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Ch 21: Heat Engines and Refrigerators
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 21: Heat Engines and RefrigeratorsProblem 46a
Chapter 21, Problem 46a

FIGURE P21.46 shows a Carnot heat engine driving a Carnot refrigerator. Determine Q2, Q3 and Q4.

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Step 1: Understand the setup of the problem. A Carnot heat engine and a Carnot refrigerator are connected. The heat engine absorbs heat \( Q_1 \) from a high-temperature reservoir at temperature \( T_1 \), does work \( W \), and rejects heat \( Q_2 \) to a lower-temperature reservoir at \( T_2 \). The refrigerator absorbs heat \( Q_3 \) from a low-temperature reservoir at \( T_3 \) and rejects heat \( Q_4 \) to a higher-temperature reservoir at \( T_4 \).
Step 2: Use the efficiency formula for the Carnot heat engine. The efficiency \( \eta \) of a Carnot engine is given by \( \eta = 1 - \frac{T_2}{T_1} \). The work done by the engine is related to the heat absorbed and rejected by \( W = Q_1 - Q_2 \). Rearrange these equations to express \( Q_2 \) in terms of \( Q_1 \), \( T_1 \), and \( T_2 \): \( Q_2 = Q_1 \cdot \frac{T_2}{T_1} \).
Step 3: Use the coefficient of performance (COP) for the Carnot refrigerator. The COP for a refrigerator is given by \( \text{COP} = \frac{Q_3}{W} = \frac{T_3}{T_4 - T_3} \). Rearrange this equation to express \( Q_3 \) in terms of \( W \), \( T_3 \), and \( T_4 \): \( Q_3 = W \cdot \frac{T_3}{T_4 - T_3} \).
Step 4: Relate the work \( W \) done by the heat engine to the work required by the refrigerator. Since the heat engine drives the refrigerator, the work output of the engine is equal to the work input to the refrigerator: \( W = Q_1 - Q_2 = Q_4 - Q_3 \). Use this relationship to express \( Q_4 \) in terms of \( Q_3 \), \( Q_1 \), and \( Q_2 \): \( Q_4 = Q_3 + (Q_1 - Q_2) \).
Step 5: Combine the equations from Steps 2, 3, and 4 to solve for \( Q_2 \), \( Q_3 \), and \( Q_4 \) in terms of the given temperatures \( T_1, T_2, T_3, T_4 \) and the heat input \( Q_1 \). Substitute \( Q_2 = Q_1 \cdot \frac{T_2}{T_1} \) and \( Q_3 = W \cdot \frac{T_3}{T_4 - T_3} \) into the equation for \( Q_4 \).

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Key Concepts

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

Carnot Cycle

The Carnot cycle is a theoretical thermodynamic cycle that provides the maximum possible efficiency for a heat engine operating between two temperature reservoirs. It consists of four reversible processes: two isothermal (heat transfer at constant temperature) and two adiabatic (no heat transfer). Understanding this cycle is crucial for analyzing the performance of both the heat engine and the refrigerator in the given problem.
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The Carnot Cycle and Maximum Theoretical Efficiency

Heat Transfer (Q)

In thermodynamics, heat transfer (denoted as Q) refers to the energy exchanged between a system and its surroundings due to a temperature difference. For the Carnot engine and refrigerator, Q₁, Q₂, Q₃, and Q₄ represent the heat absorbed or released at different stages of the cycle. Accurately determining these values is essential for calculating the efficiency and performance of the system.
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Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of the efficiency of a refrigerator or heat pump, defined as the ratio of useful heating or cooling provided to the work input. For a Carnot refrigerator, the COP can be calculated using the temperatures of the hot and cold reservoirs. Understanding COP is important for evaluating the effectiveness of the refrigerator in the context of the Carnot cycle.
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Related Practice
Textbook Question

A car's internal combustion engine can be modeled as a heat engine operating between a combustion temperature of 1500℃ and an air temperature of 20℃ with 30% of the Carnot efficiency. The heat of combustion of gasoline is 47 kJ/g. What mass of gasoline is burned to accelerate a 1500 kg car from rest to a speed of 30 m/s?

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Textbook Question

A freezer with a coefficient of performance 30% that of a Carnot refrigerator keeps the inside temperature at -22℃ in a 25℃ room. 3.0 L of water at 20℃ are placed in the freezer. How long does it take for the water to freeze if the freezer's compressor does work at the rate of 200 W while the water is freezing?

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Textbook Question

Home air conditioners in the United States have their power specified in the truly obscure units of tons, where 1 ton is the power needed to melt 1 ton (2000 lb or 910 kg) of ice in 24 hours. A modest-size house typically has a 4.0 ton air conditioner. If a 4.0 ton air conditioner has a coefficient of performance of 2.5, a typical value, at what rate in kW is heat energy removed from the house?

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Textbook Question

A typical coal-fired power plant burns 300 metric tons of coal every hour to generate 750 MW of electricity. 1 metric ton = 1000 kg. The density of coal is 1500 kg/m³ and its heat of combustion is 28 MJ/kg. Assume that all heat is transferred from the fuel to the boiler and that all the work done in spinning the turbine is transformed into electric energy. Suppose the coal is piled up in a 10 m ✕ 10 m room. How tall must the pile be to operate the plant for one day?

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Textbook Question

A Carnot heat engine operates between reservoirs at 182℃ and 0℃. If the engine extracts 25 J of energy from the hot reservoir per cycle, how many cycles will it take to lift a 10 kg mass a height of 10 m?

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Textbook Question

A Carnot engine operates between temperatures of 5℃ and 500℃. The output is used to run a Carnot refrigerator operating between -5℃ and 25℃. How many joules of heat energy does the refrigerator exhaust into the room for each joule of heat energy used by the heat engine?

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