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Lecture PowerPoints

Physics for Scientists and Engineers, with Modern

Physics, 4th edition

Giancoli

Chapter 20


Chapter 20Second Law of

Thermodynamics


• The Second Law of Thermodynamics—Introduction

• Heat Engines

• Reversible and Irreversible Processes; the Carnot Engine

• Refrigerators, Air Conditioners, and Heat Pumps

• Entropy

• Entropy and the Second Law of Thermodynamics

Units of Chapter 20


• Order to Disorder

• Unavailability of Energy; Heat Death

• Statistical Interpretation of Entropy and the Second Law

• Thermodynamic Temperature; Third Law of Thermodynamics

• Thermal Pollution, Global Warming, and Energy Resources

Units of Chapter 20


The first law of thermodynamics tells us that energy is conserved. However, the absence of the process illustrated above indicates that conservation of energy is not the whole story. If it were, movies run backwards would look perfectly normal to us!

20-1 The Second Law of Thermodynamics—Introduction


The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one:

Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object.

20-1 The Second Law of Thermodynamics—Introduction


It is easy to produce thermal energy using work, but how does one produce work using thermal energy?

This is a heat engine; mechanical energy can be obtained from thermal energy only when heat can flow from a higher temperature to a lower temperature.

20-2 Heat Engines


We will discuss only engines that run in a repeating cycle; the change in internal energy over a cycle is zero, as the system returns to its initial state.

The high-temperature reservoir transfers an amount of heat QH to the engine, where part of it is transformed into work W and the rest, QL, is exhausted to the lower temperature reservoir. Note that all three of these quantities are positive.

20-2 Heat Engines


A steam engine is one type of heat engine.

20-2 Heat Engines


The internal combustion engine is a type of heat engine as well.

20-2 Heat Engines


Why does a heat engine need a temperature difference?

Otherwise the work done on the system in one part of the cycle would be equal to the work done by the system in another part, and the net work would be zero.

20-2 Heat Engines


The efficiency of the heat engine is the ratio of the work done to the heat input:

Using conservation of energy to eliminate W, we find:

20-2 Heat Engines


20-2 Heat Engines

Example 20-1: Car efficiency.

An automobile engine has an efficiency of 20% and produces an average of 23,000 J of mechanical work per second during operation.

(a) How much heat input is required, and

(b) How much heat is discharged as waste heat from this engine, per second?


20-2 Heat Engines

No heat engine can have an efficiency of 100%. This is another way of writing the second law of thermodynamics:

No device is possible whose sole effect is to transform a given amount of heat completely into work.


The Carnot engine was created to examine the efficiency of a heat engine. It is idealized, as it has no friction. Each leg of its cycle is reversible.

The Carnot cycle consists of:

• Isothermal expansion

• Adiabatic expansion

• Isothermal compression

• Adiabatic compression

20-3 Reversible and Irreversible Processes; the Carnot Engine


From this we see that 100% efficiency can be achieved only if the cold reservoir is at absolute zero, which is impossible.

Real engines have some frictional losses; the best achieve 60–80% of the Carnot value of efficiency.

For an ideal reversible engine, the efficiency can be written in terms of the temperature:




Example 20-2: A phony claim?

An engine manufacturer makes the following claims: An engine’s heat input per second is 9.0 kJ at 435 K. The heat output per second is 4.0 kJ at 285 K. Do you believe these claims?



Automobiles run on the Otto cycle, shown here, which is two adiabatic paths alternating with two constant-volume paths. The gas enters the engine at point a and is ignited at point b. Curve cd is the power stroke, and da is the exhaust.



Example 20-3: The Otto cycle.

(a) Show that for an ideal gas as working substance, the efficiency of an Otto cycle engine is

e = 1 – (Va/Vb)1-γ

where γ is the ratio of specific heats (γ = CP/CV) and Va/Vb is the compression ratio.

(b) Calculate the efficiency for a compression ratio Va/Vb = 8.0 assuming a diatomic gas like O2 and N2.


These appliances are essentially heat engines operating in reverse.

By doing work, heat is extracted from the cold reservoir and exhausted to the hot reservoir.

