A diesel engine performs J of mechanical work and discards J of heat each cycle. How much heat must be supplied to the engine in each cycle?
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23. The Second Law of Thermodynamics
Heat Engines and the Second Law of Thermodynamics
Problem 46a
Textbook Question
FIGURE P21.46 shows a Carnot heat engine driving a Carnot refrigerator. Determine Q2, Q3 and Q4.


1
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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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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