1. Physics 101
Lecture 13

2. Thermal Physics Thermodynamics
Third Physics

3. What is Thermodynamics?
The branch of physics that studies the transformation of energy
For example, thermal energy changing into work

4. What is a Thermodynamic System?
A system is the region of universe under study
Everything in the universe except the system is known as the surroundings
A system is separated from the remainder of the universe by a real or imaginary boundary
Exchanges of work, thermal energy, or matter between system and surroundings take place across the boundary
Surroundings
System
Boundary

5. Thermodynamic System Examples
Steam engine
Gasoline engine
Solution in a test tube
Battery
Living organism
Deceased organism
Tropical cyclone
Planet
Black hole
Gas in a container

6. Thermodynamic System Types
Type of system depends on whether heat, work, or matter can cross the boundary
For example:
Isolated System
Heat, work, and matter cannot cross boundary
Open System
Matter can cross boundary
Closed System
Matter cannot cross boundary
Adiabatic Systems
Heat can not cross boundary

7. What is Thermal Equilibrium Single System
Mechanical equilibrium
Pressure ceases to change with time
Chemical equilibrium
Internal structure ceases to change with time
Chemical reaction
Diffusion (transfer of matter from one part of system to another)
Thermal equilibrium
Temperature ceases to change with time is achieved when pressure and temperature cease to change with time
Thermodynamic equilibrium
System is in mechanical, chemical, and thermal equilibrium

8. What is Thermal Equilibrium Two Systems
Thermal equilibrium is achieved when two systems in thermal contact with each other cease to exchange energy by heat

9. Zeroth Law

10. Thermodynamics Laws Zeroth Law First Law Second Law
Introduced the concept of temperature
First Law
Introduced internal energy
Describes feasible changes in the universe because energy is conserved
Second Law
Introduced entropy
Identifies from among feasible changes the ones that are spontaneous
That is, have a tendency to occur without us having to do work to drive them

11. Zeroth Law of Thermodynamics
Two systems, each separately in thermal equilibrium with a third, are in equilibrium with each other
This tells us that there is some property of each system that is the same
This property is temperature

12. First Law

13. First Law of Thermodynamics Conservation of Energy
Energy is conserved
Heat and work are both forms of energy

14. First Law of Thermodynamics Conservation of Energy
The internal energy of a system changes from an initial value Ui, to a final value of Uf due to heat Q and work W
DU = Uf – Ui = Q + W
Q is positive when the system gains heat and negative when it loses heat
W is positive when work is done on the system and negative when work is done by the system

15. First Law – Example 1
(a)
A system gains 1500 J of heat from its surroundings, and 2200 J of work is done by the system on the surroundings
What is the change in the internal energy of the system?
ΔU = (-2200) = -700 J
(b)
A system gains 1500 J of heat from its surroundings, and 2200 J of work is done on the system by the surroundings
ΔU = (2200) = 3700 J

16. First Law – Example 2a
Temperature of 3 moles of a monatomic ideal gas is reduced from Ti = 540 K to Tf = 350 K
5500 J of heat flows into the gas
(a) Find the change in the internal energy
The internal energy of the gas is given by (3/2)nRT
Ui = (3/2)nRT = (3/2)(3)(8.31)(540) = J
Uf = (3/2)nRT = (3/2)(3)(8.31)(350) = J
ΔU = Uf – Ui = – = J
(b) Find the work done
ΔU = Q + W
W = ΔU - Q = J – (5500) = J

17. First Law – Example 2b
Temperature of 3 moles of a monatomic ideal gas is reduced from Ti = 540 K to Tf = 350 K
1500 J of heat flows into the gas
(a) Find the change in the internal energy
The internal energy of the gas is given by (3/2)nRT
Ui = 3/2(3)R(540) = (9/4)(8.31)(540) = J
Uf = 3/2(3)R(350) = (9/4)(8.31)(350) = J
ΔU = Uf – Ui = – = J
(b) Find the work done
ΔU = Q + W
W = ΔU - Q = J - (1500) = J

