Lesson 6 First Law of Thermodynamics Liceo Alfano I.

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Transcript of Lesson 6 First Law of Thermodynamics Liceo Alfano I.
Lesson 6
First Law of Thermodynamics
Liceo Alfano I
Reminder• Last time: Specific heat, internal energy
• Q = ΔU = mcΔT
• This time: First law of thermodynamics
Energy Transfer• What kind of energy transfer have we been talking
about?
• Heat
• The other kind of energy transfer is work. What is work?
• Work = Force x Distance
• Ex: pushing a shopping cart forward, gravity pulling a ball down, etc.
• Remember: WORK IS ENERGY!
First Law• The First Law of Thermodynamics is:
Energy cannot be created or destroyed, only changed in form
• You’ve probably heard this before as the principle of energy conservation, but the first law adds a new part: heat
First Law
• Change in internal energy is defined as the heat transferred into the system minus the net work done by the system
• If Q is positive, heat is transferred into the system. If W is positive, there is net work done by the system
• A positive Q adds energy to the system, a positive W takes energy out
ΔU = Q  WChange in internal energy
Heat added TO the system
Work done BY the system
Concept QuestionIf I do work to a system, then W is
1)Positive
2)Negative
Concept QuestionIf I do work to a system, then W is
1)Positive
2)Negative
Concept QuestionIf I add heat to a system, then Q is
1)Positive
2)Negative
Concept QuestionIf I add heat to a system, then Q is
1)Positive
2)Negative
Concept QuestionIf a system releases heat energy, then Q is
1)Positive
2)Negative
Concept QuestionIf a system releases heat energy, then Q is
1)Positive
2)Negative
Concept QuestionIf I add 300kJ of heat to a system, then do 400kJ of work on it, then ΔU is
1)Positive
2)Negative
Concept QuestionIf I add 300kJ of heat to a system, then do 400kJ of work on it, then ΔU is
1)Positive
2)Negative
Concept QuestionIf I do 100kJ of work on a system, then the system releases 300kJ of heat, then ΔU is
1)Positive
2)Negative
Concept QuestionIf I do 100kJ of work on a system, then the system releases 300kJ of heat, then ΔU is
1)Positive
2)Negative
Human Body• Body temperature is kept constant by
transferring heat to our surroundings so Q is negative
• We generally do work on our surroundings, so W is positive
• This means ΔU is negative, so we are constantly losing energy to our surroundings
Human Body• However, when we eat, we add chemical
potential energy to our bodies
• If we eat perfectly, ΔU is zero—all that we consume is used to run our bodies
• If we eat a lot, ΔU is positive, so our bodies store the extra energy as fat
• If ΔU is negative, the body uses the fat to release heat and do work—that is how we lose weight
VideoHeat and work
Practice Problem• A system receives 1600J of heat. At the
same time 800J of work are done on the system by outside forces. Calculate the change in internal energy of the system.
Ideal Gas Work• Let’s think back to
ideal gases
• Think about the container to the right as a system.
• If I push the piston in, is W positive or negative?
Ideal Gas Work• Let’s think about how much
work I did. Remember, W=F*d
• What is the force?
• F=P*A
• What is the work?
• W=P*A*Δx=PΔV
x
PV Diagrams• When we talk about the first law with ideal gases,
we often use a PV diagram. How can we find work done using a PV diagram?
• A PV diagram can show some special processes:
• Isothermal
• Adiabatic
• Isobaric
• Isochoric
Isothermal• IsoTHERMal = constant
temperature
• What is our gas law when we have constant temperature?
• PV = constant
• As volume decreases, pressure increases, as volume increases, pressure decreases
Adiabatic• Adiabatic = no heat
in/out, so Q=0
• This happens with good insulation or when a process happens so quickly that heat cannot flow in or out
• If Q=0, what is ΔU?
• ΔU=W
Isobaric
• IsoBARic = constant pressure (bar is a measure of pressure)
• Straight line on PV diagram
• Work can be calculated as W=PΔV (as we saw before)
Isochoric• Isochoric = constant
volume (chorus takes up a lot of room?)
• Vertical line on a PV diagram
• No work done because volume doesn’t change, so W=0
Cycles• Here we have a PV
diagram of a cycle
• A cycle is simply a process that ends where it starts
• Since the cycle ends where it begins, ΔU=0
• How can we find the work done on this graph?
Videos• Otto cycle
• 2stroke engine
• Steam engine
Practice Problems• Sketch a PV diagram of this cycle: 2L of an
ideal gas at 105 Pa are cooled at constant pressure to a volume of 1L, and then expanded isothermally back to 2L, then the pressure is increased at constant volume until 105 Pa.
• 1L of air at 105 Pa is heated at constant pressure until its volume is 2L, then it is compressed isothermally back to 1L, then the pressure is decreased at constant volume back to 105 Pa. Sketch the PV diagram.
Practice Problems
The internal energy at point A is 800J. The work transfer from B to C is 300J, and from C to A is 100J. What is the total work done by the gas in this cycle? What is the final internal energy?