20-4 Refrigerators, Air Conditioners, and Heat Pumps



This figure shows more details of a typical refrigerator.


Refrigerator performance is measured by the coefficient of performance (COP):


Substituting:

For an ideal refrigerator,



Example 20-4: Making ice.

A freezer has a COP of 3.8 and uses 200 W of power. How long would it take this otherwise empty freezer to freeze an ice-cube tray that contains 600 g of water at 0°C?


A heat pump can heat a house in the winter:



A heat pump is evaluated on a different type of “Coefficient of Performance” COP:


COP (heat pump) = QH/W

vs.

COP (refrigerator) = Qc/W


Example 20-5: Heat pump.

A heat pump has a coefficient of performance of 3.0* and is rated to do work at 1500 W.

(a)How much heat can it add to a room per second?

(b)

COP = 3.0 = Qh/W =>Qh = 3.0 x W = 4500 W = 4500 Joule/Secso…4500 J per second, or ~4.3 BTU


Example 20-5: Heat pump.

A heat pump has a coefficient of performance of 3.0* and is rated to do work at 1500 W.

(b) If the heat pump were turned around to act as an air conditioner in the summer, what would you expect its coefficient of performance to be, assuming all else stays the same?

Qc = Qh – W = 4500 J – 1500 J = 3000 J

And COP (refrigerator) = Qc/W = 3000/1500 = 2


20-5 Entropy

You can’t measure the “heat” of something!

You CAN measure how much heat must be added to change something’s temperature.

Consider isothermal process (T constant):

P

V

How much heat must you add as the gas expands to keep it at T?


20-5 Entropy

P

V

How much heat must you add as the gas expands to keep it at T?

U = Qin – Wout = 0 (isothermal)

Q in = W out = nRT*ln(Vf/Vi)

Define Entropy ChangeS = Q/T

For this reversible isothermal process,S = nR*ln(Vf/Vi)


20-5 Entropy

Entropy = “S”

A *state* variable for a gas (P, V, T)

A measure of “disorder” of a system

An indicator of the likelihood (probability) of reaction directions (towards higher S)

Measured only when it *changes*


20-5 Entropy

Definition of the change in entropy S when an amount of heat Q is added:

if the process is reversible and the temperature is constant. Otherwise,


20-5 Entropy

IF the temperature is NOT constant:


20-5 Entropy

Consider Carnot cycles:

TH/TL = QH/QL

So…

QH/TH = QL/TL


20-5 Entropy

Since for any Carnot cycle QH/TH + QL/TL = 0, we see that the integral of dQ/T around a closed path is zero. This means that entropy is a state variable—the change in its value depends only on the initial and final states.


20-5 Entropy

Any reversible cycle can be written as a succession of Carnot cycles; therefore, what is true for a Carnot cycle is true of all reversible cycles.


20-5 EntropySince for any Carnot cycle QH/TH + QL/TL = 0, if we approximate any reversible cycle as an infinite sum of Carnot cycles, we see that the integral of dQ/T around a closed path is zero. This means that entropy is a state variable—the change in its value depends only on the initial and final states.


20-6 Entropy and the Second Law of Thermodynamics

Example 20-6: Entropy change when mixing water.

A sample of 50.0 kg of water at 20.00°C is mixed with 50.0 kg of water at 24.00°C. Estimate the change in entropy.



The total entropy always increases when heat flows from a warmer object to a colder one in an isolated two-body system. The heat transferred is the same, and the cooler object is at a lower average temperature than the warmer one, so the entropy gained by the cooler one is always more than the entropy lost by the warmer one.



Example 20-7: Entropy changes in a free expansion.

Consider the adiabatic free expansion of n moles of an ideal gas from volume V1 to volume V2, where V2 > V1. Calculate the change in entropy

(a) of the gas and

(b) of the surrounding environment.

(c) Evaluate ΔS for 1.00 mole, with V2 = 2.00 V1.



Example 20-8: Heat transfer.

A red-hot 2.00-kg piece of iron at temperature T1 = 880 K is thrown into a huge lake whose temperature is T2 = 280 K. Assume the lake is so large that its temperature rise is insignificant. Determine the change in entropy

(a) of the iron and

(b) of the surrounding environment (the lake).



