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Ch. 20 - Second Law of Thermodynamics
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 20, Problem 90

Rank the following five-card hands in order of increasing probability: (a) four aces and a king; (b) six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades; (c) two jacks, two queens, and an ace; and (d) any hand having no two equal-value cards (no pairs, etc.). Discuss your ranking in terms of microstates and macrostates.

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Understand the problem: We are tasked with ranking the given five-card hands in order of increasing probability. To do this, we need to calculate the probability of each hand by considering the number of microstates (specific arrangements of cards) and macrostates (general categories of hands). The probability of a hand is proportional to the number of microstates that correspond to it.
Step 1: Analyze hand (a) 'Four aces and a king': This is a very specific hand. There is only one way to have all four aces and one king in a standard 52-card deck. The number of microstates for this hand is extremely small, making it one of the rarest hands.
Step 2: Analyze hand (b) 'Six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades': This is a completely random hand with no pairs, no special combinations, and no repeated values. The number of microstates for this hand is relatively large because there are many ways to select five cards with no specific restrictions other than avoiding pairs or patterns.
Step 3: Analyze hand (c) 'Two jacks, two queens, and an ace': This is a full house, which is a specific type of hand. The number of microstates for this hand is larger than hand (a) but smaller than hand (b). To calculate the number of microstates, consider the combinations of choosing two jacks, two queens, and one ace from the deck.
Step 4: Analyze hand (d) 'Any hand having no two equal-value cards (no pairs, etc.)': This is a general category of hands that includes many possible combinations. The number of microstates for this hand is very large because it encompasses all hands that do not have pairs, three-of-a-kind, or other specific patterns. This makes it the most probable hand among the options.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Microstates

Microstates refer to the specific arrangements of a system's components that result in a particular macrostate. In the context of card hands, each unique combination of cards represents a different microstate. Understanding microstates is crucial for calculating the probability of different hands, as it allows us to count how many ways a specific hand can be formed.
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Microstates and Macrostates of a System

Macrostates

Macrostates are the overall states of a system characterized by macroscopic properties, which in this case are the types of five-card hands. Each macrostate can correspond to multiple microstates. For example, the macrostate of having 'no pairs' encompasses all hands that meet this criterion, making it essential to differentiate between macrostates when ranking the probabilities of various hands.
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Probability in Combinatorics

Probability in combinatorics involves calculating the likelihood of specific outcomes based on the total number of possible arrangements. In this scenario, we need to determine the number of ways to form each type of hand and compare these counts to assess their probabilities. This concept is fundamental for ranking the hands, as it provides a quantitative basis for understanding which hands are more or less likely to occur.
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Probability Distribution Graph
Related Practice
Textbook Question

A dehumidifier removes water vapor from air and has been referred to as a “refrigerator with an open door.” The humid air is pulled in by a fan and passes over a cold coil, whose temperature is less than the dew point, and some of the air’s water condenses. After this water is extracted, the air is warmed back to its original temperature and sent into the room. In a well-designed dehumidifier, the heat that is removed by the cooling coil mostly comes from the condensation of water vapor to liquid, and this heat is used to re-warm the air. Estimate how much water is removed in 1.0 h by an ideal dehumidifier, if the temperature of the room is 25°C, the water condenses at 8°C, and the dehumidifier does work at the rate of 550 W of electrical power.

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Textbook Question

A bowl contains many red, orange, and green jelly beans, in equal numbers. You are to make a line of 3 jelly beans by randomly taking 3 beans from the bowl. Construct a table showing the number of microstates that correspond to each macrostate.

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Textbook Question

Trees can help a bit to offset the buildup of CO₂ due to burning coal and other fossil fuels. CO₂ can be absorbed by tree foliage. Trees use the carbon to grow, and release O₂ into the atmosphere. Suppose a refrigerator uses 600 kWh of electricity per year (about 2 x 10⁹ J) from a 33% efficient coal-fired power plant. Burning 1 kg of coal releases about 2 x 10⁷ J of energy. Assume coal is all carbon, which when burned in air becomes CO₂.

(a) How much coal is burned per year to run this refrigerator?

(b) Assuming a forest can capture 1700 kg of carbon per hectare ( = 10, 000 m²) per year, estimate how many square meters of forest are needed to capture the carbon (in the form now of CO₂) emitted by the refrigerator in (a).

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Textbook Question

A bowl contains many red, orange, and green jelly beans, in equal numbers. You are to make a line of 3 jelly beans by randomly taking 3 beans from the bowl. Construct a table showing the number of microstates that correspond to each macrostate. Then, determine the probability of all 3 beans red.

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