothermal process.

●
By performing the integration in equation (3)

We
get W = μRT in (V_{f }/ V_{i}) ----------------- (4)

●
Since we have an isothermal expansion, V_{f }/V_{i}, > 1.

So
In (V_{f }/ V_{i}) > 0

●
The work done by the gas during an isothermal expansion is positive.

●
The above result in equation (4) is true for isothermal compression also. But
in an isothermal compression V_{f} / V_{i }< 1. So In (V_{f}
/ V_{i }) < 0. As a result the work done on the gas in an isothermal
compression is negative.

●
In the PV diagram the work done during the isothermal expansion is equal to the
area under the graph.

●
For an isothermal compression, the area under the PV graph is equal to the work
done on the gas which turns out to be the area with a negative sign.

**15. Explain in detail an adiabatic process.**

**Adiabatic
process:**

●
This is a process in which no heat flows into or out of the system (Q=0). But
the gas can expand by spending its internal energy or gas can be compressed
through some external work.

●
So the pressure, volume and temperature of the system may change in an
adiabatic process.

●
The equation of state for an adiabatic process is given by PV^{γ }=
Constant ** ****……… (1)****.**

●
Here γ is called adiabatic exponent (γ = Cp/C_{v}) which depends on the
nature of the gas.

●
If the gas goes from an equilibrium state (P_{1}, V_{1}) to
another equilibrium state (P_{f}, V_{f}) adiabatically then it
satisfies the relation.

P_{i}V_{i}^{γ}
= P_{f}V_{f}^{ γ} **………… (2)**

●
The PV diagram for an adiabatic process is also called adiabat.

●
The PV diagram for isothermal and adiabatic process look similar. But actually
the adiabatic curves is steeper than isothermal curve.

●
To rewrite the equation (1) in terms of T and V. From ideal gas equation, the
pressure P = μRT / V. Substituting this equation in the equation (1), we have [
μRT /V^{ }] V^{γ} = Constant or [T/V] V^{γ} = Constant
/ (μR)

●
So it can be written as TV^{γ−1} = Constant **……. (3)**

●
If the gas goes from an initial equilibrium state (T_{i} V_{i})
to final equilibrium state (T_{f} V_{f}) adiabatically then it
satisfies the relation T_{i} V_{i}^{ γ−1} = T_{f}
V_{f}^{ γ−1} **…….. (4)**

●
The equation of state for adiabatic process can also be written in terms of T
and P as T^{γ}P^{1-γ} = Constant **…… (5)**

** **

**16. Derive the work done in an adiabatic process**

●
Consider μ moles of an ideal gas enclosed in a cylinder having perfectly non
conducting walls and base.

●
A frictionless and insulating piston of cross sectional area A is fitted in the
cylinder. Let W be the work done when the system goes from the initial state (P_{i},
V_{i}, T_{i}) to the final state (P_{f} V_{f},
T_{f}) adiabatically, W = ^{Vf}ʃ_{Vi}PdV ……….(1)

●
The adiabatic process occurs quasi-statically, at every state the ideal gas law
is valid.

●
Under this conditions the adiabatic equation of state is PV^{γ} =
constant (or) P = constant / V^{γ }can be substituted in the equation
(1), we get W_{adia} = ^{Vf}ʃ_{Vi} constant / V^{γ }=
dV

=
Constant ^{Vf}ʃ_{Vi} V^{γ }dV

But,
= P_{1}V_{1} = P_{f}V_{f} = constant

W_{adia}
=1 / (1 – γ) [ (P* _{f}*V

W_{adia}
= (1 / 1–γ) [P_{f}V_{f} – P_{1}V_{1}] ……………… 2

From
ideal gas law, P_{f}V_{f }= RT_{f} and P_{1}V_{1}
= μRT_{1}

Substituting
in equation (2), we get, W_{adia} = μR/(γ–1) [T_{i}–T_{f}]
……………..(3)

●
In adiabatic expansion, work is done by the gas. i.e., W_{adia} is
positive. As T_{i }>T_{f}, the gas cools during adiabatic
expansion.

