July 17, Week 7 Today: Heat Pumps and Engines, Chapter 11 Final - - PowerPoint PPT Presentation

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July 17, Week 7 Today: Heat Pumps and Engines, Chapter 11 Final - - PowerPoint PPT Presentation

July 17, Week 7 Today: Heat Pumps and Engines, Chapter 11 Final Homework #7 now available. Due Monday at 5:00PM. Thermodynamics 21st July 2014 First Law of Thermodynamics First Law of Thermodynamics: W + Q = E th Thermodynamics 21st July


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July 17, Week 7

Thermodynamics 21st July 2014

Today: Heat Pumps and Engines, Chapter 11 Final Homework #7 now available. Due Monday at 5:00PM.

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First Law of Thermodynamics

Thermodynamics 21st July 2014

First Law of Thermodynamics: W + Q = ∆Eth

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First Law of Thermodynamics

Thermodynamics 21st July 2014

First Law of Thermodynamics: W + Q = ∆Eth There are two ways to change the thermal energy of on object - Work being done to the object (W) and heat (Q)

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First Law Signs

Thermodynamics 21st July 2014

In applying the first law of thermodynamics, we have to think about the “system” = the object that is of interest. Everything else is called the environment

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First Law Signs

Thermodynamics 21st July 2014

In applying the first law of thermodynamics, we have to think about the “system” = the object that is of interest. Everything else is called the environment

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T Steam is used to spin a turbine. (Assume the turbine’s temperature remains constant)

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T Steam is used to spin a turbine. (Assume the turbine’s temperature remains constant) − − −

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T Steam is used to spin a turbine. (Assume the turbine’s temperature remains constant) − − − An expanding gas inflates a balloon

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T Steam is used to spin a turbine. (Assume the turbine’s temperature remains constant) − − − An expanding gas inflates a balloon − ? ? ? Heat depends on how the gas expands

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T Steam is used to spin a turbine. (Assume the turbine’s temperature remains constant) − − − An expanding gas inflates a balloon − ? ? ? Heat depends on how the gas expands After 30 minutes of baking, a pan is removed from the oven and sits on a counter

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First-Law Followup

Thermodynamics 21st July 2014

Process W Q ∆Eth ∆T Steam is used to spin a turbine. (Assume the turbine’s temperature remains constant) − − − An expanding gas inflates a balloon − ? ? ? Heat depends on how the gas expands After 30 minutes of baking, a pan is removed from the oven and sits on a counter − − −

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth The system is the engine

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth The system is the engine W = −Wout

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth The system is the engine W = −Wout Q = QH − QC

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Heat Engine

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth The system is the engine W = −Wout Q = QH − QC ∆Eth = 0 (In a perfect situation, the engine simply uses heat to do work. It doesn’t absorb the heat itself.)

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth −Wout + QH − QC = 0

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth −Wout + QH − QC = 0 Wout = QH − QC

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth −Wout + QH − QC = 0 Wout = QH − QC Better Form: QH = Wout + QC

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth −Wout + QH − QC = 0 Wout = QH − QC Better Form: QH = Wout + QC Efficiency: e = Wout QH

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth −Wout + QH − QC = 0 Wout = QH − QC Better Form: QH = Wout + QC Efficiency: e = Wout QH e = QH − QC QH = 1 − QC QH

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J (a) 0%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J (a) 0% (b) 25%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J (a) 0% (b) 25% (c) 50%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J (a) 0% (b) 25% (c) 50% (d) 75%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J (a) 0% (b) 25% (c) 50% (d) 75% (e) 100%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD 100 J 75 J (a) 0% (b) 25% (c) 50% (d) 75% (e) 100%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD QH = 100 J QC = 75 J Wout (a) 0% (b) 25% (c) 50% (d) 75% (e) 100%

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Heat Engine Exercise

Thermodynamics 21st July 2014

What is the efficiency (Wout/QH) of the following heat engine? HOT COLD QH = 100 J QC = 75 J Wout = 25 J (a) 0% (b) 25% (c) 50% (d) 75% (e) 100%

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Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot

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Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot

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SLIDE 37

Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot W + Q = ∆Eth = 0

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Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot W + Q = ∆Eth = 0 W = Win

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Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot W + Q = ∆Eth = 0 W = Win Q = QC − QH

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Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot W + Q = ∆Eth = 0 W = Win Q = QC − QH Win + QC − QH = 0 ⇒ Win + QC = QH

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Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot W + Q = ∆Eth = 0 W = Win Q = QC − QH Win + QC − QH = 0 ⇒ Win + QC = QH Coefficient of Performance: COP = QC Win = QC QH − QC

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J (a) QC = 10 J, QH = 30 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J (a) QC = 10 J, QH = 30 J (b) QC = 10 J, QH = 10 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J (a) QC = 10 J, QH = 30 J (b) QC = 10 J, QH = 10 J (c) QC = 20 J, QH = 20 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J (a) QC = 10 J, QH = 30 J (b) QC = 10 J, QH = 10 J (c) QC = 20 J, QH = 20 J (d) QC = 40 J, QH = 60 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J (a) QC = 10 J, QH = 30 J (b) QC = 10 J, QH = 10 J (c) QC = 20 J, QH = 20 J (d) QC = 40 J, QH = 60 J (e) QC = 40 J, QH = 20 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD 20 J (a) QC = 10 J, QH = 30 J (b) QC = 10 J, QH = 10 J (c) QC = 20 J, QH = 20 J (d) QC = 40 J, QH = 60 J (e) QC = 40 J, QH = 20 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD Win = 20 J (d) QC = 40 J, QH = 60 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD Win = 20 J COP = QC/Win ⇒ QC = COP × Win ⇒ QC = 2 × 20 J (d) QC = 40 J, QH = 60 J

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Heat Pump Exercise

Thermodynamics 21st July 2014

If the COP (QC/Win) of the following heat pump is 2, how much heat was removed from the cold reservoir and how much heat was released to the hot reservoir? HOT COLD Win = 20 J COP = QC/Win ⇒ QC = COP × Win ⇒ QC = 2 × 20 J 40 J QH (d) QC = 40 J, QH = 60 J

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Entropy

Thermodynamics 21st July 2014

Heat engine have a maximum efficiency (that is MUCH less than 100%) because of the second law of thermodynamics

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Entropy

Thermodynamics 21st July 2014

Heat engine have a maximum efficiency (that is MUCH less than 100%) because of the second law of thermodynamics The second law is based on the concept of entropy

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Entropy

Thermodynamics 21st July 2014

Heat engine have a maximum efficiency (that is MUCH less than 100%) because of the second law of thermodynamics The second law is based on the concept of entropy Entropy - Measures the amount of disorder in a system.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (a) A gas is allowed to escape from its container.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (a) A gas is allowed to escape from its container. (b) A college student throws his clothes on the room’s floor looking for the perfect outfit.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (a) A gas is allowed to escape from its container. (b) A college student throws his clothes on the room’s floor looking for the perfect outfit. (c) Your physics instructor spills coffee on himself.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (a) A gas is allowed to escape from its container. (b) A college student throws his clothes on the room’s floor looking for the perfect outfit. (c) Your physics instructor spills coffee on himself. (d) A child neatly stacks her blocks to make a tower.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (a) A gas is allowed to escape from its container. (b) A college student throws his clothes on the room’s floor looking for the perfect outfit. (c) Your physics instructor spills coffee on himself. (d) A child neatly stacks her blocks to make a tower. (e) A piece of food warms while sitting under a heat lamp.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (a) A gas is allowed to escape from its container. (b) A college student throws his clothes on the room’s floor looking for the perfect outfit. (c) Your physics instructor spills coffee on himself. (d) A child neatly stacks her blocks to make a tower. (e) A piece of food warms while sitting under a heat lamp.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (d) A child neatly stacks her blocks to make a tower. A more ordered system, like stacked blocks, has less entropy.

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Entropy Exercise

Thermodynamics 21st July 2014

Which of the following processes would involve a decrease in entropy? In each case the system has been underlined. (d) A child neatly stacks her blocks to make a tower. A more ordered system, like stacked blocks, has less entropy. Disordered systems are more likely to occur, so entropy is also a measure of the probability that a state of a system will occur.

