Thermodynamics13-16

What you absolutely have to know aboutThermodynamics to pass the AP Physic B test!

Refrigerator – Just a Heat Engine in reverse!Is it possible to move heat in the wrong direction from a cold place to a hot place? Sure, but the 2nd Law of Thermo tells us that it won’t happen without outside intervention. Work is required! The device that accomplishes this task is called a refrigerator.

The energy-transfer diagram at the right shows how heat can be artificially moved from cold to hot. Notice something strange and very important. Due to Conservation of Energy: the heat exhausted to the hot place is actually greater than the heat removed from the cold place! This is why you can’t cool down your kitchen by leaving the refrigerator door open.

What does a Refrigerator look like on a pV diagram?On a pV diagram, a refrigerator will be a cycle that moves in a counterclockwise direction. Here is an example:

Path T U W QABBCCANet

Continued on the next page!

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Now lets compare the refrigerator on the pV diagram with the energy-transfer diagram of a refrigerator.

Notice: Work input is required to move heat out of a cold place and into a hot place.

The exhausted heat is larger than the removed heat .

Conservation of Energy tell us that

Things to remember about heat engines and refrigerators:1. Heat Engines and Refrigerators are both cycles on a pV diagram. for cycles2. Heat Engines move in a clockwise cycle on a pV diagram. 3. Refrigerators move counterclockwise on a pV diagram.4. pV diagrams show you the particulars of what is going on in a gas while the energy transfer diagram only

shows you the net effect of the energy movement.5. The efficiency of a Heat Engine will be . 6. The energy transfer diagram and efficiency equations will work equally well with units of Joules or Watts.

For example a problem on the AP exam might say: A heat engine takes in heat at a rate of 1000 W and

exhausts waist energy to the environment at a rate of 600 W. Using the equation: , you can

calculate the rate at which work is done to be 400W! Using the equation: , you can

calculate the efficiency of the heat engine to be 0.4 or 40%. As long as all of your units are the either Joules or Watts, the equations function exactly the same.

We only have one more thing to talk about…




The most perfect cycle there never was. - The Carnot cycleAs you might have guessed, making heat engines and refrigerators more efficient is a good thing. However, the 2nd Law of Thermo tells us that neither one can be perfect. Heat engines can not convert 100% of thermal energy into work. Refrigerators can’t move heat from a cold place to a hot place without inputting some work. So… just how efficient can me make them? What would the most efficient cycle look like?

The most efficient cycle is a reversible cycle. Reversibility means quite a few things but here is the bottom line: The machine would have to be frictionless with no heat transfer during its mechanical operation. Heat would have to be transfer to and from the machine in an isothermal process . The machine would have to work equally well as a heat engine running clockwise on a pV diagram or in

reverse, counterclockwise on a pV diagram, as a refrigerator!

This presents two problems:1. You can’t really build a frictionless machine.2. Isothermal processes move very slowly. So, if you could build it, it would run too slow to be of much use!Why do we worry about this imaginary machine? It shows what maximum possibly efficiency can be achieved.

The perfect cycle is called the Carnot Cycle.Here is what it looks like on a pV diagram when it is running clockwise as a heat engine:

Process Type Significance1-2 Isothermal2-3 Adiabatic3-4 Isothermal4-1 Adiabatic

Notice that a Carnot Cycle operates between two isothermals or a hot temperature and a cold temperature . The Carnot cycle is the most efficient engine that can possibly operate between any two temperatures.

The efficiency of a Carnot cycle when running clockwise as a

heat engine is:

If we go back to our heat engine on pages 11-12, we will see that its actual efficiency was 10%. That heat engine operated between a highest temperature of 600K and a lowest temp or 300K. The max possible efficiency that the heat engine could ever possibly achieve operating between those two temperatures would be:

. As you can see, our heat engine was not very good.

Trick question: What is the max possible efficiency of a heat engine operating between and ?The answer is on the last page.




Key for table on page 6:Path T U W Q

A + + - + (Big +)B 0 0 + - (same sign as W)C - - 0 - (same sign as U)D - - + - (Big -)

How to determine the sign

T is found by seeing how the path moves through the Isotherms.

T & U always have the same sign.

Move to the right = -Move to the left = +

Up or down = 0

Find U and W first. Then use the 1st Law of Thermo to calculate Q.

Key for table on page 8: (If the processes were moving to the left and upward instead of right and downward, all the signs in the table below would be revered!)

Path T U W Q#1 - - 0 -#2 - - - 0#3 0 0 - +#4 + + - + (Big +)

How to determine the sign

T is found by seeing how the path moves through the Isotherms.

T & U always have the same sign.

Move to the right = -Move to the left = +

Up or down = 0

Find U and W first. Then use the 1st Law of Thermo to calculate Q.

Key for table on page 13: (Remember that for a cycle the and that . Q = 0 for Adiabatic processes and W = 0 when the process moved up and down on the pV diagram.)

Path T U W QAB +415 K +1035 J 0 +1035 JBC -312 K -778 J +520 J -1298 JCA -103 K -257 J -257 J 0Net 0 0 +263 J -263 J

Key to question on page 15:Was your answer 0.5 or 50%? I hope not! Remember that temperature has to be in units of Kelvin not Celsius.



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Thermodynamics13-16