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Physics Chapter 8 : Topic 23
Carnot Cycle Efficiency
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The Carnot cycle establishes:
Reasoning: The Carnot cycle gives the theoretical maximum efficiency that any heat engine can achieve between two temperatures.
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The efficiency of an ideal Carnot engine depends on:
Reasoning: Carnot efficiency depends only on the absolute temperatures of the two heat reservoirs.
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The formula for Carnot engine efficiency is:
Reasoning: Carnot efficiency is given by: $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}$Where $\mathbf{T}_{\mathbf{H}}$ is hot reservoir temperature and $\mathbf{T}_{\mathbf{C}}$ is cold reservoir temperature.
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In the Carnot efficiency equation, temperatures must be measured in:
Reasoning: Thermodynamic temperature must be expressed in absolute Kelvin units.
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A Carnot engine operates between a hot reservoir at $\mathbf{500\ K}$ and a cold reservoir at $\mathbf{300\ K}$. Its efficiency is:
Reasoning: $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}\mathbf{= 1 -}\frac{\mathbf{300}}{\mathbf{500}}\mathbf{= 1 - 0.6 = 0.4}$ ; Efficiency $\mathbf{= 40\%}$
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If the temperature of the hot reservoir increases while the cold reservoir remains constant, efficiency:
Reasoning: Increasing $\mathbf{T}_{\mathbf{H}}$ increases the temperature difference, increasing Carnot efficiency.
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If the temperature of the cold reservoir increases while the hot reservoir remains constant, efficiency:
Reasoning: A higher $\mathbf{T}_{\mathbf{C}}$ reduces the fraction of heat converted into work.
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A Carnot engine will have maximum efficiency when:
Reasoning: A larger temperature difference gives greater efficiency.
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A Carnot engine has zero efficiency when:
Reasoning: $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}$If both temperatures are equal, efficiency becomes zero.
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A Carnot engine cannot achieve $\mathbf{100\%}$ efficiency because:
Reasoning: The second law requires some heat rejection to the sink.
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The efficiency of a real heat engine compared with Carnot efficiency is:
Reasoning: Real engines have friction and irreversible losses, so they cannot reach Carnot efficiency.
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The Carnot engine is considered ideal because:
Reasoning: Carnot cycle consists of reversible processes without energy losses.
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If a real engine has the same reservoir temperatures as a Carnot engine, its efficiency will be:
Reasoning: Real engines always have additional irreversible losses.
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Carnot efficiency provides:
Reasoning: No engine can exceed the efficiency predicted by the Carnot cycle.
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The efficiency of a Carnot engine does not depend on:
Reasoning: Carnot efficiency depends only on reservoir temperatures.
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A heat engine absorbs heat $\mathbf{Q}_{\mathbf{H}}$ from the hot reservoir and rejects $\mathbf{Q}_{\mathbf{C}}$. The work done is:
Reasoning: $\mathbf{W =}\mathbf{Q}_{\mathbf{H}}\mathbf{-}\mathbf{Q}_{\mathbf{C}}$ $\mathbf{\Longrightarrow}$ The difference becomes useful work.
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The efficiency of a heat engine is defined as:
Reasoning: $\mathbf{\eta =}$ $\frac{\mathbf{Work\ output}}{\mathbf{Heat\ input}}$
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The cold reservoir temperature in Carnot efficiency represents:
Reasoning: $\mathbf{T}_{\mathbf{C}}$ is the temperature of the sink where waste heat is rejected.
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A Carnot engine receives $\mathbf{1000\ J}$ heat and has efficiency $\mathbf{30\%}$. The work output is:
Reasoning: $\mathbf{W = \eta}\mathbf{Q}_{\mathbf{H}}\mathbf{= 0.3}\left( \mathbf{100}\mathbf{0} \right)\mathbf{= 300\ J}$
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For a Carnot engine, increasing the temperature difference between reservoirs:
Reasoning: Greater temperature difference increases the fraction of heat converted into work.
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Why is a Carnot engine important in thermodynamics?
Reasoning: It provides a standard for comparing real heat engines.
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Two Carnot engines operate between the same temperatures. Their efficiencies are:
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A Carnot engine operating between $\mathbf{400\ K}$ and $\mathbf{200\ K}$ has efficiency:
Reasoning: $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}\mathbf{= 1 -}\frac{\mathbf{200}}{\mathbf{400}}\mathbf{= 0.5}$ $\mathbf{\Longrightarrow}$ Efficiency $\mathbf{= 50\%}$
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To improve Carnot efficiency, one should:
Reasoning: A larger temperature difference increases efficiency.
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The efficiency of a Carnot engine depends on temperature because:
Reasoning: The available work depends on the difference between source and sink temperatures.
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A Carnot engine between $\mathbf{600\ K}$ and $\mathbf{300\ K}$ is twice as efficient as one operating between:
Reasoning: For $\ \mathbf{400\ K\ }$ and $\ \mathbf{200\ K}$ $\mathbf{\Longrightarrow}$ $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}\mathbf{= 1 -}\frac{\mathbf{200}}{\mathbf{400}}\mathbf{= 50\%}$Same temperature ratio gives same efficiency.
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The maximum possible efficiency of a heat engine occurs in:
Reasoning: Reversible processes have the highest theoretical efficiency.
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If $\mathbf{T}_{\mathbf{C}}$ approaches zero Kelvin, Carnot efficiency approaches:
Reasoning: $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}\mathbf{= 1 -}\frac{\mathbf{0}}{\mathbf{T}_{\mathbf{H}}}\mathbf{= 1 - 0 =}\mathbf{1}$
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A Carnot engine cannot exceed its efficiency limit because:
Reasoning: The second law sets the maximum possible efficiency.
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The main conclusion of Carnot’s theorem is:
Reasoning: Carnot’s theorem establishes the upper limit of heat engine efficiency.
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