Latent Heat

1. Concept Overview

The Latent Heat of a substance is the amount of heat required to change the state of unit mass of the substance without any change in its temperature.

When a substance undergoes a phase change (like melting, boiling, or condensation), its temperature remains constant, but it still absorbs or releases heat energy.
This heat is known as latent heat because it is hidden (not visible as temperature rise).

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2. Explanation and Mathematical Derivation

Let,

• [m] = mass of the substance,
• [L] = specific latent heat,
• [Q] = heat absorbed or released during the change of state.

Then, the heat absorbed or released is given by:

[
Q = m , L
]

Thus,

[
\boxed{L = \dfrac{Q}{m}}
]

Unit: joule per kilogram (J·kg⁻¹)

There are mainly two important types of latent heat:

1. Latent Heat of Fusion (Lₓ):
   The amount of heat required to convert 1 kg of a solid into liquid at its melting point, without changing temperature.
   Example: Ice → Water

   [
   Q = m , L_f
   ]

2. Latent Heat of Vaporization (Lᵥ):
   The amount of heat required to convert 1 kg of a liquid into vapor at its boiling point, without change of temperature.
   Example: Water → Steam

   [
   Q = m , L_v
   ]

Graphical Understanding:
During phase change, temperature remains constant while heat continues to be absorbed.
The flat portions of the heating curve correspond to latent heat processes.

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3. Dimensions and Units

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4. Key Features

1. Temperature remains constant during change of state.
2. The heat energy supplied is used in breaking molecular bonds.
3. Different substances have different latent heats.
4. Measured experimentally using calorimetry.
5. Expressed in J·kg⁻¹, sometimes also in cal·g⁻¹.
6. Depends on pressure and nature of the substance.
7. When vapor condenses or liquid freezes, same amount of heat is released.
8. Explains why steam burns are more severe than boiling water burns — due to large latent heat of vaporization.

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5. Important Formulas to Remember

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6. Conceptual Questions with Solutions

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7. FAQ / Common Misconceptions

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8. Practice Questions with Step-by-Step Solutions

Q1. Find the amount of heat required to melt 200 g of ice at 0°C.
Given: [L_f] [= 3.36 \times 10^5 \text{J·kg}^{-1}]

Solution:
[Q = m L_f] [= 0.2 \times 3.36 \times 10^5] [= 6.72 \times 10^4 \text{J}]
Q = 67.2 kJ

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Q2. Find the heat required to convert 1 kg of water at 100°C into steam.
Given: [L_v] [= 2.26 \times 10^6 \text{J·kg}^{-1}]

Solution:
[Q = m L_v] [= 1 \times 2.26 \times 10^6] [= 2.26 \times 10^6 \text{J}]
Q = 2.26 MJ

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Q3. 100 g of steam condenses to form water at 100°C. Find the heat released.
Solution:
[Q = m L_v] [= 0.1 \times 2.26 \times 10^6] [= 2.26 \times 10^5 \text{J}]
Q = 226 kJ released

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Q4. Find latent heat if 500 J heat changes 10 g of ice into water.
Solution:
[L = \dfrac{Q}{m}] [= \dfrac{500}{0.01}] [= 5.0 \times 10^4 \text{J·kg}^{-1}]
L = 5 × 10⁴ J·kg⁻¹

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Q5. How much ice at 0°C can be melted by 6.72 × 10⁴ J of heat?
Solution:
[m = \dfrac{Q}{L_f}] [= \dfrac{6.72 \times 10^4}{3.36 \times 10^5}] [= 0.2 \text{kg}]
Mass of ice melted = 0.2 kg