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Physics: Thermal Physics - Thermal Insulation, Heat Engines, and Cooling Systems

Grade 8IB

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

🔑Concepts

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Thermal Insulation: This refers to the process of reducing heat transfer between objects in thermal contact or in range of radiative influence. Materials with low thermal conductivity (kk), such as wool, fiberglass, and trapped air, are used to minimize energy loss.

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Heat Engines: A device that converts thermal energy into mechanical work. It operates by taking heat from a high-temperature source (QHQ_H), performing useful work (WW), and exhausting the remaining heat to a low-temperature sink (QCQ_C).

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Efficiency (η\eta): In thermal physics, efficiency is the ratio of useful work output to the total heat input. According to the Second Law of Thermodynamics, no heat engine can be 100%100\% efficient as some heat must always be rejected to the environment.

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Cooling Systems: These are 'reverse heat engines' (like refrigerators or air conditioners). They use external work (WW) to move heat from a cold reservoir (QCQ_C) to a hot reservoir (QHQ_H).

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Key Components of a Cooling System: The cycle involves an evaporator (where refrigerant absorbs heat and evaporates), a compressor (which increases pressure and temperature), and a condenser (where heat is released to the surroundings and the gas liquefies).

📐Formulae

Work Done(W)=QH−QC\text{Work Done} (W) = Q_H - Q_C

Efficiency(η)=WQH\text{Efficiency} (\eta) = \frac{W}{Q_H}

Percentage Efficiency=(QH−QCQH)×100%\text{Percentage Efficiency} = \left( \frac{Q_H - Q_C}{Q_H} \right) \times 100\%

Heat Transfer(Q)=mcΔT\text{Heat Transfer} (Q) = mc\Delta T

💡Examples

Problem 1:

A heat engine absorbs 8000 J8000\text{ J} of heat from a hot reservoir and performs 2000 J2000\text{ J} of useful work. Calculate the efficiency of the engine and the amount of heat rejected to the cold reservoir.

Solution:

QH=8000 JQ_H = 8000\text{ J}, W=2000 JW = 2000\text{ J}.

  1. To find rejected heat (QCQ_C): QC=QH−WQ_C = Q_H - W QC=8000−2000=6000 JQ_C = 8000 - 2000 = 6000\text{ J}

  2. To find Efficiency (η\eta): η=WQH×100%\eta = \frac{W}{Q_H} \times 100\% η=20008000×100%=25%\eta = \frac{2000}{8000} \times 100\% = 25\%

Explanation:

The engine converts 25%25\% of the input thermal energy into useful mechanical work, while the remaining 6000 J6000\text{ J} is lost as waste heat to the environment.

Problem 2:

A cooling system extracts 1500 J1500\text{ J} of heat from a refrigerator compartment while requiring 500 J500\text{ J} of electrical work. How much heat is released into the room?

Solution:

QC=1500 JQ_C = 1500\text{ J}, W=500 JW = 500\text{ J}.

In a cooling system, the heat released to the hot reservoir (QHQ_H) is the sum of the work done and the heat extracted: QH=QC+WQ_H = Q_C + W QH=1500+500=2000 JQ_H = 1500 + 500 = 2000\text{ J}

Explanation:

According to the principle of conservation of energy, the total energy expelled by the condenser is the sum of the heat energy removed from the cold interior and the energy provided by the compressor.