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 - W Change in internal energy Heat added TO the system Work done BY the system

Concept Question If I do work to a system, then W is 1) Positive 2) Negative

Concept Question If I do work to a system, then W is 1) Positive 2) Negative

Concept Question If I add heat to a system, then Q is 1) Positive 2) Negative

Concept Question If I add heat to a system, then Q is 1) Positive 2) Negative

Concept Question If a system releases heat energy, then Q is 1) Positive 2) Negative

Concept Question If a system releases heat energy, then Q is 1) Positive 2) Negative

Concept Question If I add 300kJ of heat to a system, then do 400kJ of work on it, then ΔU is 1) Positive 2) Negative

Concept Question If I add 300kJ of heat to a system, then do 400kJ of work on it, then ΔU is 1) Positive 2) Negative

Concept Question If I do 100kJ of work on a system, then the system releases 300kJ of heat, then ΔU is 1) Positive 2) Negative

Concept Question If 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

Video Heat 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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 2-stroke engine Steam engine

Practice Problems Sketch a PV diagram of this cycle: 2L of an ideal gas at 10 5 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 10 5 Pa. 1L of air at 10 5 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 10 5 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?