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Thermal Physics - Melting, boiling and evaporation

Grade 11IGCSEPhysics

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

🔑Concepts

Melting: The process by which a solid changes into a liquid. This occurs at a constant temperature called the melting point. During melting, the thermal energy supplied is used to overcome the intermolecular forces of attraction in the solid lattice rather than increasing the kinetic energy of the particles.

Boiling: A rapid change of state from liquid to gas occurring throughout the bulk of the liquid at a specific temperature called the boiling point. The temperature remains constant at TboilingT_{boiling} until all the liquid has evaporated.

Evaporation: The change from liquid to gas that occurs at the surface of a liquid at any temperature below the boiling point. Molecules with higher-than-average kinetic energy escape the surface, leading to a decrease in the average kinetic energy of the remaining molecules, which causes a cooling effect.

Factors affecting Evaporation: Evaporation rate increases with higher surface area, higher temperature, and increased air movement (wind) over the surface.

Specific Latent Heat (LL): The amount of thermal energy required to change the state of 1 kg1\text{ kg} of a substance without a change in temperature. It is measured in J/kgJ/kg.

Specific Latent Heat of Fusion (LfL_f): The energy required to change 1 kg1\text{ kg} of a substance from solid to liquid at its melting point.

Specific Latent Heat of Vaporization (LvL_v): The energy required to change 1 kg1\text{ kg} of a substance from liquid to gas at its boiling point.

📐Formulae

Q=mLfQ = m L_f

Q=mLvQ = m L_v

P=EtP = \frac{E}{t}

E=P×tE = P \times t

💡Examples

Problem 1:

Calculate the energy required to melt 0.25 kg0.25\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=mLfQ = m L_f Q=0.25 kg×3.34×105 J/kgQ = 0.25\text{ kg} \times 3.34 \times 10^5\text{ J/kg} Q=83,500 JQ = 83,500\text{ J}

Explanation:

To find the energy required for a phase change (melting), we use the mass of the substance multiplied by its specific latent heat of fusion. Since the temperature is already at the melting point (0C0^\circ C), no energy is used for a temperature increase.

Problem 2:

An electric heater with a power rating of 2.0 kW2.0\text{ kW} is used to boil water. If the water is already at 100C100^\circ C, how much mass is turned into steam in 5 minutes5\text{ minutes}? (Specific latent heat of vaporization Lv=2.26×106 J/kgL_v = 2.26 \times 10^6\text{ J/kg})

Solution:

First, calculate total energy: E=P×t=2000 W×(5×60) s=600,000 JE = P \times t = 2000\text{ W} \times (5 \times 60)\text{ s} = 600,000\text{ J} Then, calculate mass: m=QLv=600,000 J2.26×106 J/kgm = \frac{Q}{L_v} = \frac{600,000\text{ J}}{2.26 \times 10^6\text{ J/kg}} m0.265 kgm \approx 0.265\text{ kg}

Explanation:

Energy supplied by the heater is found by multiplying power (in Watts) by time (in seconds). This energy is then equated to the latent heat formula Q=mLvQ = m L_v to solve for mass mm converted to steam.

Melting, boiling and evaporation - Revision Notes & Key Formulas | IGCSE Grade 11 Physics