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Thermal Physics - Thermal properties and temperature

Grade 12A LevelPhysics

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

🔑Concepts

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The kinetic molecular model states that all matter consists of tiny particles in constant motion. In solids, particles vibrate about fixed positions; in liquids, they move past each other; in gases, they move randomly at high speeds.

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Temperature is a measure of the average kinetic energy of the particles in a substance. The absolute temperature scale is measured in Kelvin (KK), where 0 K0 \text{ K} (Absolute Zero) is the temperature at which particles have minimum internal energy.

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Internal energy is the sum of the random distribution of kinetic and potential energies associated with the molecules of a system.

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Specific Heat Capacity (cc) is defined as the energy required per unit mass to raise the temperature of a substance by 1∘C1^{\circ}\text{C} (or 1 K1 \text{ K}). It is measured in J kg−1 ∘C−1\text{J kg}^{-1\text{ }\circ}\text{C}^{-1}.

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Specific Latent Heat (LL) is the energy required per unit mass to change the state of a substance without any change in temperature. Latent heat of fusion (LfL_f) refers to the solid-liquid transition, while latent heat of vaporization (LvL_v) refers to the liquid-gas transition.

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Thermal expansion occurs when the temperature of a substance increases, causing the particles to move or vibrate more vigorously, which increases the average distance between them.

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The fixed points for the Celsius scale are the melting point of pure ice (0∘C0^{\circ}\text{C}) and the boiling point of pure water at standard atmospheric pressure (100∘C100^{\circ}\text{C}).

📐Formulae

ΔQ=mcΔθ\Delta Q = mc\Delta\theta

Q=mLfQ = mL_f

Q=mLvQ = mL_v

T(K)=θ(∘C)+273.15T(K) = \theta(^{\circ}\text{C}) + 273.15

C=mcC = mc

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

💡Examples

Problem 1:

A 2.0 kg2.0 \text{ kg} block of copper is heated from 20∘C20^{\circ}\text{C} to 150∘C150^{\circ}\text{C}. Calculate the thermal energy supplied to the block. (Specific heat capacity of copper c=390 J kg−1 ∘C−1c = 390 \text{ J kg}^{-1\text{ }\circ}\text{C}^{-1})

Solution:

Δθ=150∘C−20∘C=130∘C\Delta \theta = 150^{\circ}\text{C} - 20^{\circ}\text{C} = 130^{\circ}\text{C} ΔQ=mcΔθ=2.0×390×130=101,400 J=101.4 kJ\Delta Q = mc\Delta\theta = 2.0 \times 390 \times 130 = 101,400 \text{ J} = 101.4 \text{ kJ}

Explanation:

The energy required depends on the mass of the object, its specific heat capacity, and the magnitude of the temperature change using the formula ΔQ=mcΔθ\Delta Q = mc\Delta\theta.

Problem 2:

How much energy is required to completely evaporate 0.5 kg0.5 \text{ kg} of water already at 100∘C100^{\circ}\text{C}? (Specific latent heat of vaporization of water Lv=2.26×106 J kg−1L_v = 2.26 \times 10^6 \text{ J kg}^{-1})

Solution:

Q=mLvQ = mL_v Q=0.5×2.26×106=1.13×106 J=1.13 MJQ = 0.5 \times 2.26 \times 10^6 = 1.13 \times 10^6 \text{ J} = 1.13 \text{ MJ}

Explanation:

Since the water is already at its boiling point, no energy is used to increase temperature; all energy is used to overcome intermolecular forces to change state from liquid to gas.

Problem 3:

An electric heater with a power rating of 1.5 kW1.5 \text{ kW} is used to heat 3.0 kg3.0 \text{ kg} of a liquid. If the temperature rises by 20∘C20^{\circ}\text{C} in 2 minutes2 \text{ minutes}, calculate the specific heat capacity of the liquid.

Solution:

P=1500 W,t=120 sP = 1500 \text{ W}, t = 120 \text{ s} Q=P×t=1500×120=180,000 JQ = P \times t = 1500 \times 120 = 180,000 \text{ J} c=QmΔθ=180,0003.0×20=180,00060=3000 J kg−1 ∘C−1c = \frac{Q}{m\Delta\theta} = \frac{180,000}{3.0 \times 20} = \frac{180,000}{60} = 3000 \text{ J kg}^{-1\text{ }\circ}\text{C}^{-1}

Explanation:

First, find the total energy supplied using power and time (Q=PtQ=Pt). Then, rearrange the specific heat formula to solve for cc.