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Physics: Thermal Physics - Condensation, Evaporation, and Thermal Expansion

Grade 8IB

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

🔑Concepts

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Evaporation: A surface phenomenon where molecules with higher kinetic energy EkE_k escape from the liquid surface into the gaseous state. This occurs at temperatures below the boiling point.

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Cooling Effect of Evaporation: As high-energy particles leave the liquid, the average kinetic energy of the remaining particles decreases. Since T∝EavgT \propto E_{avg}, the temperature of the liquid drops.

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Factors Affecting Evaporation: Rate increases with higher temperature TT, larger surface area AA, higher wind speed, and lower humidity.

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Condensation: The process where gas particles lose thermal energy, their kinetic energy decreases, and intermolecular forces pull them together to form a liquid. This release of energy is known as latent heat.

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Thermal Expansion: When a substance is heated, its particles vibrate or move more vigorously, increasing the average distance between them. This results in an increase in volume VV.

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Relative Expansion: Gases expand the most for a given temperature change, followed by liquids, and then solids (Expansiongas>Expansionliquid>ExpansionsolidExpansion_{gas} > Expansion_{liquid} > Expansion_{solid}).

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Bimetallic Strip: A device consisting of two different metals (e.g., brass and iron) bonded together. Upon heating, the metal with the higher coefficient of expansion α\alpha expands more, causing the strip to bend.

📐Formulae

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

ΔV=βV0ΔT\Delta V = \beta V_0 \Delta T

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15

💡Examples

Problem 1:

A steel railway track has a length L0=10.0L_0 = 10.0 m at 20∘C20^\circ C. If the temperature rises to 45∘C45^\circ C, calculate the increase in length ΔL\Delta L. (Given αsteel=12×10−6/∘C\alpha_{steel} = 12 \times 10^{-6} /^\circ C)

Solution:

ΔT=45∘C−20∘C=25∘C\Delta T = 45^\circ C - 20^\circ C = 25^\circ C ΔL=(12×10−6/∘C)×10.0 m×25∘C\Delta L = (12 \times 10^{-6} /^\circ C) \times 10.0 \text{ m} \times 25^\circ C ΔL=0.003 m\Delta L = 0.003 \text{ m}

Explanation:

The linear expansion formula is used to find the change in length based on the original length, the change in temperature, and the specific expansion coefficient of the material.

Problem 2:

Why do you feel cold when you step out of a swimming pool on a windy day, even if the air temperature is warm?

Solution:

The water on the skin evaporates rapidly due to the wind. Evaporation requires energy, which is taken from the skin in the form of heat.

Explanation:

The process of evaporation removes the fastest-moving (hottest) molecules from the surface of the liquid. This lowers the average kinetic energy of the remaining liquid on the skin, creating a cooling sensation.

Problem 3:

Calculate the final temperature in Kelvin if a liquid is heated from 25∘C25^\circ C to 75∘C75^\circ C.

Solution:

TCelsius=75∘CT_{Celsius} = 75^\circ C TKelvin=75+273.15T_{Kelvin} = 75 + 273.15 TKelvin=348.15 KT_{Kelvin} = 348.15 \text{ K}

Explanation:

To convert Celsius to Kelvin, we add 273.15273.15 to the Celsius value as the Kelvin scale starts at absolute zero.