Heat engines

zUse heat to do work in a cyclic process: same process is repeated over and over again

Reservoirs zKeep the same temperature zSupply the working substance with energy so that its temperature remains or becomes equal to T reservoir zAll processes are “slow” and reversible from the reservoir’s point of view

The Carnot cycle zConsists of four slow steps: z1  2: isothermal z2  3: adiabatic z3  4: isothermal z4  1: adiabatic

The first step   U=0 as  T=0 (nearly true for non-ideal gas) z   S total =0, hence this step is reversible

Four questions  In the second step, Q, W,  U,  S are a) positive b) zero c) negative d) dependent on the gas used

Comments on the Carnot cycle zIt is customary to define Q C as positive even though heat flows out of the system zHeat loss to cold reservoir necessary to bring system back to original state   S cycle =0, hence   S hot reservoir +  S cold reservoir = 0

Work and efficiency zWork done during cycle: W = Q H – Q C z zThis is independent of working substance!

Reversibility zThere is no net change in entropy: the entropy of gas and environment are unchanged zHence the entire Carnot cycle is reversible, and it can be run as a “fridge” z“Reversible” means: we can reverse the cycle and restore the original state without any residual changes outside the system

Maximum efficiency zImagine we have some heat engine converting heat into work, and a Carnot engine running backwards zThe Carnot engine takes heat Q C from the cold reservoir, work W is done on the gas and the engine gives Q H to the hot reservoir

Maximum efficiency II   other =  Carnot : no net work done, no net heat transfer from either reservoir   other >  Carnot : Say the other engine delivers Q C to cold reservoir. Then it takes in Q H + , which is used to do work W + . However: This violates second law of thermodynamics

Carnot engine is most efficient  If there was a more efficient engine, net work  could be done by extracting heat from the hot reservoir without any other changes anywhere. zThus disorderly thermal motion would be converted into orderly mechanical motion zEntropy would decrease, not allowed by 2 nd law

The Otto cycle zFour steps: z1  2: slow adiabatic z2  3: fast isochoric z3  4: slow adiabatic z4  1: fast isochoric

Temperature and heat zThe temperature T 2 (and T 4 ) can be evaluated: zHeat added: 4  1: 2  3:

Entropy in the Otto cycle zStep 4  1:  gas: fast process, calculate  S as if it were slow yreservoir: slow process, calculate as usual zHence

Entropy in the Otto cycle II zLikewise in step 2  3: zWe conclude for the entire cycle zSince zThe cycle is thus not reversible

PS225 – Thermal Physics topics zThe atomic hypothesisThe atomic hypothesis zHeat and heat transferHeat and heat transfer zKinetic theoryKinetic theory zThe Boltzmann factorThe Boltzmann factor zThe First Law of ThermodynamicsThe First Law of Thermodynamics zSpecific HeatSpecific Heat zEntropyEntropy zHeat enginesHeat engines zPhase transitionsPhase transitions