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Physics - Thermal physics (States of matter and Heat transfer)

Grade 9IGCSE

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

🔑Concepts

The Kinetic Molecular Theory: Matter is made of particles in constant motion. In solids, particles vibrate about fixed positions; in liquids, they slide over each other; in gases, they move randomly at high speeds.

Brownian Motion: The random movement of microscopic particles (like smoke or pollen) in a fluid, providing evidence for the kinetic theory as they are bombarded by invisible atoms or molecules.

Thermal Expansion: Most substances expand when heated because their particles move faster and push further apart. This is quantified by the change in volume or length relative to the change in temperature ΔT\Delta T.

Internal Energy: The sum of the total microscopic kinetic energy and potential energy of the particles within a system.

Specific Heat Capacity (cc): The energy required per unit mass to raise the temperature of a substance by 1C1^\circ C. It is measured in J/(kgC)J/(kg\cdot^\circ C).

Latent Heat (LL): The energy absorbed or released during a change of state (fusion or vaporization) without a change in temperature. LfL_f is for melting/freezing, and LvL_v is for boiling/condensing.

Conduction: The transfer of thermal energy through a substance by the vibration of atoms and the movement of delocalized electrons (primarily in metals).

Convection: The transfer of heat in fluids (liquids and gases) caused by the upward movement of less dense, warmer regions and the downward movement of denser, cooler regions.

Radiation: The transfer of energy by infrared electromagnetic waves. It does not require a medium and can travel through a vacuum. Dull black surfaces are the best emitters and absorbers, while shiny silver surfaces are the best reflectors.

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

T(K)=θ(C)+273T(K) = \theta(^\circ C) + 273

P=QtP = \frac{Q}{t}

💡Examples

Problem 1:

Calculate the energy required to heat 0.5 kg0.5\text{ kg} of water from 20C20^\circ C to 100C100^\circ C. The specific heat capacity of water is 4200 J/(kgC)4200\text{ J/(kg}\cdot^\circ C).

Solution:

Q=0.5 kg×4200 J/(kgC)×(10020)C=168,000 JQ = 0.5\text{ kg} \times 4200\text{ J/(kg}\cdot^\circ C) \times (100 - 20)^\circ C = 168,000\text{ J}

Explanation:

We use the specific heat capacity formula Q=mcΔTQ = mc\Delta T, where m=0.5m = 0.5, c=4200c = 4200, and ΔT=80\Delta T = 80.

Problem 2:

How much thermal energy is needed to melt 2 kg2\text{ kg} of ice at 0C0^\circ C? The specific latent heat of fusion of ice is 3.34×105 J/kg3.34 \times 10^5\text{ J/kg}.

Solution:

Q=mL=2 kg×3.34×105 J/kg=6.68×105 JQ = mL = 2\text{ kg} \times 3.34 \times 10^5\text{ J/kg} = 6.68 \times 10^5\text{ J}

Explanation:

Since the state is changing from solid to liquid at a constant temperature, we use the latent heat formula Q=mLQ = mL.

Problem 3:

A metal rod is heated at one end. Explain why the other end eventually becomes hot.

Solution:

Thermal energy is transferred via conduction. As particles at the hot end gain kinetic energy, they vibrate more vigorously and collide with neighboring particles, transferring energy. In metals, free electrons also move through the lattice, rapidly transferring energy to the cooler end.

Explanation:

This describes the microscopic process of conduction in solids, specifically highlighting the role of molecular vibrations and delocalized electrons.

Thermal physics (States of matter and Heat transfer) Revision - Grade 9 Science IGCSE