Chapter 15

ThermodynamicsThermodynamics  The name we give to the study of processes in which energy is transferred as heat and as work  There are 4 Laws of Thermodynamics: 0, 1, 2, 3  1 & 2 have the most meaning and will be studied the most

Heat and Work- Both forms of Energy (J) Heat- a transfer of energy due to a difference in temperature Heat- a transfer of energy due to a difference in temperature Work- a transfer of energy NOT due to temperature difference Work- a transfer of energy NOT due to temperature difference System- object on set of objects we wish to consider System- object on set of objects we wish to consider Environment- everything else NOT in our system Environment- everything else NOT in our system Closed System- no mass enter or leaves our system (energy may be exchanged w/ environment) Closed System- no mass enter or leaves our system (energy may be exchanged w/ environment)

Open System- mass may enter or leave (also energy) Open System- mass may enter or leave (also energy) Isolated System- No energy of any form leaves the system Isolated System- No energy of any form leaves the system Continued…

1 st Law of Thermodynamics ΔU = Q – W ΔU = Q – W –Derived from the law of conservation of energy U= internal energy of a system (J) U= internal energy of a system (J) Q = heat (J) Q = heat (J) W = Net Work (J) W = Net Work (J)

We would expect internal energy to INCREASE if work were done or if heat were added. We would expect internal energy to INCREASE if work were done or if heat were added. Likewise internal energy DECREASES if heat flows out or if work were done by the system Likewise internal energy DECREASES if heat flows out or if work were done by the system –Signs +/- +Q heat added +Q heat added -Q heat lost -Q heat lost -W work ON system -W work ON system +W Work BY system +W Work BY system

ΔU = Q – W applies for a closed system, it also applies to an open system if we take into account the change in internal energy due to the increase or decrease in the amount of matter ΔU = Q – W applies for a closed system, it also applies to an open system if we take into account the change in internal energy due to the increase or decrease in the amount of matter

For an isolated system, No work is done and NO heat enters or leaves, W = Q = O therefore ΔU = O For an isolated system, No work is done and NO heat enters or leaves, W = Q = O therefore ΔU = O Properties of a system are U, P, V, T, mass, n(# of moles)…Work and Heat are NOT properties of a system but they are involved in a system change! Properties of a system are U, P, V, T, mass, n(# of moles)…Work and Heat are NOT properties of a system but they are involved in a system change!

Example An amount of heat equal to 2500J is added to a system, and 1800J of work is done on the system, what is the change in internal energy of the system? An amount of heat equal to 2500J is added to a system, and 1800J of work is done on the system, what is the change in internal energy of the system?

Q= 2500J Work on System= -1800J 1800 is done on the system: (-) Q= 2500J Work on System= -1800J 1800 is done on the system: (-) ΔU = 2500J - (-1800J) ΔU = 4300J ΔU = 2500J - (-1800J) ΔU = 4300J It seems intuitive to add them but keep up with signs because not every problem will be intuitive

Example What would be the internal energy change if 2500J of heat is added to the system and 1800J of work is done by the system. What would be the internal energy change if 2500J of heat is added to the system and 1800J of work is done by the system.

Heat is being added so Q = +2500J. but now work is being done by the system ΔU = 2500J – 1800J = 700J 700J ΔU is much less here because work was output. READ PAGE

Simple Systems Prefix – ISO (same) Prefix – ISO (same) Isothermal- same temperature Isothermal- same temperature Isobaric- same pressure Isobaric- same pressure Isochoric- same volume (isovolumetric) Isochoric- same volume (isovolumetric)

Pressure volume diagrams… Isothermal Isothermal P V A’ A B B’ A’- B’ Lower Constant Temp Curve is called an isotherm A’- B’ Lower Constant Temp Curve is called an isotherm

Isobaric P V

Isochoric V P

1 st Law is seen T = constant ΔU = O Q= W P= constant W = PΔV V= constant W= O

Adiabatic process is one which NO heat is allowed to flow into or out of the system Q=0 Adiabatic process is one which NO heat is allowed to flow into or out of the system Q=0 1.Happens fast, no time to transfer (I.e. engine) 2.Happens in a very well insulated container Slow adiabatic expansion follows curve AC since Q=0Slow adiabatic expansion follows curve AC since Q=0 AB isothermal AC adiabatic A B C

ΔU = -W U decreases as gas expands therefore temperature decreases as well ΔU = -W U decreases as gas expands therefore temperature decreases as well Adiabatic compression follows curve AC, work is done ON gas, as U increases therefore temperature increases Adiabatic compression follows curve AC, work is done ON gas, as U increases therefore temperature increases In diesel engines air/fuel is compressed so fast that fuel explodes spontaneously due to increase in U and T In diesel engines air/fuel is compressed so fast that fuel explodes spontaneously due to increase in U and T