18. Thermal Processes isothermal isobaric adiabatic isovolumic (isochoric)
Temperature of system is constant
isobaric
Pressure of system is constant
adiabatic
No heat transferred to or from the system
isovolumic (isochoric)
Volume of the system is constant

19. Isobaric Process
W = Fd (cosq = +1 or -1)
A = area
W = (F/A)Ad = P(Vi – Vf)
W = -P(Vf – Vi)
If Vf is greater than Vi, the system does work on the surroundings
The convention is that W is negative and the formula gives a negative result

20. Isobaric Process – Example 1
0.001 kg of water is placed in the cylinder
Pressure is maintained at 2.0 x 105 Pa
Temperature of the water is raised by 31 0C
Water is in the liquid phase and expands 1.0 x 10-8 m3
Specific heat capacity of water is 4186 J/kgC0
Determine the change in the internal energy of the water
DU = Q + W
Q = cm(Tf – Ti)
W = -PDV
DU = cm(Tf – Ti) - PDV
DU = 4186(0.001)(31) – (2.0 x 105)(1.0 x 10-8)
DU = – = J

21. Isobaric Process – Example 2
0.001 kg of water is placed in the cylinder
Pressure is maintained at 2.0 x 105 Pa
Temperature of the water is raised by 31 0C
Water is in the gas phase and expands 7.1 x 10-5 m3
Specific heat capacity is 2020 J/kgC0
Determine the change in the internal energy of the water
DU = Q + W
Q = cm(Tf – Ti)
W = -PDV
DU = cm(Tf – Ti) - PDV
DU = 2020(0.001)(31) – (2.0 x 105)(7.1 x 10-5)
DU = 62.6 – 14.2 = 48.4 J

22. Isochoric Process No Volume Change
W = -PDV = 0
Because no change in volume
DU = Q + W = Q

23. Adiabatic Process No Heat Transferred
Q = 0
DU = + W

24. Adiabatic Expansion / Compress Ideal Gas
Q = 0
DU = Q + W = +W
As a gas expands adiabatically, it does negative work
W is negative in the equation
Ti – Tf is positive
Final temperature of gas must be less than initial temperature
Internal energy of gas is reduced to provide necessary energy to do work
Internal energy decreases

25. Isothermal Expansion / Compression Ideal Gas
As lid moves up the pressure changes
Work is
Internal energy of ideal gas is proportional to T
Internal energy remains constant throughout an isothermal process
Change in internal energy is zero
DU = 0 = Q + W
Q = -W
Energy of the work originates in hot water
If the gas is compressed isothermally, heat flows into the gas from the water
If the gas is expanded isothermally, heat flows into the water from the gas

26. Isothermal Process - Example
Two moles of monatomic argon gas expand isothermally at 298 K, from in initial volume of Vi = m3 to a final volume of Vf = m3
(a) Find the work done by the gas
(b) Find the change in the internal energy of the gas
U = (3/2)nRT
DU = (3/2)nRDT = 0
(c) Find the heat supplied to the gas
Q = -W = J

27. Second Law

28. Second Law of Thermodynamics
“The second law is of central importance in the whole of science, and hence in our rational understanding of the universe, because it provides a foundation for understanding why any change occurs. Thus, not only is it a basis for understanding why engines run and chemical reactions occur … “
Peter Atkins, Four Laws that Drive the Universe

29. Second Law of Thermodynamics Kelvin Statement
“No cyclic process is possible in which heat is taken from a hot source and converted completely into work.”
Some of the energy supplied by the hot source must be lost into the surroundings as heat

30. Second Law of Thermodynamics Clausius Statement
“Heat does not pass from a body at low temperature to one at high temperature without an accompanying change elsewhere.”
Heat can be transferred in the “wrong” direction, but to achieve that transfer work must be done.