The fact that after every interaction the entropy of the system plus the environment increases is another way of putting the second law of thermodynamics:

The entropy of an isolated system never decreases. It either stays constant (reversible processes) or increases (irreversible processes).


Entropy is a measure of the disorder of a system. This gives us yet another statement of the second law:

Natural processes tend to move toward a state of greater disorder.

Example: If you put milk and sugar in your coffee and stir it, you wind up with coffee that is uniformly milky and sweet. No amount of stirring will get the milk and sugar to come back out of solution.

20-7 Order to Disorder


Another example: When a tornado hits a building, there is major damage. You never see a tornado approach a pile of rubble and leave a building behind when it passes.

Thermal equilibrium is a similar process—the uniform final state has more disorder than the separate temperatures in the initial state.

20-7 Order to Disorder


Another consequence of the second law:

In any natural process, some energy becomes unavailable to do useful work.

If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the heat death of the universe.

20-8 Unavailability of Energy; Heat Death


Entropy vs. Enthalpy

Enthalpy = Internal Energy PLUS Energy in the surrounding system:

The enthalpy of a system is defined as:H = U + pV

And the change in enthalpy = H = U + (pV)

For constant pressure processes:H = U + pV = U +Work = Q!



Enthalpy is sometimes described as the heat content of a system under a given pressure, Such a visualization assumes no energy exchange with the environment other than heat or expansion work. Given such restrictions, it can be shown that:

The enthalpy is the total amount of energy which the system can emit through heat,

Adding or removing energy through heat is the only way to change the enthalpy, and

The amount of change in enthalpy is equal to the amount of energy added through heat.



ENTHALPHY: Is the energy content of a process (chemical, thermodynamic, mechanical, etc) that can be recovered. It is also described as useful energy.

ENTROPY: Is the energy content of a process (chemical, thermodynamic, mechanical, etc) that CANNOT be recovered. It is also described as chaos.


20-9 Statistical Interpretation of Entropy and the Second Law

Microstate: a particular configuration of atomsMacrostate: a particular set of macroscopic variablesThis example uses coin tosses:



Similarly, the most probable distribution of velocities in a gas is Maxwellian:

The most probable state is the one with the greatest disorder, or the greatest entropy. With k being Boltzmann’s constant and W the number of microstates,



Example 20-9: Free expansion—statistical determination of entropy.

Determine the change in entropy for the adiabatic free expansion of one mole of a gas as its volume doubles. Assume W, the number of microstates for each macrostate, is the number of possible positions.



In this form, the second law of thermodynamics does not forbid processes in which the total entropy decreases; it just makes them exceedingly unlikely.


20-10 Thermodynamic Temperature; Third Law of Thermodynamics

Since the ratio of heats exchanged between the hot and cold reservoirs in a Carnot engine is equal to the ratio of temperatures, we can define a temperature scale using the triple point of water:

T = (273.16K)(Q/Qtp).

Here, Q and Qtp are the heats exchanged by a Carnot engine with reservoirs at temperatures T and Ttp.


20-10 Thermodynamic Temperature; Third Law of Thermodynamics

there is no way to achieve a temperature of absolute zero. This is the third law of thermodynamics:

It is not possible to reach absolute zero in any finite number of processes.

Also, since the maximum efficiency of a heat engine is


20-11 Thermal Pollution, Global Warming, and Energy Resources

Over 90% of the energy used in the U.S. is generated using heat engines to drive turbines and generators—even nuclear power plants use the energy generated from fission heat water for a steam engine. The thermal output QL of all these heat engines contributes to warming of the atmosphere and water. This is an inevitable consequence of the second law of thermodynamics.


Summary of Chapter 20

• Heat engine changes heat into useful work

• Efficiency: work/heat input

• Maximum efficiency: 1 – TL/TH

• Refrigerators and air conditioners are heat engines, reversed; COP = heat removed/work

• Heat pump: COP = heat delivered/work

• Second law of thermodynamics: Natural processes always tend to increase entropy


Summary of Chapter 20

• Change in entropy gives direction to “arrow of time”

• As time goes on, energy becomes degraded.

• Heat engines cause thermal pollution.

• Entropy change in reversible process:

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