●
In adiabatic compression, work is done on the gas. i.e., W_{adia} is
negative.

As
T_{i }>T_{f}, the temperature of the gas increases during
adiabatic compression.

**17. Explain the isobaric process and derive the work done in this process**

**Isobaric
process:**

●
This is a thermodynamic process that occurs at constant pressure. Even though
pressure is constant in this process, temperature, volume and internal energy
are not constant. From the ideal gas equation, we have

V
= [ μR / P ] T ………
1

Here μR / P = Constant

●
In an isobaric process the temperature is directly proportional to volume.

V
∝ T (Isobaric process) ……….. (2)

●
This implies that for a isobaric proces, the V-T graph is a straight line
passing through the origin.

●
If a gas goes from a state (V_{i} , T_{i}) to (V_{f}, T_{f})
at constant pressure, then the system satisfies the following equation

*T _{f
}*/ V

●
The work done in an isobaric process: Work done by the gas W = ^{vf}ʃ_{vi}
PdV

●
In an isobaric process, the pressure is constant W = P ^{vf}ʃ_{vi}
dV , W = P [V_{f} − V_{i} ] = P ΔV ……….. (3)

●
Where ΔV denotes change in the volume. If ΔV is negative, W is also negative.
This implies that the work is done by the gas.

●
The equation (3) can also be rewritten using the ideal gas equation.

●
From ideal gas equation PV = μRT and V = μRT / P

●
Substituting this in equation (3) we get, W = μRT_{f} (1 − T* _{i}*/T

●
In the PV diagram, area under the isobaric curve is equal to the work done in
isobaric process.

●
The first law of thermodynamics for isobaric process is given by ΔU = Q − PΔV.

**18. Explain in detail the isochoric process.**

**Isochoric
Process:**

This
is a thermodynamic process in which the volume of the system is kept constant.

But
pressure, temperature and internal energy continue to be variables.

The
pressure - volume graph for an isochoric process is a vertical line parallel to
pressure axis.

●
The equation of state for an isochoric process is given by P = (μR/V)T.

Whrere,
(μR/V)= Constant

●
The pressure is directly proportional to temperature.

●
This implies that the P-T graph for an isochoric process is a straight line
passing through origin.

●
If a gas goes from state (P_{i}, T_{i}) to (P_{f}, T_{f})
at constant volume, then the system satisfies the following equation = P_{i}
/ T_{i }= P_{f} / T_{f}

●
For an isochoric process, ΔV = 0 and W = 0. Then the first law becomes.

ΔU
= Q

●
Implying that the heat supplied is used to increase only the internal energy.

Aa
a result the temperature increases and pressure also increases.

●
Suppose a system loser heat to the surroundings through conducting walls by
keeping the volume constant, then its internal energy decreases.

●
As a result the temperature decreases; the pressure also decreases.

**19. What are the limitations of the first law of thermodynamics?**

**Limitations
of first law of thermodynamics:**

●
The first law of thermodynamics explains well the inter convertibility of heat
and work.

But
it does not indicate the direction of change.

**For
example:**

●
When a hot object is in contact with a cold object, heat always flows from the
hot object to cold object but not in the reverse direction.

●
According to first law, it is possible for the energy to flow from hot object to
cold object or from cold object to hot object.

●
The direction of heat flow is always from higher temperature to lower
temperature.

●
When brakes are applied, a car stops due to friction and the work done against
friction is converted into heat.

●
But this heat is not reconverted to the kinetic energy of the car.

●
So the first law is not sufficient to explain many of natural phenomena.

**20. Explain the heat engine and obtain its efficiency.**

●
Heat engine is a device which takes heat as input and converts this heat in to
work by undergoing a cyclic process.