31. Second Law of Thermodynamics Textbook Statement
“Heat flows spontaneously from a substance at a higher temperature to a substance at a low temperature and does not flow spontaneously in the reverse direction”

32. Second Law of Thermodynamics Heat Engine
Device that uses heat to perform work
Heat is supplied to the engine at a high temperature from the hot reservoir
Part of the input heat is used to perform work
Remainder of the input heat is rejected to the cold reservoir, which has a temperature lower than the input temperature

33. Second Law of Thermodynamics Heat Engine
The symbol QH refers to the input heat, and the subscript H indicates the hot reservoir
The symbol QC stands for the rejected heat, and the subscript C denotes the cold reservoir
The symbol W refers to the work done
The vertical bars enclosing each of these three symbols are included to emphasize that we are concerned here with the absolute values, or magnitudes
Thus |QH| indicates the magnitude of the input heat
|QH|, |QC|, and |W| refer to magnitudes only, they never have negative values when they appear in equations

34. Heat Engine Efficiency (e)
To be highly efficient, a heat engine must produce a relatively large amount of work from as little input heat as possible
This is a fraction
Multiply by 100 to get a percentage

35. Heat Engine Alternative Efficiency Formulas
Conservation of energy gives
Gives alternative efficiency formulas

36. Heat Engine Efficiency Example
An automobile engine has an efficiency of 22.0% and produces 2510 J of Work
How much heat is rejected by the engine?

37. Second Law of Thermodynamics Reversible Process
A reversible process is one in which both the system and its surroundings can be returned to exactly the states they were in before the process occurred
A reversible process needs to take place slowly and not to be far from equilibrium
A process is not reversible if:
Process involves friction
Process involves spontaneous flow of heat from hot substance to cold substance

38. Second Law of Thermodynamics Carnot’s Principle
No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures
All reversible engines operating between same temperatures have same efficiency

39. Carnot Engine
Carnot engine has the maximum possible efficiency for its operating conditions because the processes occurring within it are reversible
Irreversible processes, such as friction, cause real engines to operate at less than maximum efficiency, for they reduce the ability to use heat to perform work

40. Carnot Engine Thermodynamic Temperature Scale
Carnot’s principle implies that the efficiency of a reversible engine can depend only on the temperatures of the hot and cold reservoirs
This observation led Lord Kelvin to propose a thermodynamic temperature scale (Kelvin)
He proposed that the thermodynamic temperatures of the cold and hot reservoirs be defined such that their ratio is equal to

41. Carnot Engine Efficiency
Temperatures are expressed in Kelvin

42. Carnot Energy Efficiency Example
Water near the surface of a tropical ocean has a temperature of K
Water 700 m beneath the surface has a temperature of K
It has been proposed that the warm water be used as the hot reservoir and the cool water as the cold reservoir of a heat engine
Find the maximum possible efficiency for such an engine

43. Second Law of Thermodynamics Entropy Statement
“The entropy of the universe increases in the course of any spontaneous change”
Entropy Change = DS = Q/T

44. Entropy Change Carnot Engine
From efficiency of a Carnot Engine
Rewrite the first equation
Change in entropy due to heat lost from hot temperature reservoir
Change in entropy due to heat gained by the low temperature reservoir
Total entropy change for reversible Carnot Engine is 0

45. Entropy Change Reversible Process
Total change in entropy is zero for a Carnot engine
It can be proved that when any reversible process occurs, the change in entropy of the universe is zero
DSuniverse = 0 J/K for a reversible process
Reversible processes do not alter the total entropy of the universe
The entropy of one part of the universe may change because of a reversible process, but if so, the entropy of another part changes in the opposite way by the same amount

46. Entropy Change Irreversible Process
Any irreversible process increases the entropy of the universe

47. Entropy Change Irreversible Process Example 1
1200 J of heat flows spontaneously through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K
Determine the amount by which this irreversible process changes the entropy of the universe

48. Entropy Change Irreversible Process Example 2
0.02 kg of ice at 0 0C melts into water with no change in temperature
By how much does the entropy of the 0.02 kg mass change in this process?