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Thermal Properties of Matter - Latent Heat

Grade 11CBSEPhysics

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

🔑Concepts

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Latent heat is the heat energy absorbed or released by a substance during a change in its physical state (phase) that occurs without a change in temperature.

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The heat energy supplied during a phase change is used to overcome the intermolecular forces of attraction, increasing the potential energy of the molecules rather than their kinetic energy; hence, the temperature remains constant.

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Specific Latent Heat of Fusion (LfL_f) is the amount of heat required to change a unit mass of a substance from solid to liquid at its melting point. For water, Lf≈3.33×105 J kg−1L_f \approx 3.33 \times 10^5 \text{ J kg}^{-1} or 80 cal g−180 \text{ cal g}^{-1}.

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Specific Latent Heat of Vaporization (LvL_v) is the amount of heat required to change a unit mass of a substance from liquid to gas at its boiling point. For water, Lv≈22.6×105 J kg−1L_v \approx 22.6 \times 10^5 \text{ J kg}^{-1} or 540 cal g−1540 \text{ cal g}^{-1}.

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On a heating curve (Temperature vs. Time graph), phase changes are represented by horizontal plateaus where dTdt=0\frac{dT}{dt} = 0.

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The total heat QQ required for a process involving both temperature change and phase change is the sum of sensible heat (mcΔTmc\Delta T) and latent heat (mLmL).

📐Formulae

Q=mLQ = mL

L=QmL = \frac{Q}{m}

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

1 cal g−1=4186 J kg−11 \text{ cal g}^{-1} = 4186 \text{ J kg}^{-1}

💡Examples

Problem 1:

Calculate the amount of heat required to convert 2 kg2 \text{ kg} of ice at 0∘C0^\circ\text{C} to water at 20∘C20^\circ\text{C}. Given Lf=3.34×105 J kg−1L_f = 3.34 \times 10^5 \text{ J kg}^{-1} and specific heat of water sw=4186 J kg−1K−1s_w = 4186 \text{ J kg}^{-1} \text{K}^{-1}.

Solution:

Total heat Q=Qmelting+QheatingQ = Q_{melting} + Q_{heating}.

  1. Heat for melting: Q1=mLf=2×3.34×105=6.68×105 JQ_1 = m L_f = 2 \times 3.34 \times 10^5 = 6.68 \times 10^5 \text{ J}.
  2. Heat to raise temperature from 0∘C0^\circ\text{C} to 20∘C20^\circ\text{C}: Q2=mswΔT=2×4186×(20−0)=1.6744×105 JQ_2 = m s_w \Delta T = 2 \times 4186 \times (20 - 0) = 1.6744 \times 10^5 \text{ J}. Total heat Q=6.68×105+1.6744×105=8.3544×105 JQ = 6.68 \times 10^5 + 1.6744 \times 10^5 = 8.3544 \times 10^5 \text{ J}.

Explanation:

The process involves two stages: first, the phase change from solid to liquid at a constant 0∘C0^\circ\text{C} (Latent Heat), and second, the temperature increase of the resulting water (Sensible Heat).

Problem 2:

Why does steam at 100∘C100^\circ\text{C} cause more severe burns than boiling water at 100∘C100^\circ\text{C}?

Solution:

Steam at 100∘C100^\circ\text{C} contains more heat energy per unit mass than boiling water at the same temperature due to the Latent Heat of Vaporization (LvL_v). When steam hits the skin, it first releases 22.6×105 J kg−122.6 \times 10^5 \text{ J kg}^{-1} of energy just to condense into water at 100∘C100^\circ\text{C}, and then continues to release heat as it cools down.

Explanation:

The additional energy stored as latent heat in steam is released during the phase change (condensation) on the skin, resulting in much higher thermal damage compared to liquid water which only releases sensible heat.

Latent Heat - Revision Notes & Key Formulas | CBSE Class 11 Physics