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Thermodynamics - Second Law of Thermodynamics

Grade 11ICSEPhysics

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

🔑Concepts

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The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time; it can only remain constant in ideal reversible processes.

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Kelvin-Planck Statement: It is impossible to construct a heat engine that operates in a cycle and produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work. This implies that no heat engine can have 100%100\% efficiency.

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Clausius Statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a cooler body to a hotter body without the input of external work.

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Entropy (SS): A thermodynamic property that serves as a measure of the molecular disorder or randomness of a system. For a reversible process, the change in entropy is given by dS=dQTdS = \frac{dQ}{T}.

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Carnot Engine: A theoretical ideal engine that operates on the Carnot cycle (two isothermal and two adiabatic processes). It provides the maximum possible efficiency between two temperatures T1T_1 (source) and T2T_2 (sink).

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Reversible and Irreversible Processes: A reversible process is one that can be retraced in the opposite direction such that the system and surroundings return to their original states. All natural processes are irreversible due to friction, turbulence, or heat transfer across finite temperature differences.

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Coefficient of Performance (COP): For a refrigerator or heat pump, it is the ratio of the heat extracted (or delivered) to the work input required.

📐Formulae

η=WQ1=Q1−Q2Q1=1−Q2Q1\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}

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

ΔS=∫dQrevT\Delta S = \int \frac{dQ_{rev}}{T}

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

βideal=T2T1−T2\beta_{ideal} = \frac{T_2}{T_1 - T_2}

Work Done (W)=Q1−Q2\text{Work Done } (W) = Q_1 - Q_2

💡Examples

Problem 1:

A Carnot engine absorbs 1000 J1000 \text{ J} of heat from a reservoir at 627∘C627^\circ\text{C} and rejects heat to a sink at 27∘C27^\circ\text{C}. Calculate the efficiency of the engine and the amount of heat rejected.

Solution:

Convert temperatures to Kelvin: T1=627+273=900 KT_1 = 627 + 273 = 900 \text{ K}, T2=27+273=300 KT_2 = 27 + 273 = 300 \text{ K}. Efficiency η=1−T2T1=1−300900=1−13=23≈0.667\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.667 or 66.7%66.7\%. Heat rejected Q2Q_2: Since Q2Q1=T2T1\frac{Q_2}{Q_1} = \frac{T_2}{T_1}, Q2=Q1×T2T1=1000×300900=333.33 JQ_2 = Q_1 \times \frac{T_2}{T_1} = 1000 \times \frac{300}{900} = 333.33 \text{ J}.

Explanation:

Efficiency depends only on the absolute temperatures of the source and sink. The ratio of heat exchange is proportional to the ratio of absolute temperatures in a Carnot cycle.

Problem 2:

A refrigerator maintains food at 2∘C2^\circ\text{C} while the room temperature is 32∘C32^\circ\text{C}. Find the Coefficient of Performance (COP) of the refrigerator assuming it is an ideal Carnot refrigerator.

Solution:

Convert temperatures to Kelvin: T2=2+273=275 KT_2 = 2 + 273 = 275 \text{ K} (Sink/Inside), T1=32+273=305 KT_1 = 32 + 273 = 305 \text{ K} (Source/Room). COP(β)=T2T1−T2=275305−275=27530≈9.17COP (\beta) = \frac{T_2}{T_1 - T_2} = \frac{275}{305 - 275} = \frac{275}{30} \approx 9.17.

Explanation:

The COP of a refrigerator measures how efficiently it removes heat from a cold space. A higher COP indicates a more efficient cooling process.

Problem 3:

Calculate the change in entropy when 10 g10 \text{ g} of ice at 0∘C0^\circ\text{C} is converted into water at the same temperature. (Latent heat of fusion L=80 cal/gL = 80 \text{ cal/g})

Solution:

Heat required Q=m×L=10×80=800 calQ = m \times L = 10 \times 80 = 800 \text{ cal}. Temperature in Kelvin T=0+273=273 KT = 0 + 273 = 273 \text{ K}. Change in entropy ΔS=QT=800273≈2.93 cal/K\Delta S = \frac{Q}{T} = \frac{800}{273} \approx 2.93 \text{ cal/K}.

Explanation:

Even though the temperature remains constant during a phase change, entropy increases because heat is absorbed, leading to increased molecular disorder as the solid turns into a liquid.

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