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The Second Law of Thermodynamics

Thermodynamics 21st July 2014

The Second Law of Thermodynamics - The entropy of an isolated system never decreases. It increases until the system reaches equilibrium and then stays constant.

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The Second Law of Thermodynamics

Thermodynamics 21st July 2014

The Second Law of Thermodynamics - The entropy of an isolated system never decreases. It increases until the system reaches equilibrium and then stays constant. The second law does NOT imply that the entropy never decreases. The system must be isolated. If there is an external force doing work on the system, its entropy can decrease.

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The Second Law of Thermodynamics

Thermodynamics 21st July 2014

The Second Law of Thermodynamics - The entropy of an isolated system never decreases. It increases until the system reaches equilibrium and then stays constant. The second law does NOT imply that the entropy never decreases. The system must be isolated. If there is an external force doing work on the system, its entropy can decrease. On the macroscopic scale, any process that obeys the second law will be irreversible. (Irreversible processes increase entropy.)

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics?

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (a) A freshly-baked pie sits on the counter and cools.

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (a) A freshly-baked pie sits on the counter and cools. (b) The wind gathers up the leaves in your front yard and blows them into a neat pile.

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (a) A freshly-baked pie sits on the counter and cools. (b) The wind gathers up the leaves in your front yard and blows them into a neat pile. (c) Liquid water freezes to form ice.

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (a) A freshly-baked pie sits on the counter and cools. (b) The wind gathers up the leaves in your front yard and blows them into a neat pile. (c) Liquid water freezes to form ice. (d) A man piles logs into a neat pyramid.

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (a) A freshly-baked pie sits on the counter and cools. (b) The wind gathers up the leaves in your front yard and blows them into a neat pile. (c) Liquid water freezes to form ice. (d) A man piles logs into a neat pyramid. (e) An acorn grows into a very tall tree.

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (a) A freshly-baked pie sits on the counter and cools. (b) The wind gathers up the leaves in your front yard and blows them into a neat pile. (c) Liquid water freezes to form ice. (d) A man piles logs into a neat pyramid. (e) An acorn grows into a very tall tree.

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Second Law Exercise

Thermodynamics 21st July 2014

Which of the following processes violates the second law of thermodynamics? (b) The wind gathers up the leaves in your front yard and blows them into a neat pile. This is the “reverse” of an everyday event. The wind can scatter the leaves from a neat pile, but it cannot gather them.

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Heat Engine II

Thermodynamics 21st July 2014

Heat Engine - Device that uses the transfer of heat from a higher temperature to lower temperature to extract work W + Q = ∆Eth −Wout + QH − QC = 0 Wout = QH − QC Better Form: QH = Wout + QC Efficiency: e = Wout QH e = QH − QC QH = 1 − QC QH

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SLIDE 76

Heat Pump

Thermodynamics 21st July 2014

Heat Pump - Device that does work in order to move heat from cold to hot W + Q = ∆Eth = 0 W = Win Q = QC − QH Win + QC − QH = 0 ⇒ Win + QC = QH Coefficient of Performance: COP = QC Win = QC QH − QC

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Maximum Efficiencies

Thermodynamics 21st July 2014

Using the mathematical version of the second law, we can find equation for the maximum efficiency of heat engine and COP of a heat pump.

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Maximum Efficiencies

Thermodynamics 21st July 2014

Using the mathematical version of the second law, we can find equation for the maximum efficiency of heat engine and COP of a heat pump.

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Maximum Efficiencies

Thermodynamics 21st July 2014

Using the mathematical version of the second law, we can find equation for the maximum efficiency of heat engine and COP of a heat pump. emax = 1 − TC TH

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Maximum Efficiencies

Thermodynamics 21st July 2014

Using the mathematical version of the second law, we can find equation for the maximum efficiency of heat engine and COP of a heat pump. emax = 1 − TC TH

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Maximum Efficiencies

Thermodynamics 21st July 2014

Using the mathematical version of the second law, we can find equation for the maximum efficiency of heat engine and COP of a heat pump. emax = 1 − TC TH COPmax = TC TH − TC