English

Latent Heat

Advertisements

Topics

Estimated time: 18 minutes
  • Introduction
  • Definition: Latent Heat
  • Definition: Latent Heat of Fusion
  • Definition: Latent Heat of Vaporization
  • Formula: Latent Heat
  • Experiment
  • Reference Data Table
  • Example
  • Key Points: Latent Heat
Maharashtra State Board: Class 11

Introduction

The word latent comes from Latin latere, meaning "to lie hidden." When a substance changes its physical state — solid to liquid, or liquid to gas — it absorbs or releases a certain quantity of heat energy. This energy transfer occurs without any change in temperature. Because a thermometer shows no change, the heat appears to be "hidden."

Maharashtra State Board: Class 11
CISCE: Class 10

Definition: Latent Heat

The heat energy absorbed (or liberated) in change of phase is not externally manifested by any rise or fall in temperature, it is called the latent heat.

OR

Latent heat is the quantity of heat energy required to change the state of unit mass of a substance from one phase to another, at constant temperature and constant pressure.

Maharashtra State Board: Class 10, 11

Definition: Latent Heat of Fusion

The quantity of heat required to convert unit mass of a substance from its solid state to the liquid state, at its melting point, without any change in its temperature, is called its latent heat of fusion (Lf).

OR

The heat energy absorbed at constant temperature during the transformation of solid into liquid is called the latent heat of fusion. The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.

Maharashtra State Board: Class 11

Definition: Latent Heat of Vaporization

The quantity of heat required to convert unit mass of a substance from its liquid state to vapour state, at its boiling point without any change in its temperature is called its latent heat of vapourization (Lv).

Maharashtra State Board: Class 11

Formula: Latent Heat

Q = m × L

where,

Q = Heat energy absorbed or released during phase change
m = Mass of the substance undergoing phase change
L = Specific Latent Heat (characteristic of the substance & process)

SI Units = J kg⁻¹

Maharashtra State Board: Class 10

Experiment

Procedure

  • Place pieces of ice in a glass beaker. Insert a thermometer bulb into the ice and record the initial temperature.
  • Mount the beaker on a stand and apply heat uniformly using a spirit burner.
  • Record the temperature using the thermometer at regular 1-minute intervals.
  • Stir the ice-water mixture continuously as melting begins to maintain thermal equilibrium.
  • Continue heating even after all ice has melted — do not stop until all water converts to steam.
  • Plot a graph of temperature (y-axis) versus time (x-axis) to visualise the phase transitions.

Ice in beaker with thermometer and burner — experiment setup

Fig.: Experimental setup: ice in a beaker, thermometer, and burner on a stand

Observations Table:

Phase Temperature Range Temp. Change? What Is Happening?
Pre-A Below 0°C → 0°C Yes ↑ Ice is warming up; the kinetic energy of ice molecules increases.
A→B 0°C (constant) No — Flat Ice melting. Heat breaks rigid bonds. Latent Heat of Fusion at work.
B→C 0°C → 100°C Yes ↑ Liquid water warming; kinetic energy rises steadily.
C→D 100°C (constant) No — Flat Water boiling to steam. Heat completely separates molecules. Latent Heat of Vaporisation at work.
D→E Above 100°C Yes ↑ Steam (gas phase) heating further; the kinetic energy of gas molecules rises.

The Heating Curve

Temperature vs time heating curve for water — showing phase change plateaus at 0°C and 100°C

Fig. 3 — Temperature vs. Time graph for heating ice → water → steam (at 1 atm). The flat segments A→B and C→D are where latent heat is released.

Maharashtra State Board: Class 11

Reference Data Table

All values at one standard atmospheric pressure (1 atm). Values vary at different pressures.

Substance Melting Point (°C) Lf (×10⁵ J/kg) Boiling Point (°C) Lv (×10⁵ J/kg)
Water/Ice  0 3.33 100 22.6
Gold 1063 0.645 2660 15.8
Lead 328 0.25 1744 8.67
Ethyl Alcohol −114 1.0 78 8.5
Mercury −39 0.12 357 2.7
Nitrogen −210 0.26 −196 2.0
Oxygen −219 0.14 −183 2.1
Maharashtra State Board: Class 11

Example

Problem: When 0.1 kg of ice at 0°C is mixed with 0.32 kg of water at 35°C in a container, the resulting temperature of the mixture is 7.8°C. Calculate the latent heat of fusion of ice.

Given: swater = 4186 J kg⁻¹ K⁻¹

Given Data:

Mass of ice (mice) = 0.1 kg
Mass of water (mw) = 0.32 kg
Temp. of ice (Tice) = 0°C
Temp. of water (Tw) = 35°C
Final temp. (TF) = 7.8°C
Specific heats = 4186 J kg⁻¹K⁻¹

Step 1: Heat Lost by Hot Water
The warm water cools from 35°C to 7.8°C, losing heat to the ice.
Qlost = mw × s × (Tw − TF) = 0.32 × 4186 × (35 − 7.8) = 0.32 × 4186 × 27.2 = 36,434.9 J
Step 2: Heat Gained by Ice (Two Stages)
Ice gains heat in two stages: (a) to melt at 0°C, then (b) to warm the meltwater to 7.8°C.
Q₁ = mice × Lf (heat to melt ice) = 0.1 × Lf Q₂ = mice × s × (TF − 0) (heat to warm melt-water) = 0.1 × 4186 × 7.8 = 3265.08 J
Step 3: Apply Principle of Calorimetry
Heat lost by water = Heat gained by ice (melting + warming)
36,434.9 = 0.1 × Lf + 3265.08 0.1 × Lf = 36,434.9 − 3265.08 = 33,169.82 Lf = 33,169.82 ÷ 0.1 = 3.317 × 10⁵ J kg⁻¹
∴ Lf ≈ 3.32 × 10⁵ J kg⁻¹   (Standard value = 3.33 × 10⁵ J kg⁻¹)
Maharashtra State Board: Class 11

Key Points: Latent Heat

  • Formula: Q = mL. Specific latent heat L has SI unit J kg⁻¹.
  • Temperature stays constant during any phase change. Heat energy goes into breaking or forming intermolecular bonds, not into raising kinetic energy.
  • Latent Heat of Fusion (water): Lf = 3.33 × 10⁵ J kg⁻¹ = 80 cal/g. Heat needed to melt 1 kg of ice at 0°C.
  • Latent Heat of Vaporisation (water): Lv = 22.6 × 10⁵ J kg⁻¹ = 540 cal/g. Heat is needed to convert 1 kg of water to steam at 100°C.
  • Lv ≫ Lf because vaporisation requires complete molecular separation and work against atmospheric pressure during expansion.
  • All latent heat values depend on atmospheric pressure. Standard values quoted at 1 atm. Increasing pressure raises the boiling point (pressure cooker effect).

Test Yourself

Related QuestionsVIEW ALL [17]

Advertisements
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×