1. Physics 101 Lecture 11

2. Thermal Physics Thermodynamics

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 System Surroundings 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  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 and 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 U i, to a final value of U f due to heat Q and work W   U = U f – U i = 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 = 1500 + (-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 What is the change in the internal energy of the system?  ΔU = 1500 + (2200) = 3700 J

16. First Law – Example 2a Temperature of 3 moles of a monatomic ideal gas is reduced from T i = 540 K to T f = 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  U i = (3/2)nRT = (3/2)(3)(8.31)(540) = 20193.3 J  U f = (3/2)nRT = (3/2)(3)(8.31)(350) = 13088.3 J  ΔU = U f – U i = 13088.3 – 20193.3 = -7105 J (b) Find the work done  ΔU = Q + W  W = ΔU - Q = -7100 J – (5500) = -12600 J

17. First Law – Example 2b Temperature of 3 moles of a monatomic ideal gas is reduced from T i = 540 K to T f = 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  U i = 3/2(3)R(540) = (9/4)(8.31)(540) = 20193.3 J  U f = 3/2(3)R(350) = (9/4)(8.31)(350) = 13088.3 J  ΔU = U f – U i = 13088.3 – 20193.3 = -7105 J (b) Find the work done  ΔU = Q + W  W = ΔU - Q = -7100 J - (1500) = -8600 J

18. Thermal Processes isothermal  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 (cos  = +1 or -1)  A = area  W = (F/A)Ad = P(V i – V f )  W = -P(V f – V i ) If V f is greater than V i, 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 10 5 Pa Temperature of the water is raised by 31 0 C Water is in the liquid phase and expands 1.0 x 10 -8 m 3 Specific heat capacity of water is 4186 J/kgC 0 Determine the change in the internal energy of the water   U = Q + W  Q = cm(T f – T i )  W = -P  V   U = cm(T f – T i ) - P  V   U = 4186(0.001)(31) – (2.0 x 10 5 )(1.0 x 10 -8 )   U = 129.766 – 0.002 = 129.764 J

21. Isobaric Process – Example 2 0.001 kg of water is placed in the cylinder Pressure is maintained at 2.0 x 10 5 Pa Temperature of the water is raised by 31 0 C Water is in the gas phase and expands 7.1 x 10 -5 m 3 Specific heat capacity is 2020 J/kgC 0 Determine the change in the internal energy of the water   U = Q + W  Q = cm(T f – T i )  W = -P  V   U = cm(T f – T i ) - P  V   U = 2020(0.001)(31) – (2.0 x 10 5 )(7.1 x 10 -5 )   U = 62.6 – 14.2 = 48.4 J

22. Isochoric Process No Volume Change  W = -P  V = 0  Because no change in volume   U = Q + W = Q

23. Adiabatic Process No Heat Transferred  Q = 0   U = + W

24. Adiabatic Expansion / Compress Ideal Gas Q = 0  U = Q + W = +W As a gas expands adiabatically, it does negative work W is negative in the equation T i – T f 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. Second Law

26. 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

27. 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.

28. 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 paid into the surroundings as heat

29. 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”

30. 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

31. Second Law of Thermodynamics Heat Engine The symbol Q H refers to the input heat, and the subscript H indicates the hot reservoir The symbol Q C 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 |Q H | indicates the magnitude of the input heat |Q H |, |Q C |, and |W| refer to magnitudes only, they never have negative values when they appear in equations

32. 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

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

34. 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?

35. 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

36. 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

37. 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

38. 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

39. Carnot Engine Efficiency Temperatures are expressed in Kelvin

40. Carnot Energy Efficiency Example Water near the surface of a tropical ocean has a temperature of 298.2 K Water 700 m beneath the surface has a temperature of 280.2 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

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

42. 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

43. 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   S universe = 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

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

45. 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

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