F = PA F = PA W = Fd W = Fd W = Pad W = Pad W= PΔV W= PΔV Concepts pg. 447 Concepts pg. 447 Read Example 15-4 pg. 448 Read Example 15-4 pg. 448 area x dist. = volume

In BDA only 1 leg does work, BD  W = PΔV (2x10 5 n/m 2 )(2x10 -3 m x10 -3 m 3 ) W= -1.6 x 10 3 J negative means work done on the gas  There is NO temperature change from B to A (along isotherm) so there is NO change in internal energy ΔU = 0 U = Q-W 0 = Q-W Q= W Q= 1.6 x 10 3 J 1600J flows out of the gas, total heat loss for process BDA In BDA only 1 leg does work, BD  W = PΔV (2x10 5 n/m 2 )(2x10 -3 m x10 -3 m 3 ) W= -1.6 x 10 3 J negative means work done on the gas  There is NO temperature change from B to A (along isotherm) so there is NO change in internal energy ΔU = 0 U = Q-W 0 = Q-W Q= W Q= 1.6 x 10 3 J 1600J flows out of the gas, total heat loss for process BDA

Problems Example problems 15-5, 15-6 (pg. 448) Example problems 15-5, 15-6 (pg. 448) Class Work pg #’s 1, 3, 5, 7 Class Work pg #’s 1, 3, 5, 7 Homework pg #’s 2, 4, 6, 8 Homework pg #’s 2, 4, 6, 8

 ΔU = Q-W = x 10 3 J - (-1.6 x 10 3 J) = -1.8x 10 3 J A C B P (Atm) V(L)

5. A B C Vo/ 2 Vo0 Po/ 2 Po V P

7A Adiabatic Process No Heat Flow Q=0 B 1 st Law ΔU = Q-W = 0 - (-1350) = +1350J C Internal Energy of ideal gas depends only on the temperature, U= 3/2nRT Increase in U means…INCREASE IN T

2 nd Law of Thermodynamics Speaks to the direction of heat flow Speaks to the direction of heat flow There is a natural direction that does not reverse itself There is a natural direction that does not reverse itself –Example: you will never see a broken glass cup on the floor reassemble itself, accelerate up to a table top and come to rest. There is an irreversible direction to some process. There is an irreversible direction to some process.

Just to look at the 1 st law of thermodynamics there is NO direction expressed or implied to which way heat flows. Just to look at the 1 st law of thermodynamics there is NO direction expressed or implied to which way heat flows. The second law states direction The second law states direction “Heat flows naturally from a hot object to a cold object, heat will NOT flow spontaneously from a cold object to a hot object.” – Clausius “Heat flows naturally from a hot object to a cold object, heat will NOT flow spontaneously from a cold object to a hot object.” – Clausius 2 nd law came about partly due to the study of heat engines 2 nd law came about partly due to the study of heat engines

Heat Engines Any device that changes thermal energy into mechanical work (steam engine, automobile engine) Any device that changes thermal energy into mechanical work (steam engine, automobile engine) –Fig 15-9 pg. 451 –Steam engine –reciprocating –Turbine –Internal combustion engine  4 cycle pg. 452

Overhead FIG PV diagram for different process FIG PV diagram for different process FIG 15-9 Energy transfers for a heat engine FIG 15-9 Energy transfers for a heat engine FIG Steam Engine FIG Steam Engine FIG the 4 cycle internal combustion engine (otto cycle) FIG the 4 cycle internal combustion engine (otto cycle)

Why is a temperature difference required?  If the steam were at the same temperature on both sides of the piston or turbine were the same, that would mean the pressure was the same on both sides and then NO work would be done

Efficiency of Heat Engine ( e) e= W/Q H e= W/Q H –W = work output (J) –what you get from engine –Q H = what you put in Since energy is conserved, heat input Q H must equal work done plus heat that flows out at the low temp (Q L ) Since energy is conserved, heat input Q H must equal work done plus heat that flows out at the low temp (Q L )

Q H = W + Q L W = Q H – Q L e= W/Q H = Q H - Q L / Q H = 1 - Q L /Q H ** to write as % must still multiply by 100**

Example A 20% efficient car engine produces an average of 23,000J of mechanical work per second. How much heat is discharged per second from this engine?

