Lesson 6First Law of ThermodynamicsLiceo Alfano I

ReminderLast time: Specific heat, internal energyQ = U = mcTThis time: First law of thermodynamics

Energy TransferWhat kind of energy transfer have we been talking about?HeatThe other kind of energy transfer is work. What is work?Work = Force x DistanceEx: pushing a shopping cart forward, gravity pulling a ball down, etc.Remember: WORK IS ENERGY!

First LawThe First Law of Thermodynamics is: Energy cannot be created or destroyed, only changed in formYouve probably heard this before as the principle of energy conservation, but the first law adds a new part: heat

First LawChange in internal energy is defined as the heat transferred into the system minus the net work done by the systemIf Q is positive, heat is transferred into the system. If W is positive, there is net work done by the systemA positive Q adds energy to the system, a positive W takes energy outU = Q - WChange in internal energyHeat added TO the systemWork done BY the system

Concept QuestionIf I do work to a system, then W isPositiveNegative

Concept QuestionIf I do work to a system, then W isPositiveNegative

Concept QuestionIf I add heat to a system, then Q isPositiveNegative

Concept QuestionIf I add heat to a system, then Q isPositiveNegative

Concept QuestionIf a system releases heat energy, then Q isPositiveNegative

Concept QuestionIf a system releases heat energy, then Q isPositiveNegative

Concept QuestionIf I add 300kJ of heat to a system, then do 400kJ of work on it, then U isPositiveNegative

Concept QuestionIf I add 300kJ of heat to a system, then do 400kJ of work on it, then U isPositiveNegative

Concept QuestionIf I do 100kJ of work on a system, then the system releases 300kJ of heat, then U isPositiveNegative

Concept QuestionIf I do 100kJ of work on a system, then the system releases 300kJ of heat, then U isPositiveNegative

Human BodyBody temperature is kept constant by transferring heat to our surroundings so Q is negativeWe generally do work on our surroundings, so W is positiveThis means U is negative, so we are constantly losing energy to our surroundings

Human BodyHowever, when we eat, we add chemical potential energy to our bodiesIf we eat perfectly, U is zeroall that we consume is used to run our bodiesIf we eat a lot, U is positive, so our bodies store the extra energy as fatIf U is negative, the body uses the fat to release heat and do workthat is how we lose weight

VideoHeat and work

Practice ProblemA 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 WorkLets think back to ideal gasesThink about the container to the right as a system.If I push the piston in, is W positive or negative?

Ideal Gas WorkLets think about how much work I did. Remember, |W|=F*dWhat is the force?F=P*AWhat is the work?W=P*A*x=PVDx

PV DiagramsWhen 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:IsothermalAdiabaticIsobaricIsochoric

IsothermalIsoTHERMal = constant temperatureWhat is our gas law when we have constant temperature?PV = constantAs volume decreases, pressure increases, as volume increases, pressure decreases

AdiabaticAdiabatic = no heat in/out, so Q=0This happens with good insulation or when a process happens so quickly that heat cannot flow in or outIf Q=0, what is U?U=-W

IsobaricIsoBARic = constant pressure (bar is a measure of pressure)Straight line on PV diagramWork can be calculated as W=PV (as we saw before)

IsochoricIsochoric = constant volume (chorus takes up a lot of room?)Vertical line on a PV diagramNo work done because volume doesnt change, so W=0

CyclesHere we have a PV diagram of a cycleA cycle is simply a process that ends where it startsSince the cycle ends where it begins, U=0How can we find the work done on this graph?

VideosOtto cycle2-stroke engineSteam engine

Practice ProblemsSketch 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 ProblemsThe 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?