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Thermodynamics - Heat Engines and Refrigerators

Grade 11ICSEPhysics

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

🔑Concepts

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A Heat Engine is a device that converts heat energy into mechanical work through a cyclic process. It consists of a high-temperature source T1T_1, a low-temperature sink T2T_2, and a working substance.

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The Efficiency (ηη) of a heat engine is the ratio of the net work done (WW) to the heat absorbed (Q1Q_1) from the source. It is expressed as η = rac{W}{Q_1}.

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A Carnot Engine is a theoretical, ideal heat engine that operates on the Carnot cycle, consisting of two isothermal and two adiabatic processes. Its efficiency depends solely on the temperatures of the source and sink.

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The Second Law of Thermodynamics (Kelvin-Planck statement) states that it is impossible to construct an engine that, operating in a cycle, will produce no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.

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A Refrigerator or Heat Pump is essentially a heat engine running in reverse. It extracts heat Q2Q_2 from a cold body at T2T_2 and rejects heat Q1Q_1 to a hot body at T1T_1 by performing external work WW.

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The Coefficient of Performance (ββ) for a refrigerator is the ratio of heat extracted from the cold reservoir to the work done on the system. For an ideal refrigerator, β = rac{T_2}{T_1 - T_2}.

📐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}

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

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}

β=1−ηη\beta = \frac{1 - \eta}{\eta}

💡Examples

Problem 1:

A Carnot engine works between 600 K600\text{ K} and 300 K300\text{ K}. Find its efficiency. If the engine absorbs 1000 J1000\text{ J} of heat from the source, how much work does it perform?

Solution:

Given T1=600 KT_1 = 600\text{ K}, T2=300 KT_2 = 300\text{ K}, and Q1=1000 JQ_1 = 1000\text{ J}. Efficiency η=1−T2T1=1−300600=1−0.5=0.5\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 or 50%50\%. Work done W=η×Q1=0.5×1000 J=500 JW = \eta \times Q_1 = 0.5 \times 1000\text{ J} = 500\text{ J}.

Explanation:

The efficiency is calculated using the absolute temperature ratio. Since the engine is 50%50\% efficient, it converts half of the absorbed heat into work.

Problem 2:

A refrigerator maintains its contents at 2∘C2^\circ\text{C} while the room temperature is 32∘C32^\circ\text{C}. Calculate the maximum possible Coefficient of Performance (COP).

Solution:

Convert temperatures to Kelvin: T2=2+273=275 KT_2 = 2 + 273 = 275\text{ K} and T1=32+273=305 KT_1 = 32 + 273 = 305\text{ K}. β=T2T1−T2=275305−275=27530≈9.17\beta = \frac{T_2}{T_1 - T_2} = \frac{275}{305 - 275} = \frac{275}{30} \approx 9.17.

Explanation:

The COP of a refrigerator increases as the temperature difference between the sink and the source decreases.

Heat Engines and Refrigerators - Revision Notes & Key Formulas | ICSE Class 11 Physics