**A
heat engine has three parts:** (a) Hot reservoir, (b) Working
substance, (c) Cold reservoir.

**Hot
reservoir (or) source: **It supplies heat to the engine. It is always
maintained at a high temperature T_{H}

**Working
substance:**

●
It is substance like gas or water, which converts the heat supplied into work.

●
A simple example of a heat engine is a steam engine. In olden days steam
engines were used to drive trains.

●
The working substance in these is water which absorbs heat from the burning of
coal.

●
The heat converts the water into steam.

●
This steam does work by rotating the wheels of the train, thus making the train
move.

**Cold
reservoir (or) Sink:** The heat engine ejects some amount of heat (Q_{L})
in to cold reservoir after it doing work. It is always maintained at a low
temperature T_{L}.

●
**For example:** In the automobiles engine, the cold reservoir is the
surroundings at room temperature. The automobile ejects heat to these
surroundings through a silencer.

●
The heat engine works in a cyclic process. After a cyclic process it returns to
the same state.

●
Since the heat engine returns to the same state after it ejects heat, the
change in the internal energy of the heat engine is zero.

● The efficiency of the heat engine is defined as the ratio of the work done (output) to the heat absorbed (input) in one cyclic process.

●
Let the working substance absorb heat Q_{H} units from the source and
reject Q_{L} units to the sink after doing work W units.

Input
heat = Work done + ejected heat

Q_{H}
= W + Q_{L}

W
= Q_{H} − Q_{L}

Then
the efficiency of heat engine η = outPut / Input = W/Q_{H }= (Q_{H}−
Q_{L}) / Q_{H}

η = Output / Input = W / Q_{H }= 1 – (Q_{L}/Q_{H})

●
Q_{H}, Q_{L} and W all are taken as positive, a sign convention
followed in this expression.

Since
Q_{L} < Q_{H}, the efficiency (η) always less than 1.

●
This implies that heat absorbed is not completely converted into work.

●
The second law of thermodynamics placed fundamental restrictions on converting
heat completely into work.

**21. Explain in detail Carnot heat engine.**

●
A reversible heat engine operating in a cycle between two temperatures in a
particular way is called a Carnot Engine.

●
The Carnot engine has four parts which are given below.

**Source:** It is
the source of heat maintained at constant high temperature T_{H}. Any amount
of heat can be extracted from it, without changing its temperature.

**Sink:** It is a
cold body maintained at a constant low temperature T_{L}. It can absorb
any amount of heat.

**Insulating
stand:** It is made of perfectly non-conducting material. Heat is not conducted
through this stand.

**Working
substance:** It is an ideal gas enclosed in a cylinder with perfectly
non-conducting walls and perfectly conducting bottom. A non-conducting and
frictionless piston is fitted in it.

**Carnot's
cycle:**

●
The working substance is subjected to four successive reversible process called
Carnot's cycle.

●
Let the initial pressure, volume of the working substance be P_{1} ,V_{1}.

**Step
A to B:** Quali-static isothermal expansion from (P_{1},V_{1},
T_{H}) to (P_{2}, V_{2}, T_{H}).

●
The cylinder is placed on the source. The heat (Q_{H}) flows from
source to the working substance (ideal gas) through the bottom of the cylinder.
Since the process is isothermal, the internal energy of the working substance
will not change.

●
The input heat increases the volume of the gas. The piston is allowed to move,
out very slowly (quasi-statically).

●
W_{1} is the work done by the gas in expanding from volume V_{1}
to volume V_{2} with a decrease of pressure from P_{1} to P_{2}.

●
This is represented by the P-V diagram along the path AB.

●
Then the work done by the gas (working substance) is given by

Q_{H}
= W_{A→B} = ^{v2}ʃ_{v1 }PdV

●
Since the process occurs quasi-statically, the gas is in equlibrium with the
source till it reaches the final state. The work down in the isothermal expansion
is given by the equation.

W_{A→B}
= μRT_{H} In ( V_{2}/V_{1} ) = Area under the curve AB.

