 # Thermodynamic Processes Illustrate how the 1 st law of thermodynamics is a statement of energy conservation Calculate heat, work, and the change in internal.

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Thermodynamic Processes Illustrate how the 1 st law of thermodynamics is a statement of energy conservation Calculate heat, work, and the change in internal energy by applying the 1 st law of thermo Apply 1 st law of thermo to cyclic processes

How does the potential energy vary as the car rolls up and down the track? The PE depends only on the car’s elevation.

How would a 4 th bar representing ME relate to the U bars? The new ME bar should equal KE + PE in each case. Thus, it will get shorter from (b) to (e) as the U bar gets taller by the same amount.

The First Law of Thermodynamics Closed system: Δ U=Q-W U Internal energy: all the energy of the molecules Q : energy transferred to/from system as heat + for heat added; - for heat lost W: energy transferred to/from system as work + work done by the system; - work on system

Signs for heat and work in a system are summarized as follows…

1 st law of thermodynamics mathematically Δ U = U f - U i Energy conservation requires that the total change in internal energy from its initial to its final equilibrium conditions be EQUAL to the transfer of energy as BOTH HEAT and WORK. According to the 1 st law of thermo, a systems internal energy can be changed by transferring energy as either work, heat, or a combination of the two.

KNOWTHISCHARTFORTEST

Thermodynamic Processes Isothermal: delta T=0; delta U=0, Q=W Adiabatic: Q=0; delta U=-W ISOBARIC (constant P) ISOVOLUMETRIC Delta P=0 delta V=0 W=Fd=P(ad)=P delta V W=0, Q=deltaU Q=delta U+W=delta U+P delta V

CYCLIC PROCESSES A thermodynamic process in which a system returns to the same conditions under which it started. Final internal energy = initial internal energy The change in internal energy of a system is ZERO in a cyclic process  U net = 0 + Q net = W net Q net = W net

CYCLIC PROCESS A cyclic process represents an isothermal process in that all energy is transferred as work and heat. Energy is transferred as heat from the cold interior of the refrigerator to the even colder evaporating refrigerant. (Q cold or Q c ) Energy is also transferred as heat from the hot condensing refrigerant to the relatively colder air outside the refrigerator. (Q hot or Q h )

Cyclic Process Q net = Q h – Q c = W net W net = Q h – Q c Where Q h > Q c The colder you want the inside of a refrigerator to be, the greater the net energy transferred as heat (Qh – Qc) must be.

Cyclic Process * A refrigerator performs work to create a temperature difference between its closed interior and its environment. *Transfers energy from a body at low temperature to one at a high temperature. *Uses work performed by an electric motor to compress the refrigerant.

Cyclic Process In each of the 4 steps of a Refrigeration cycle, energy is transferred to or from the refrigerant either by heat or by work.

OPEN YOUR TEXTBOOK TO PAGE 414 Cyclic Process Frige

Refrigerator Cyclic Process 1: Electrically-run compressor does work on the Freon gas, increasing the pressure of the gas. High Pressure and High Temperature 2: High Pressure Freon Gas released into external heat enters exchange coil on the outside of the refrigerator 3: Heat flows from High Temperature gas to the lower-temperature air of the room surrounding the coil. This heat loss causes the Freon to condense to a liquid releasing heat to outside of the refrigerator. (space behind the frige) 4: As Freon passes through the expansion valve it expands and evaporates now at low pressure and low temperature 5: Evaporating Freon absorbs heat from inside the refrigerator. This causes the refrigerator to cool as its heat is absorbed by the evaporating Freon in the internal coils. Thus, temperature inside refrigerator is reduced 6: When all Freon changes to gas, the CYLCE REPEATS

Cyclic Process The fact that the refrigerant’s final internal energy is the same as its initial internal energy is very important! – (consistent with 1 st law of thermodynamics) This is true for all machines that use heat to do work or that do work to create temperature differences

Heat Engine Efficiency (x100%) e=W/QH =(QH-QL)/QH =1- QL/QH e<1

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