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Physics for Scientists and Engineers, 6e. Chapter 22 – Heat Engines, Entropy, and the Second Law of Thermodynamics. The energy input to an engine is 3.00 times greater than the work it performs. What is its thermal efficiency? . 3.00 1.00 0.333 impossible to determine.

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## Physics for Scientists and Engineers, 6e

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**Physics for Scientists and Engineers, 6e**Chapter 22 – Heat Engines, Entropy, and the Second Law of Thermodynamics**The energy input to an engine is 3.00 times greater than**the work it performs. What is its thermal efficiency? • 3.00 • 1.00 • 0.333 • impossible to determine**The energy input to an engine is 3.00 times greater than**the work it performs. What fraction of the energy input is expelled to the cold reservoir? • 0.333 • 0.667 • 1.00 • impossible to determine**The work represents one third of the input energy. The**remainder, two thirds, must be expelled to the cold reservoir.**The energy entering an electric heater by electrical**transmission can be converted to internal energy with an efficiency of 100%. By what factor does the cost of heating your home change when you replace your electric heating system with an electric heat pump that has a COP of 4.00? Assume that the motor running the heat pump is 100% efficient. • 4.00 • 2.00 • 0.500 • 0.250**The COP of 4.00 for the heat pump means that you are**receiving four times as much energy as the energy entering by electrical transmission. With four times as much energy per unit of energy from electricity, you need only one fourth as much electricity.**Three engines operate between reservoirs separated in**temperature by 300 K. The reservoir temperatures are as follows: Engine A: Th= 1 000 K, Tc= 700 K; Engine B: Th= 800 K, Tc= 500 K; Engine C: Th= 600 K, Tc= 300 K. Rank the engines in order of theoretically possible efficiency, from highest to lowest. • A, B, C • A, C, B • B, C, A • B, A, C • C, A, B • C, B, A**C, B, A. Although all three engines operate over a 300-K**temperature difference, the efficiency depends on the ratio of temperatures, not the difference.**Suppose that you select four cards at random from a**standard deck of playing cards and end up with a macrostate of four deuces. How many microstates are associated with this macrostate? • 1 • 2 • 3 • 4**Suppose you pick up two cards at random from a standard**deck of playing cards and end up with a macrostate of two aces. How many microstates are associated with this macrostate? • 3 • 4 • 5 • 6**Six microstates – club-diamond, club-heart, club-spade,**diamond-heart, diamond-spade, heart-spade. The macrostate of two aces is more probable than that of four deuces in Quick Quiz 22.5 because there are six times as many microstates for this particular macrostate compared to the macrostate of four deuces. Thus, in a hand of poker, two of a kind is less valuable than four of a kind.**Which of the following is true for the entropy change of a**system that undergoes a reversible, adiabatic process? • ΔS < 0 • ΔS = 0 • ΔS > 0**Because the process is reversible and adiabatic, Qr = 0;**therefore, ΔS = 0.**An ideal gas is taken from an initial temperature Tito a**higher final temperature Tfalong two different reversible paths: Path A is at constant pressure; Path B is at constant volume. The relation between the entropy changes of the gas for these paths is • ΔSA > ΔSB • ΔSA = ΔSB • ΔSA < ΔSB**From the first law of thermodynamics, for these two**reversible processes, Qr = ΔEint – W. During the constant-volume process, W = 0, while the work W is nonzero and negative during the constant-pressure expansion. Thus, Qr is larger for the constant-pressure process, leading to a larger value for the change in entropy. In terms of entropy as disorder, during the constant-pressure process, the gas must expand. The increase in volume results in more ways of locating the molecules of the gas in a container, resulting in a larger increase in entropy.**The entropy change in an adiabatic process must be zero**because Q = 0. • True • False**The determining factor for the entropy change is Qr, not Q.**If the adiabatic process is not reversible, the entropy change is not necessarily zero because a reversible path between the same initial and final states may involve energy transfer by heat.

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