**Step
B to C:** Quasi-static adiabatic expansion from (P_{2}, V_{2},
T_{H}) to (P_{3}, V_{3}, T_{H})

●
The cylinder is placed on the insulating stand and the piston is allowed to
move out.

●
As the gas expands adiabatically from volume V_{2} to volume V_{3}
the pressure falls from P_{2} to P_{3}.

●
The temperature falls to T_{L}. This adiabatic expansion is represented
by curve BC in the P-V diagram.

●
This adiabatic process also occurs quasi-statically and implying that this
process is reversible and the ideal gas is in equilibrium throughout the
process.

●
The work done by the gas in an adiabatic expansion is given by,

W_{B→C}
= ^{v3}ʃ_{v2 }PdV = μR/(γ−1) [ T_{H} – T_{L}] =
Area under the curve BC.

**Step
C - D :** Quasi-static isothermal compression from (P_{3}, V_{3},
T_{1},) to (P_{4}, V_{4}, T_{L}).

●
The cylinder is placed on the sink and the gas is isothermally compressed until
the pressure and volume become P_{4} and V_{4} respectively.

●
This is represented by the curve CD in the PV diagram.

●
Let W_{C→D} be the work done on the gas.

●
According to first law of thermodynamics.

W_{C→D}
= ^{v4}ʃ^{ }_{v3 }PdV = μRT_{L }In ( V_{4}/V_{3
}) = − μRT_{L }In ( V_{3 }/ V_{4}) = − Area under
curve CD

V_{3}
is greater than V_{4}. So the work done is negative, implying work is
done on the gas.

**Step
D-A:**
Quasi-static adiabatic compression from (P_{4}, V_{4}, T_{1})
to (P_{1}, V_{1}, T_{H}).

●
The cylinder is placed on the insulating stand again and the gas is compressed
adiabatically till it attains the initial pressure P_{1}, volume V_{1}
and temperature T_{H}. This is shown by the curved DA in the P-V
diagram.

W_{B→C}
^{v1}ʃ_{v4 }PdV = -μR / (γ −1) [ T_{L} – T_{H}]
= Area under the curve DA

●
In the adiabatic compression also work is done on the gas so it is negative.
Let 'W' be the net work done by the working substance in one cycle.

∴W_{A→B}
+ W_{B→C }+ W_{C→D} + W_{D→A} since W_{D→A} = W_{B→C}

=
W_{A→B} + W_{C→D}

●
The net work done by the Carnot engine in one cycle W = |W|_{A→B} − |W|_{C→D}

●
The network done by the working substance in one cycle is equal to the area
(enclosed by ABCD) of the P-V diagram.

●
After, one cycle the working substance returns to the initial temperature T_{H}.
This implies that the change in internal energy of the working substance after
one cycle is zero.

**22. Derive the expression for Carnot engine efficiency.**

**Efficiency
of a Carnot engine:**

●
Efficiency is defined as the ratio of work done by the working substance in one
cycle to the amount of heat extracted from the source.

η
= Work done / Heat extracted = W/ Q_{H} ……..(1)

●
From the first law of thermodynamics, W = Q_{H} − Q_{L}

η
= [ Q_{H} - Q_{L }] / Q_{H} = 1 [Q_{L }/ Q_{H}] ………..(2)

●
Applying isothermal condition, we get,

Q_{H
}= μRT_{H} In (V_{2} / V_{1}); QL = μRT_{L}
In (V_{3} / V_{4 })_{ }………..(3)

●
The amount of (Q_{L}) ejected into the sink, (Q_{L}/Q_{H })
= [T_{L }In (V_{3} / V_{4} ) ] / [ T_{H} In (V_{2}
/ V_{1}) ]_{ }………..(4)

●
By applying adiabatic conditions, we get, T_{H} V_{2}^{γ−1}
= T_{L} V_{3}^{γ−1}

