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Thermodynamics - Second Law of Thermodynamics

Grade 11CBSEPhysics

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

🔑Concepts

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Limitations of the First Law: The First Law of Thermodynamics establishes the conservation of energy but does not specify the direction of heat flow or the feasibility of a process.

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Kelvin-Planck Statement: No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work. This implies that a heat engine cannot have 100%100\% efficiency.

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Clausius Statement: No process is possible whose sole result is the transfer of heat from a colder object to a hotter object without the assistance of external work.

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Reversible and Irreversible Processes: A reversible process can be retraced in the reverse direction such that the system and surroundings return to their original states. Most natural processes are irreversible due to dissipative forces like friction.

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Carnot Engine: An ideal heat engine that operates on the Carnot cycle, consisting of two isothermal and two adiabatic processes. It provides the maximum possible efficiency for any engine operating between two temperatures T1T_1 (source) and T2T_2 (sink).

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Carnot's Theorem: (i) No engine operating between two given temperatures can be more efficient than a reversible Carnot engine. (ii) The efficiency of a Carnot engine is independent of the nature of the working substance.

📐Formulae

η=WQ1=1−Q2Q1\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}

ηCarnot=1−T2T1\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1}

β=Q2W=Q2Q1−Q2\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}

βCarnot=T2T1−T2\beta_{\text{Carnot}} = \frac{T_2}{T_1 - T_2}

Q1T1=Q2T2 (For a Carnot cycle)\frac{Q_1}{T_1} = \frac{Q_2}{T_2} \text{ (For a Carnot cycle)}

💡Examples

Problem 1:

A Carnot engine operates between a source at 500 K500\text{ K} and a sink at 300 K300\text{ K}. If the engine absorbs 1000 J1000\text{ J} of heat from the source, calculate its efficiency and the work done per cycle.

Solution:

Given: T1=500 KT_1 = 500\text{ K}, T2=300 KT_2 = 300\text{ K}, Q1=1000 JQ_1 = 1000\text{ J}. Efficiency η=1−T2T1=1−300500=1−0.6=0.4\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 or 40%40\%. Work done W=η×Q1=0.4×1000 J=400 JW = \eta \times Q_1 = 0.4 \times 1000\text{ J} = 400\text{ J}.

Explanation:

The efficiency is determined solely by the absolute temperatures of the source and sink. The work output is the product of efficiency and the heat input.

Problem 2:

A refrigerator extracts 200 J200\text{ J} of heat from a cold reservoir at 250 K250\text{ K} and exhausts heat to a room at 300 K300\text{ K}. Calculate the minimum work required and the coefficient of performance (COP).

Solution:

Given: T2=250 KT_2 = 250\text{ K}, T1=300 KT_1 = 300\text{ K}, Q2=200 JQ_2 = 200\text{ J}. β=T2T1−T2=250300−250=25050=5\beta = \frac{T_2}{T_1 - T_2} = \frac{250}{300 - 250} = \frac{250}{50} = 5. Since β=Q2W\beta = \frac{Q_2}{W}, we have W=Q2β=2005=40 JW = \frac{Q_2}{\beta} = \frac{200}{5} = 40\text{ J}.

Explanation:

The Coefficient of Performance (COP) measures the efficiency of a refrigerator. For an ideal refrigerator, it depends on the temperature difference between the interior and the exterior.

Second Law of Thermodynamics - Revision Notes & Key Formulas | CBSE Class 11 Physics