Q L = ? Q L / Q H = 1 - e =.80 Q L =.8Q H Q H = W/e 23000J/.20 = 1.15 x 10 5 J Q L =.80Q H =.8(1.15 x 10 5 J) = 9.2 x 10 4 J =92KW Q L = ? Q L / Q H = 1 - e =.80 Q L =.8Q H Q H = W/e 23000J/.20 = 1.15 x 10 5 J Q L =.80Q H =.8(1.15 x 10 5 J) = 9.2 x 10 4 J =92KW

The Ideal Engine is the Carnot engine One does not really exist. This concept proves that 100% efficient engines cannot exist and efficiencies can be determined by a ratio of temperatures One does not really exist. This concept proves that 100% efficient engines cannot exist and efficiencies can be determined by a ratio of temperatures

Carnot (Ideal) Efficiency e ideal = T H - T L / T H = 1 - T L /T H Pg. 454 – 455 example 15-9, 15-10

For a 100% efficient engine to happen T L would need to be absolute 0 … 0 Kelvin For a 100% efficient engine to happen T L would need to be absolute 0 … 0 Kelvin 100% efficiency is a violation of the 3 rd law of thermodynamics- “absolute zero is unattainable” 100% efficiency is a violation of the 3 rd law of thermodynamics- “absolute zero is unattainable” For a 100% efficient engine to happen T L would need to be absolute 0 … 0 Kelvin For a 100% efficient engine to happen T L would need to be absolute 0 … 0 Kelvin 100% efficiency is a violation of the 3 rd law of thermodynamics- “absolute zero is unattainable” 100% efficiency is a violation of the 3 rd law of thermodynamics- “absolute zero is unattainable”

Overhead Fig The Carnot Cycle

2 nd Law Restrained Kelvin- Plank Statement Kelvin- Plank Statement No device is possible whose sole effect is to transform a given amount of heat completely into work No device is possible whose sole effect is to transform a given amount of heat completely into work

Refrigerators, Air Conditioners & Heat Pumps The above three operate on the same principle just in the reverse direction as a heat engine. The above three operate on the same principle just in the reverse direction as a heat engine. They transfer heat out of an environment to make it cold They transfer heat out of an environment to make it cold By doing work, heat is taken from a low temp, T L (inside the fridge) and a greater amount of heat is exhausted at a high temp, T H (the room), you can feel this warmth and that’s why behind the fridge is so warm. By doing work, heat is taken from a low temp, T L (inside the fridge) and a greater amount of heat is exhausted at a high temp, T H (the room), you can feel this warmth and that’s why behind the fridge is so warm.

The work (W) is normally done by a compressor. The work (W) is normally done by a compressor. A perfect refrigerator is NOT possible A perfect refrigerator is NOT possible – See Clausius statement of 2 nd Law Heat Pump: these can heat a house in winter (Q L outside to Q H inside) by doing work. True heat pumps can be turned around and used to cool a house in the summer Heat Pump: these can heat a house in winter (Q L outside to Q H inside) by doing work. True heat pumps can be turned around and used to cool a house in the summer

OVERHEAD FIG Typical Refrigerator system FIG Typical Refrigerator system

Entropy This is the term that defines the general statement of 2 nd Law of Thermodynamics. This is the term that defines the general statement of 2 nd Law of Thermodynamics. Entropy is a function of the state of a system Entropy is a function of the state of a system Entropy is a measure of the disorder or order of a system Entropy is a measure of the disorder or order of a system ΔS = Q/T T in Kelvin

The total entropy of any system plus that of its environment increases as a result of any natural process. (2 nd Law, General Statement) The total entropy of any system plus that of its environment increases as a result of any natural process. (2 nd Law, General Statement) This is quite different that the other laws of physics. Most are equalities: This is quite different that the other laws of physics. Most are equalities: – V = d/t p= mv PE = mgh Entropy is NOT conserved it ALWAYS increases

Phase Changes and Entropy During a phase change, if energy flows out (water freezes) entropy decreases. Even though the Waters entropy decreases, the entropy of the universe has still increased. (think about the electrical energy used to cool the water)

Clockwise and Counter clockwise On a PV graph, when a complete cycle is shown, Clockwise means that heat is removed from the gas (substance). Counterclockwise would indicate heat is added to the gas (substance).

Natural Processes tend to move toward a state of greater disorder. ORDERDISORDER Pepper Salt Shake it up!

Falls CRASH!

With heat it goes like this… High heat time same heat time same heat same average KE same average KE Low Heat Order Disorder (could do work) (cant do work)

Disorder can be associated with randomness Disorder can be associated with randomness Order associates with information Order associates with information Heat death Energy Resources