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Thermal Properties of Matter - Change of State

Grade 11CBSEPhysics

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

🔑Concepts

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Change of State: The transition of a substance from one physical state (solid, liquid, or gas) to another is called a change of state. This process occurs at a constant temperature.

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Melting and Fusion: The change of state from solid to liquid is called melting or fusion. The constant temperature at which this occurs at standard atmospheric pressure is the normal melting point.

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Vaporization: The change of state from liquid to gas is called vaporization. The constant temperature at which a liquid boils at standard atmospheric pressure is the normal boiling point.

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Latent Heat (LL): The amount of heat per unit mass transferred during a change of state at constant temperature. It is defined as L=QmL = \frac{Q}{m}, where QQ is the heat supplied and mm is the mass.

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Latent Heat of Fusion (LfL_f): The heat required to change a unit mass of a substance from solid to liquid state at its melting point. For water, Lf≈3.33×105 J/kgL_f \approx 3.33 \times 10^5 \text{ J/kg} or 80 cal/g80 \text{ cal/g}.

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Latent Heat of Vaporization (LvL_v): The heat required to change a unit mass of a substance from liquid to gas state at its boiling point. For water, Lv≈2.26×106 J/kgL_v \approx 2.26 \times 10^6 \text{ J/kg} or 540 cal/g540 \text{ cal/g}.

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Regelation: The phenomenon where ice melts under increased pressure and refreezes when the pressure is removed. This happens because the melting point of ice decreases with an increase in pressure.

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Triple Point: The unique temperature and pressure at which the solid, liquid, and gaseous phases of a substance coexist in equilibrium. For water, the triple point is 273.16 K273.16 \text{ K} at a pressure of 6.11×10−3 atm6.11 \times 10^{-3} \text{ atm}.

📐Formulae

Q=mLQ = mL

Lf=QfusionmL_f = \frac{Q_{fusion}}{m}

Lv=QvaporizationmL_v = \frac{Q_{vaporization}}{m}

Qtotal=mcΔT+mLQ_{total} = mc\Delta T + mL

💡Examples

Problem 1:

Calculate the amount of heat required to convert 0.5 kg0.5 \text{ kg} of ice at 0∘C0^\circ\text{C} to water at 0∘C0^\circ\text{C}. Given Lf=3.33×105 J/kgL_f = 3.33 \times 10^5 \text{ J/kg}.

Solution:

Using the formula for latent heat: Q=mLfQ = mL_f. Given m=0.5 kgm = 0.5 \text{ kg} and Lf=3.33×105 J/kgL_f = 3.33 \times 10^5 \text{ J/kg}. Q=0.5×3.33×105Q = 0.5 \times 3.33 \times 10^5 Q=1.665×105 JQ = 1.665 \times 10^5 \text{ J}

Explanation:

Since there is no change in temperature, only the latent heat of fusion is required to break the molecular bonds of the solid ice.

Problem 2:

How much total heat is needed to turn 10 g10 \text{ g} of ice at 0∘C0^\circ\text{C} into water at 20∘C20^\circ\text{C}? (Take Lf=80 cal/gL_f = 80 \text{ cal/g} and swater=1 cal/g ∘Cs_{water} = 1 \text{ cal/g } ^\circ\text{C})

Solution:

Step 1: Heat to melt ice (Q1Q_1): Q1=mLf=10×80=800 calQ_1 = m L_f = 10 \times 80 = 800 \text{ cal} Step 2: Heat to raise water temperature from 0∘C0^\circ\text{C} to 20∘C20^\circ\text{C} (Q2Q_2): Q2=msΔT=10×1×(20−0)=200 calQ_2 = ms\Delta T = 10 \times 1 \times (20 - 0) = 200 \text{ cal} Step 3: Total Heat (QQ): Q=Q1+Q2Q = Q_1 + Q_2 800+2001000\begin{array}{r} 800 \\ + 200 \\ \hline 1000 \end{array} Total Heat Q=1000 calQ = 1000 \text{ cal}.

Explanation:

The total heat is the sum of the latent heat required for the phase change (solid to liquid) and the sensible heat required to increase the temperature of the resulting water.