Ch_20_Study_Problem_Solutions - Ch 20 Study Problem...

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1 Ch. 20 Study Problem Solutions 20.1 A pot is half-filled with water, and a lid is placed on the pot, forming a tight seal so that no water vapor can escape. The pot is heated on a stove, forming water vapor inside the pot. The heat is then turned off and the water vapor condenses back to liquid. Is this cycle reversible or irreversible? As long as this cycle is carried out slowly enough that the temperature of the water and the vapor are always at nearly the same uniform temperature, the cycle involving the liquid water and vapor is (nearly) reversible. The reason is that the cycle can be reversed at any point of the warming or cooling parts of the cycle by slightly reducing or increasing the heat input to the pot, respectively. 20. 2 Suppose you want to increase the efficiency of a Carnot heat engine. Would it be better to increase T H or to decrease T C by an equal amount? Why? Let T C = 300 K, T H = 600 K and ± T = 100 K. The initial maximum (ideal) efficiency (the Carnot efficiency) of a reversible heat engine operating between two temperature reservoirs is e initial = 1 ² T C T H = 1 ± 300 600 = 0.500 = 50.0% . If T H is increased by the amount ± T, then e = 1 ² T C T H + ± T = 1 ± 300 600 + 100 = 0.571 = 57.1% . (1) On the other hand, if T C is decreased by ± T, then e = 1 ² T C ² ± T T H = 1 ± 300 ± 100 600 = 0.667 = 66.7% . (2) Thus it is clearly better to decrease T C than to increase T H by an equal amount. 20.3 Real heat engines, like the gasoline engine in a car, always have some friction between their moving parts, although lubricants keep the friction to a minimum. Would a heat engine with completely frictionless parts be 100% efficient? Why or why not? Does the answer depend on whether or not the engine runs on the Carnot cycle? Again, why or why not? The definition of the thermal efficiency of a heat engine is e = W |Q H | = 1 ² |Q C | |Q H | .
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2 In order for e = 1, one would need the thermal energy U contained in the exhaust gas to be U = Q C = 0. The thermal energy of an ideal gas is, apart from a possible additive constant, given by U = n C V T where T is the absolute temperature in Kelvins. Thus in order for Q C to be zero, the temperature of the cold reservoir T C = 0 K. It is not possible theoretically for the cold reservoir to be at exactly zero Kelvin. This discussion applies to any type of cycle that the engine runs on including the Carnot cycle. Thus it is impossible for any heat engine to be exactly 100% efficient.
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Ch_20_Study_Problem_Solutions - Ch 20 Study Problem...

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