●
By dividing the above the two equations, we get, (V_{2} / V_{1})^{
γ−1} = (V_{3} / V_{4} )^{ γ−1}

●
Which implies that, V_{2}/V_{1} = V_{3}/V_{4} ………. (5)

●
Substituting equation (5) in (4), we get, Q_{L}/Q_{H }= T_{L}/T_{H}

●
The efficiency η = 1 − T_{L}/T_{H}

_{}

**23. Explain the second law of thermodynamics in terms of entropy.**

●
The quantity Q_{H} / T_{H} is equal to Q_{L }/ Q_{H}
the quantity is called entropy.

●
It is a very important thermodynamic property of system.

●
It is also state variable Q_{H} / T_{H} is the entropy received
by the Carnot engine from hot reservoir and Q_{L}/ T_{L} is
entropy given out by the Carnot engine to the cold reservoir.

●
For reveersible engines (Carnot Engine) both entropies should be same.

●
So that the change in entropy of the Carnot engine in one cycle is zero. But
for all practical engines like diesel and petrol engines which are not reversible
engines, they satisfy the relation Q_{L}/T_{L} > Q_{H}/T_{H}

●
"For all the processes that occur in nature (irreversible process), the
entropy always increases. For reversible process entropy will not change".

●
Entropy determines the direction in which natural process should occur.

●
Because entropy increases when heat flows from hot object to cold object. If
heat were to flow from a cold to a hot object.

●
Entropy will decrease leading to violation of second law thermodynamics.

●
Entropy is also called 'measure of disorder'. All natural process occur such
that the disorder should always increases.

●
Consider a bottle with a gas inside. When the gas molecules are inside the
bottle it has less disorder. Once it spreads into the entire room it leads to
more disorder.

●
In order words when the gas is inside the bottle the entropy is less and once
the gas spreads into entire room, the entropy increases. From the second law of
thermodynamics, entropy always increases.

●
If the air molecules go back in to the bottle, the entropy should decrease,
which is not allowed by the second law of thermodynamics.

●
The same explanation applies to a drop of ink diffusing into water. Once the
drop of ink spreads, its entropy is increased.

●
The diffused ink can never become a drop again. So the natural processes occur
in such a way that entropy should increase for all irreversible process.

**24. Explain in detail the working of a refrigerator.**

**Refrigerator:** A
refrigerator is a Carnot's engine working in the reverse order.

**Working
principle:**

●
The working substance (gas) absorbs a quantity of heat Q_{1} from the
cold body (sink) at a lower temperature T_{L}.

●
A certain amount of work W is done on the working substance by the compressor
and a quantity of heat Q_{H} is rejected to the hot body (source) ie.
the atmosphere at T_{H}.

●
From the first law of thermodynamics.

●
Q_{L} + W = Q_{H} AS a
result, the cold reservoir (refrigerator) further cools down and the
surroundings (kitchen or atmosphere) gets hotter.

**Co-effeicient
of performance (COP) (β):**

●
Measure of the efficiency of a regrigerator.

●
The ratio of heat extracted from the cold body (sink) to the enternal work done
by the compressor W.

COP
= β = Q_{L}/W

β
= Q_{L} / [Q_{H} − Q_{L}]

β
= 1 / [ (Q_{H} − Q_{L}) − 1]

●
Q_{H} / Q_{L} = T_{H} / T_{L}

β
= T_{L }/ [T_{H} − T_{L}]

●
Generater the COP, the better is the condition of the refrigerator.

●
A typical refrigerator has COP around 5 to 6

●
Lesser the difference in the temperature of the cooling chamber and the
atmosphere, COP is higher.

●
Without external work heat cannot flow from cold object to hot object.

Tags : Heat and Thermodynamics | Physics , 11th Physics : UNIT 8: Heat and Thermodynamics

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11th Physics : UNIT 8: Heat and Thermodynamics : Long Questions and Answer | Heat and Thermodynamics | Physics

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