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Thermodynamics - Carnot Engine

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Carnot engine is a theoretical thermodynamic cycle proposed by Nicolas Léonard Sadi Carnot. It provides the maximum possible efficiency that a heat engine can achieve operating between two temperatures.

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The engine operates on the Carnot cycle, which consists of four reversible processes: (1) Reversible Isothermal Expansion, (2) Reversible Adiabatic Expansion, (3) Reversible Isothermal Compression, and (4) Reversible Adiabatic Compression.

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Efficiency η\eta is defined as the ratio of net work done WW to the heat absorbed Q1Q_1 from the high-temperature reservoir (source).

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According to Carnot's theorem, all reversible engines operating between the same two temperatures have the same efficiency, and no irreversible engine can be more efficient than a reversible one.

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The temperatures T1T_1 (source) and T2T_2 (sink) must always be expressed in Kelvin (K) for all thermodynamic calculations.

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The efficiency of a Carnot engine is always less than 100%100\% because T2T_2 cannot be absolute zero (0 K0 \text{ K}) in practice.

📐Formulae

η=WQ1\eta = \frac{W}{Q_1}

η=Q1−Q2Q1=1−Q2Q1\eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}

η=1−T2T1\eta = 1 - \frac{T_2}{T_1}

Q1T1=Q2T2\frac{Q_1}{T_1} = \frac{Q_2}{T_2}

W=Q1−Q2W = Q_1 - Q_2

💡Examples

Problem 1:

A Carnot engine works between a source at 227∘C227^{\circ} \text{C} and a sink at 127∘C127^{\circ} \text{C}. Calculate the efficiency of the engine.

Solution:

First, convert temperatures from Celsius to Kelvin: Source temperature T1=227+273=500 KT_1 = 227 + 273 = 500 \text{ K} Sink temperature T2=127+273=400 KT_2 = 127 + 273 = 400 \text{ K}

Using the efficiency formula: η=1−T2T1\eta = 1 - \frac{T_2}{T_1} η=1−400500\eta = 1 - \frac{400}{500} η=1−0.8=0.2\eta = 1 - 0.8 = 0.2

Percentage efficiency: η%=0.2×100=20%\eta \% = 0.2 \times 100 = 20 \%

Explanation:

The efficiency is determined solely by the absolute temperatures of the reservoirs. Converting to Kelvin is a mandatory step.

Problem 2:

A Carnot engine absorbs 1000 J1000 \text{ J} of heat from a reservoir at 600 K600 \text{ K} and rejects heat to a sink at 300 K300 \text{ K}. Find the work done by the engine and the heat rejected.

Solution:

Given Q1=1000 JQ_1 = 1000 \text{ J}, T1=600 KT_1 = 600 \text{ K}, and T2=300 KT_2 = 300 \text{ K}. Efficiency η=1−300600=1−0.5=0.5\eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5.

Work done WW: W=η×Q1W = \eta \times Q_1 W=0.5×1000=500 JW = 0.5 \times 1000 = 500 \text{ J}

Heat rejected Q2Q_2: 1000−500500\begin{array}{r} 1000 \\ - 500 \\ \hline 500 \end{array} Q2=Q1−W=500 JQ_2 = Q_1 - W = 500 \text{ J}

Explanation:

Since the efficiency is 0.50.5 (or 50%50\%), half of the input heat is converted into useful work, and the remaining half is rejected to the sink.