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Physics - Thermal Physics

Grade 10IGCSE

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

🔑Concepts

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Kinetic Molecular Model: Matter is made of particles. In solids, particles vibrate about fixed positions; in liquids, they move past each other; in gases, they move randomly at high speeds. Temperature is a measure of the average kinetic energy of the particles.

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Thermal Expansion: When a substance is heated, its particles move faster and take up more space, leading to expansion. Solids expand slightly, liquids more, and gases the most. This is used in liquid-in-glass thermometers.

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Specific Heat Capacity (cc): The energy required to raise the temperature of 1 kg1\text{ kg} of a substance by 1 ∘C1\text{ }^\circ\text{C}. It is expressed in units of J/(kg⋅∘C)J/(kg\cdot^\circ C).

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Melting and Boiling (Latent Heat): During a state change, energy is used to break intermolecular bonds rather than increasing kinetic energy. Therefore, temperature remains constant during melting (latent heat of fusion) and boiling (latent heat of vaporization).

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Thermal Energy Transfer: Conduction (collision of particles and electron diffusion in solids), Convection (density changes in fluids causing circulation currents), and Radiation (infra-red electromagnetic waves that do not require a medium).

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Gas Laws (Boyle's Law): For a fixed mass of gas at constant temperature, the pressure (pp) is inversely proportional to the volume (VV). Thus, p∝1Vp \propto \frac{1}{V}.

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Absolute Zero: The temperature at which particles have minimum kinetic energy, defined as 0 K0\text{ K} or −273 ∘C-273\text{ }^\circ\text{C}.

📐Formulae

E=mcΔθE = mc\Delta\theta

E=mLE = mL

p1V1=p2V2p_1V_1 = p_2V_2

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

P=FAP = \frac{F}{A}

💡Examples

Problem 1:

How much thermal energy is needed to heat 2.5 kg2.5\text{ kg} of aluminum from 20 ∘C20\text{ }^\circ\text{C} to 60 ∘C60\text{ }^\circ\text{C}? (Specific heat capacity of aluminum c=900 J/(kg⋅∘C)c = 900\text{ J/(kg}\cdot^\circ\text{C)}).

Solution:

E=mcΔθ=2.5×900×(60−20)=2.5×900×40=90,000 JE = mc\Delta\theta = 2.5 \times 900 \times (60 - 20) = 2.5 \times 900 \times 40 = 90,000\text{ J}

Explanation:

To find the energy, we use the specific heat capacity formula. The temperature change Δθ\Delta\theta is 40 ∘C40\text{ }^\circ\text{C}.

Problem 2:

Calculate the energy required to melt 0.2 kg0.2\text{ kg} of ice at 0 ∘C0\text{ }^\circ\text{C}. (Specific latent heat of fusion of ice Lf=334,000 J/kgL_f = 334,000\text{ J/kg}).

Solution:

E=mLf=0.2×334,000=66,800 JE = mL_f = 0.2 \times 334,000 = 66,800\text{ J}

Explanation:

Since the state is changing (melting) at a constant temperature, we use the specific latent heat formula E=mLE = mL.

Problem 3:

A gas syringe contains 50 cm350\text{ cm}^3 of gas at a pressure of 1.2×105 Pa1.2 \times 10^5\text{ Pa}. If the plunger is pushed in so the volume becomes 20 cm320\text{ cm}^3 at constant temperature, what is the new pressure?

Solution:

p1V1=p2V2⇒(1.2×105)×50=p2×20⇒p2=6.0×10620=3.0×105 Pap_1V_1 = p_2V_2 \Rightarrow (1.2 \times 10^5) \times 50 = p_2 \times 20 \Rightarrow p_2 = \frac{6.0 \times 10^6}{20} = 3.0 \times 10^5\text{ Pa}

Explanation:

Applying Boyle's Law (p1V1=p2V2p_1V_1 = p_2V_2), the decrease in volume leads to a proportional increase in pressure, assuming temperature remains constant.

Thermal Physics - Revision Notes & Key Formulas | IGCSE Grade 10 Science