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 12.1 Heat and Temperature 12.2 The Kinetic Theory of Gases & the Ideal Gas Law 12.3 The Laws of Thermodynamics 12.4 Heat Engines

 12.5 Key Formulas 12.6 Practice Questions 12.7 Explanations
Heat Engines
A heat engine is a machine that converts heat into work. Heat engines are important not only because they come up on SAT II Physics, but also because a large number of the machines we use—most notably our cars—employ heat engines.
A heat engine operates by taking heat from a hot place, converting some of that heat into work, and dumping the rest in a cooler heat reservoir. For example, the engine of a car generates heat by combusting gasoline. Some of that heat drives pistons that make the car do work on the road, and some of that heat is dumped out the exhaust pipe.
Assume that a heat engine starts with a certain internal energy U, intakes heat from a heat source at temperature , does work , and exhausts heat into a the cooler heat reservoir with temperature . With a typical heat engine, we only want to use the heat intake, not the internal energy of the engine, to do work, so . The First Law of Thermodynamics tells us:
To determine how effectively an engine turns heat into work, we define the efficiency, e, as the ratio of work done to heat input:
Because the engine is doing work, we know that > 0, so we can conclude that > . Both and are positive, so the efficiency is always between 0 and 1:
Efficiency is usually expressed as a percentage rather than in decimal form. That the efficiency of a heat engine can never be 100% is a consequence of the Second Law of Thermodynamics. If there were a 100% efficient machine, it would be possible to create perpetual motion: a machine could do work upon itself without ever slowing down.
Example
 80 J of heat are injected into a heat engine, causing it to do work. The engine then exhausts 20 J of heat into a cool reservoir. What is the efficiency of the engine?
If we know our formulas, this problem is easy. The heat into the system is = 80 J, and the heat out of the system is = 20 J. The efficiency, then, is: 1 – 20 ⁄ 80 = 0.75 = 75%.
 Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II PhysicsStrategies for Taking SAT II PhysicsVectorsKinematicsDynamicsWork, Energy, and PowerSpecial Problems in MechanicsLinear MomentumRotational MotionCircular Motion and GravitationThermal PhysicsElectric Forces, Fields, and PotentialDC CircuitsMagnetismElectromagnetic InductionWavesOpticsModern PhysicsPhysics GlossaryPractice Tests Are Your Best Friends
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